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Chapter 5

Quadratic Equations

Class - 10 Concise Mathematics Selina

Exercise 5(A)

Question 1(a)

4x2 - 9 = 0 implies x is equal to :

  1. 32\dfrac{3}{2}

  2. 94\dfrac{9}{4}

  3. −32-\dfrac{3}{2}

  4. ±32\pm \dfrac{3}{2}

Answer

Given,

⇒ 4x2 - 9 = 0

⇒ 4x2 = 9

⇒ x2 = 94\dfrac{9}{4}

⇒ x = 94=±32\sqrt{\dfrac{9}{4}} = \pm \dfrac{3}{2}

Hence, Option 4 is the correct option.

Question 1(b)

(x - 3)(x + 5) = 0 gives x equal to :

  1. 3

  2. 3 or 5

  3. 3 or -5

  4. 3 and -5

Answer

Given,

⇒ (x - 3)(x + 5) = 0

⇒ (x - 3) or (x + 5) = 0

⇒ x - 3 = 0 or x + 5 = 0

⇒ x = 3 or x = -5.

Hence, Option 3 is the correct option.

Question 1(c)

If 4 is a root of the equation x2 + kx - 4 = 0; the value of k is :

  1. 3

  2. -3

  3. 2

  4. -2

Answer

Since, 4 is the root of the equation x2 + kx - 4 = 0.

∴ It will satisfy the equation x2 + kx - 4 = 0.

∴ 42 + 4k - 4 = 0

⇒ 16 + 4k - 4 = 0

⇒ 4k + 12 = 0

⇒ 4k = -12

⇒ k = -124\dfrac{12}{4} = -3.

Hence, Option 2 is the correct option.

Question 1(d)

The equation 2x2 - 3x + k = 0 is satisfied by x = 2; the value of k is :

  1. -2

  2. 2

  3. 4

  4. 3

Answer

Given,

x = 2 satisfies the equation 2x2 - 3x + k = 0.

∴ 2(2)2 - 3(2) + k = 0

⇒ 2(4) - 6 + k = 0

⇒ 8 - 6 + k = 0

⇒ k + 2 = 0

⇒ k = -2.

Hence, Option 1 is the correct option.

Question 1(e)

If x2 - 7x = 0; the value of x is :

  1. 0 and 7

  2. 7

  3. 0

  4. 0 or 7

Answer

Given,

⇒ x2 - 7x = 0

⇒ x(x - 7) = 0

⇒ x = 0 or x - 7 = 0

⇒ x = 0 or x = 7.

Hence, Option 4 is the correct option.

Question 2(i)

Find which of the following equations are quadratic :

5x2 - 8x = -3(7 - 2x)

Answer

5x2 - 8x = -3(7 - 2x)

⇒ 5x2 - 8x = -21 + 6x

⇒ 5x2 - 8x - 6x + 21 = 0

⇒ 5x2 - 14x + 21 = 0 which is of the form ax2 + bx + c = 0.

∴ Given equation is a quadratic equation.

Question 2(ii)

Find which of the following equations are quadratic :

(x - 4)(3x + 1) = (3x - 1)(x + 2)

Answer

(x - 4)(3x + 1) = (3x - 1)(x + 2)

⇒ 3x2 + x - 12x - 4 = 3x2 + 6x - x - 2

⇒ 3x2 - 3x2 - 11x - 5x - 4 + 2 = 0

⇒ -16x - 2 = 0

⇒ 16x + 2 = 0 which is not of the form ax2 + bx + c = 0.

∴ Given equation is not a quadratic equation.

Question 2(iii)

Find which of the following equations are quadratic :

7x3 - 2x2 + 10 = (2x - 5)2

Answer

7x3 - 2x2 + 10 = (2x - 5)2

⇒ 7x3 - 2x2 + 10 = 4x2 + 25 - 20x

⇒ 7x3 - 2x2 - 4x2 + 10 - 25 + 20x = 0

⇒ 7x3 - 6x2 - 15 + 20x = 0 which is not of the form ax2 + bx + c = 0.

∴ Given equation is not a quadratic equation.

Question 3(i)

Is x = 5 a solution of the equation x2 - 2x - 15 = 0 ?

Answer

Substituting, x = 5 in x2 - 2x - 15 = 0 we get,

⇒ 52 - 2(5) - 15 = 0

⇒ 25 - 10 - 15 = 0

⇒ 25 - 25 = 0

⇒ 0 = 0

Since, L.H.S. = R.H.S.

Hence, proved that x = 5 is a solution of the equation x2 - 2x - 15 = 0.

Question 3(ii)

Is x = -3 a solution of the equation 2x2 - 7x + 9 = 0 ?

Answer

Substituting, x = -3 in 2x2 - 7x + 9 = 0 we get,

⇒ 2.(-3)2 - 7(-3) + 9 = 0

L.H.S.,

⇒ 2.(9) + 21 + 9

⇒ 18 + 21 + 9

⇒ 48

Since, L.H.S. ≠ R.H.S.

Hence, proved that x = -3 is not a solution of the equation 2x2 - 7x + 9 = 0.

Question 4

If 23\sqrt{\dfrac{2}{3}} is a solution of equation 3x2 + mx + 2 = 0, find the value of m.

Answer

Since, 23\sqrt{\dfrac{2}{3}} is a solution of equation 3x2 + mx + 2 = 0.

⇒3.(23)2+m(23)+2=0⇒3×23+2+m(23)=0⇒2+2+m(23)=0⇒4+m(23)=0⇒m(23)=−4⇒m=−4×32⇒m=−223⇒m=−26.\Rightarrow 3.\Big(\sqrt{\dfrac{2}{3}}\Big)^2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) + 2 = 0 \\[1em] \Rightarrow 3 \times \dfrac{2}{3} + 2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 2 + 2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 4 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow m\Big(\sqrt{\dfrac{2}{3}}\Big) = -4 \\[1em] \Rightarrow m = -4 \times \sqrt{\dfrac{3}{2}} \\[1em] \Rightarrow m = -2\sqrt{2}\sqrt{3} \\[1em] \Rightarrow m = -2\sqrt{6}.

Hence, value of m is −26.-2\sqrt{6}.

Question 5

23\dfrac{2}{3} and 1 are the solutions of equation mx2 + nx + 6 = 0. Find the values of m and n.

Answer

Since, 23\dfrac{2}{3} is a solution of equation mx2 + nx + 6 = 0.

Substituting 23\dfrac{2}{3} in mx2 + nx + 6 = 0,

⇒m(23)2+n×23+6=0⇒m×49+2n3+6=0⇒4m9+2n3+6=0⇒4m+6n+549=0⇒4m+6n+54=0⇒2(2m+3n+27)=0⇒2m+3n+27=0⇒2m+3n=−27.......(i)\Rightarrow m\Big(\dfrac{2}{3}\Big)^2 + n \times \dfrac{2}{3} + 6 = 0 \\[1em] \Rightarrow m \times \dfrac{4}{9} + \dfrac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \dfrac{4m}{9} + \dfrac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \dfrac{4m + 6n + 54}{9} = 0 \\[1em] \Rightarrow 4m + 6n + 54 = 0 \\[1em] \Rightarrow 2(2m + 3n + 27) = 0 \\[1em] \Rightarrow 2m + 3n + 27 = 0 \\[1em] \Rightarrow 2m + 3n = -27 .......(i)

Since, 1 is a solution of equation mx2 + nx + 6 = 0.

Substituting 1 in mx2 + nx + 6 = 0,

⇒m(1)2+n(1)+6=0⇒m+n+6=0⇒m=−(n+6)........(ii)\Rightarrow m(1)^2 + n(1) + 6 = 0 \\[1em] \Rightarrow m + n + 6 = 0 \\[1em] \Rightarrow m = -(n + 6) ........(ii)

Sustituting above value of m in eq. 1 we get,

⇒2.−(n+6)+3n=−27⇒−2(n+6)+3n=−27⇒−2n−12+3n=−27⇒n−12=−27⇒n=−27+12⇒n=−15.\Rightarrow 2.-(n + 6) + 3n = -27 \\[1em] \Rightarrow -2(n + 6) + 3n = -27 \\[1em] \Rightarrow -2n - 12 + 3n = -27 \\[1em] \Rightarrow n - 12 = -27 \\[1em] \Rightarrow n = -27 + 12 \\[1em] \Rightarrow n = -15.

Substituting value of n in (ii) we get,

⇒m=−(−15+6)⇒m=−(−9)⇒m=9.\Rightarrow m = -(-15 + 6) \\[1em] \Rightarrow m = -(-9) \\[1em] \Rightarrow m = 9.

Hence, the value of m = 9 and n = -15.

Exercise 5(B)

Question 1(a)

The roots of the quadratic equation x2 - 6x - 7 = 0 are :

  1. -1 and 7

  2. 1 and 7

  3. -1 or 7

  4. 1 and -7

Answer

Given,

⇒ x2 - 6x - 7 = 0

⇒ x2 - 7x + x - 7 = 0

⇒ x(x - 7) + 1(x - 7) = 0

⇒ (x + 1)(x - 7) = 0

⇒ x + 1 = 0 or x - 7 = 0

⇒ x = -1 or x = 7.

Hence, Option 3 is the correct option.

Question 1(b)

The roots of the quadratic equation x(x + 8) + 12 = 0 are :

  1. 6 or 2

  2. -6 or -2

  3. 6 and -2

  4. -6 and -2

Answer

Given,

⇒ x(x + 8) + 12 = 0

⇒ x2 + 8x + 12 = 0

⇒ x2 + 2x + 6x + 12 = 0

⇒ x(x + 2) + 6(x + 2) = 0

⇒ (x + 6)(x + 2) = 0

⇒ x + 6 = 0 or x + 2 = 0

⇒ x = -6 or x = -2.

Hence, Option 2 is the correct option.

Question 1(c)

If one root of equation (p - 3)x2 + x + p = 0 is 2, the value of p is :

  1. -2

  2. 2

  3. ±2

  4. 1 and 2

Answer

Given,

One root of equation (p - 3)x2 + x + p = 0 is 2.

∴ x = 2 satisfies the equation.

∴ (p - 3)(2)2 + 2 + p = 0

⇒ 4(p - 3) + 2 + p = 0

⇒ 4p - 12 + 2 + p = 0

⇒ 5p - 10 = 0

⇒ 5p = 10

⇒ p = 105\dfrac{10}{5} = 2.

Hence, Option 2 is the correct option.

Question 1(d)

If x+1xx + \dfrac{1}{x} = 2.5, the value of x is :

  1. 4

  2. 5 and 15\dfrac{1}{5}

  3. 2 or 12\dfrac{1}{2}

  4. 2 and 12\dfrac{1}{2}

Answer

Given,

⇒x+1x=2.5⇒x2+1x=2510⇒10(x2+1)=25x⇒10x2+10=25x⇒10x2−25x+10=0⇒10x2−20x−5x+10=0⇒10x(x−2)−5(x−2)=0⇒(x−2)(10x−5)=0⇒x−2=0 or 10x−5=0⇒x=2 or 10x=5⇒x=2 or x=510=12.\Rightarrow x + \dfrac{1}{x} = 2.5 \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{25}{10} \\[1em] \Rightarrow 10(x^2 + 1) = 25x \\[1em] \Rightarrow 10x^2 + 10 = 25x \\[1em] \Rightarrow 10x^2 - 25x + 10 = 0 \\[1em] \Rightarrow 10x^2 - 20x - 5x + 10 = 0 \\[1em] \Rightarrow 10x(x - 2) - 5(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(10x - 5) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } 10x - 5 =0 \\[1em] \Rightarrow x = 2 \text{ or } 10x = 5 \\[1em] \Rightarrow x = 2 \text{ or } x = \dfrac{5}{10} = \dfrac{1}{2}.

Hence, Option 3 is the correct option.

Question 1(e)

For quadratic equation 2x+5x2x + \dfrac{5}{x} = 5 :

  1. x ≠ 0

  2. x = 1

  3. x = 5

  4. x = 2

Answer

For quadratic equation :

2x + 5x\dfrac{5}{x} = 5

If x = 0, 5x\dfrac{5}{x} will not be defined.

∴ x ≠ 0.

Hence, Option 1 is the correct option.

Question 2

Solve the following equation by factorisation:

(2x - 3)2 = 49

Answer

⇒ (2x - 3)2 = 72

⇒ (2x - 3)2 - 72 = 0

As, (a2 - b2) = (a + b)(a - b)

⇒ (2x - 3 + 7)(2x - 3 - 7) = 0

⇒ (2x + 4)(2x - 10) = 0

⇒ 2x + 4 = 0 or 2x - 10 = 0      [Using Zero-product rule]

⇒ 2x = -4 or 2x = 10

⇒ x = -2 or x = 5.

Hence, x = -2 or x = 5.

Question 3

Solve the following equation by factorisation:

(x + 1)(2x + 8) = (x + 7)(x + 3)

Answer

⇒ (x + 1)(2x + 8) = (x + 7)(x + 3)

⇒ 2x2 + 8x + 2x + 8 = x2 + 3x + 7x + 21

⇒ 2x2 - x2 + 10x + 8 = 10x + 21

⇒ x2 + 10x - 10x = 21 - 8

⇒ x2 = 13

⇒ x = ±13\pm \sqrt{13}

Hence, x = +13+ \sqrt{13} or x = −13- \sqrt{13}.

i.e., x = 3.61 or x = -3.61

Question 4

Solve the following equation by factorisation:

4(2x - 3)2 - (2x - 3) - 14 = 0

Answer

4(2x - 3)2 - (2x - 3) - 14 = 0

⇒ 4(4x2 + 9 - 12x) - 2x + 3 - 14 = 0

⇒ 16x2 + 36 - 48x - 2x - 11 = 0

⇒ 16x2 - 50x + 25 = 0

⇒ 16x2 - 40x - 10x + 25 = 0

⇒ 8x(2x - 5) - 5(2x - 5) = 0

⇒ (8x - 5)(2x - 5) = 0

⇒ 8x - 5 = 0 or 2x - 5 = 0      [Using Zero-product rule]

⇒ 8x = 5 or 2x = 5

⇒ x = 58\dfrac{5}{8} or x = 52\dfrac{5}{2}.

Hence, x = 58 or x=52\dfrac{5}{8} \text{ or } x = \dfrac{5}{2}.

Question 5

Solve the following equation by factorisation:

2x2 - 9x + 10 = 0, when :

(i) x ∈ N

(ii) x ∈ Q.

Answer

2x2 - 9x + 10 = 0

⇒ 2x2 - 4x - 5x + 10 = 0

⇒ 2x(x - 2) - 5(x - 2) = 0

⇒ (2x - 5)(x - 2) = 0

⇒ 2x - 5 = 0 or x - 2 = 0      [Using Zero-product rule]

⇒ 2x = 5 or x = 2

⇒ x = 52\dfrac{5}{2} or x = 2.

(i) Since, x ∈ N

Hence, value of x = 2.

(ii) Since, x ∈ Q

Hence, value of x = 52\dfrac{5}{2} or 2.

Question 6

Solve the following equation by factorisation:

x−3x+3+x+3x−3=212\dfrac{x - 3}{x + 3} + \dfrac{x + 3}{x - 3} = 2\dfrac{1}{2}

Answer

Given,

⇒x−3x+3+x+3x−3=212⇒(x−3)(x−3)+(x+3)(x+3)(x+3)(x−3)=52⇒x2−3x−3x+9+x2+3x+3x+9x2+3x−3x−9=52⇒x2−6x+9+x2+6x+9x2−9=52⇒2x2+18x2−9=52⇒2(2x2+18)=5(x2−9)⇒4x2+36=5x2−45⇒5x2−4x2=36+45⇒x2=81⇒x=81=±9.\phantom {\Rightarrow} \dfrac{x - 3}{x + 3} + \dfrac{x + 3}{x - 3} = 2\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{(x - 3)(x - 3) + (x + 3)(x + 3)}{(x + 3)(x - 3)} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{x^2 - 3x - 3x + 9 + x^2 + 3x + 3x + 9}{x^2 + 3x - 3x - 9} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{x^2 - 6x + 9 + x^2 + 6x + 9}{x^2 - 9} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{2x^2 + 18}{x^2 - 9} = \dfrac{5}{2} \\[1em] \Rightarrow 2(2x^2 + 18) = 5(x^2 - 9) \\[1em] \Rightarrow 4x^2 + 36 = 5x^2 - 45 \\[1em] \Rightarrow 5x^2 - 4x^2 = 36 + 45 \\[1em] \Rightarrow x^2 = 81 \\[1em] \Rightarrow x = \sqrt{81} = \pm 9.

Hence, value of x = +9 or -9.

Question 7

Solve the following equation by factorisation:

4x+2−1x+3=42x+1\dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1}

Answer

Given,

⇒4x+2−1x+3=42x+1⇒4(x+3)−(x+2)(x+2)(x+3)=42x+1⇒4x+12−x−2x2+3x+2x+6=42x+1⇒3x+10x2+5x+6=42x+1⇒(3x+10)(2x+1)=4(x2+5x+6)⇒6x2+3x+20x+10=4x2+20x+24⇒6x2−4x2+23x−20x+10−24=0⇒2x2+3x−14=0⇒2x2+7x−4x−14=0⇒x(2x+7)−2(2x+7)=0⇒(x−2)(2x+7)=0⇒(x−2)=0 or 2x+7=0⇒x=2 or 2x=−7⇒x=2 or x=−72.\Rightarrow \dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{4(x + 3) - (x + 2)}{(x + 2)(x + 3)} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{4x + 12 - x - 2}{x^2 + 3x + 2x + 6} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{3x + 10}{x^2 + 5x + 6} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow (3x + 10)(2x + 1) = 4(x^2 + 5x + 6) \\[1em] \Rightarrow 6x^2 + 3x + 20x + 10 = 4x^2 + 20x + 24 \\[1em] \Rightarrow 6x^2 - 4x^2 + 23x - 20x + 10 -24 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 14 = 0 \\[1em] \Rightarrow 2x^2 + 7x - 4x - 14 = 0 \\[1em] \Rightarrow x(2x + 7) - 2(2x + 7) = 0 \\[1em] \Rightarrow (x - 2)(2x + 7) = 0 \\[1em] \Rightarrow (x - 2) = 0 \text{ or } 2x + 7 = 0 \\[1em] \Rightarrow x = 2 \text{ or } 2x = -7 \\[1em] \Rightarrow x = 2 \text{ or } x = -\dfrac{7}{2}.

Hence, value of x = 2 or -72\dfrac{7}{2}.

Question 8

Solve the following equation by factorisation:

5x−2−3x+6=4x\dfrac{5}{x - 2} - \dfrac{3}{x + 6} = \dfrac{4}{x}

Answer

Given,

⇒5(x+6)−3(x−2)(x−2)(x+6)=4x⇒5x+30−3x+6x2+6x−2x−12=4x⇒2x+36x2+4x−12=4x⇒x(2x+36)=4(x2+4x−12)⇒2x2+36x=4x2+16x−48⇒4x2−2x2+16x−36x−48=0⇒2x2−20x−48=0⇒2(x2−10x−24)=0⇒x2−10x−24=0⇒x2−12x+2x−24=0⇒x(x−12)+2(x−12)=0⇒(x+2)(x−12)=0⇒x=−2 or x=12.\Rightarrow \dfrac{5(x + 6) - 3(x - 2)}{(x - 2)(x + 6)} = \dfrac{4}{x} \\[1em] \Rightarrow \dfrac{5x + 30 - 3x + 6}{x^2 + 6x - 2x - 12} = \dfrac{4}{x} \\[1em] \Rightarrow \dfrac{2x + 36}{x^2 + 4x - 12} = \dfrac{4}{x} \\[1em] \Rightarrow x(2x + 36) = 4(x^2 + 4x - 12) \\[1em] \Rightarrow 2x^2 + 36x = 4x^2 + 16x - 48 \\[1em] \Rightarrow 4x^2 - 2x^2 + 16x - 36x - 48 = 0 \\[1em] \Rightarrow 2x^2 - 20x - 48 = 0 \\[1em] \Rightarrow 2(x^2 - 10x - 24) = 0 \\[1em] \Rightarrow x^2 - 10x - 24 = 0 \\[1em] \Rightarrow x^2 - 12x + 2x - 24 = 0 \\[1em] \Rightarrow x(x - 12) + 2(x - 12) = 0 \\[1em] \Rightarrow (x + 2)(x - 12) = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 12.

Hence, value of x = -2 or 12.

Question 9

Solve the following equation by factorisation:

(1+1x+1)(1−1x−1)=78\Big(1 + \dfrac{1}{x + 1}\Big)\Big(1 - \dfrac{1}{x - 1}\Big) = \dfrac{7}{8}

Answer

Given,

⇒(1+1x+1)(1−1x−1)=78⇒x+1+1x+1×x−1−1x−1=78⇒x+2x+1×x−2x−1=78⇒x2−4x2−1=78⇒8(x2−4)=7(x2−1)⇒8x2−32=7x2−7⇒8x2−7x2=−7+32⇒x2=25⇒x=25⇒x=±5.\Rightarrow \Big(1 + \dfrac{1}{x + 1}\Big)\Big(1 - \dfrac{1}{x - 1}\Big) = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x + 1 + 1}{x + 1} \times \dfrac{x - 1 - 1}{x - 1} = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x + 2}{x + 1} \times \dfrac{x - 2}{x - 1} = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x^2 - 4}{x^2 - 1} = \dfrac{7}{8} \\[1em] \Rightarrow 8(x^2 - 4) = 7(x^2 - 1) \\[1em] \Rightarrow 8x^2 - 32 = 7x^2 - 7 \\[1em] \Rightarrow 8x^2 - 7x^2 = - 7 + 32 \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x = \sqrt{25} \\[1em] \Rightarrow x = \pm 5.

Hence, value of x = -5 or +5.

Question 10

Find the quadratic equation, whose solution set is :

(i) {3, 5}

(ii) {-2, 3}

Answer

(i) Since, {3, 5} is solution set.

It means 3 and 5 are roots of the equation,

∴ x = 3 or x = 5

⇒ x - 3 = 0 or x - 5 = 0

⇒ (x - 3)(x - 5) = 0

⇒ (x2 - 5x - 3x + 15) = 0

⇒ x2 - 8x + 15 = 0.

Hence, quadratic equation with solution set {3, 5} = x2 - 8x + 15 = 0.

(ii) Since, {-2, 3} is solution set.

It means -2 and 3 are roots of the equation,

∴ x = -2 or x = 3

⇒ x + 2 = 0 or x - 3 = 0

⇒ (x + 2)(x - 3) = 0

⇒ (x2 - 3x + 2x - 6) = 0

⇒ x2 - x - 6 = 0.

Hence, quadratic equation with solution set {-2, 3} = x2 - x - 6 = 0.

Question 11(i)

Solve : x3+36−x=2(6+x)15\dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15}; (x ≠ 6)

Answer

Given,

⇒x3+36−x=2(6+x)15⇒x(6−x)+93(6−x)=12+2x15⇒6x−x2+918−3x=12+2x15⇒15(6x−x2+9)=(12+2x)(18−3x)⇒90x−15x2+135=216−36x+36x−6x2⇒90x−15x2+135=216−6x2⇒15x2−6x2−90x+216−135=0⇒9x2−90x+81=0⇒9(x2−10x+9)=0⇒x2−10x+9=0⇒x2−9x−x+9=0⇒x(x−9)−1(x−9)=0⇒(x−1)(x−9)=0⇒x=1 or x=9.\Rightarrow \dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15} \\[1em] \Rightarrow \dfrac{x(6 - x) + 9}{3(6 - x)} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow \dfrac{6x - x^2 + 9}{18 - 3x} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 6x^2 \\[1em] \Rightarrow 15x^2 - 6x^2 - 90x + 216 - 135 = 0 \\[1em] \Rightarrow 9x^2 - 90x + 81 = 0 \\[1em] \Rightarrow 9(x^2 - 10x + 9) = 0 \\[1em] \Rightarrow x^2 - 10x + 9 = 0 \\[1em] \Rightarrow x^2 - 9x - x + 9 = 0 \\[1em] \Rightarrow x(x - 9) - 1(x - 9) = 0 \\[1em] \Rightarrow (x - 1)(x - 9) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 9.

Hence, x = 1 or x = 9.

Question 11(ii)

Solve the equation 9x2 + 3x4+2\dfrac{3x}{4} + 2 = 0, if possible, for real values of x.

Answer

Given,

9x2 + 3x4+2\dfrac{3x}{4} + 2 = 0

⇒36x2+3x+84=0\Rightarrow \dfrac{36x^2 + 3x + 8}{4} = 0

⇒ 36x2 + 3x + 8 = 0

Comparing 36x2 + 3x + 8 = 0 with ax2 + bx + c = 0 we get,

a = 36, b = 3 and c = 8

D = b2 - 4ac = (3)2 - 4(36)(8) = 9 - 1152 = -1143.

Since, D < 0 hence, roots are imaginary.

There is no possibility of real roots.

Question 12

Find the value of x, if a + 7 = 0; b + 10 = 0 and 12x2 = ax - b.

Answer

Given,

⇒ a + 7 = 0 and b + 10 = 0

⇒ a = -7 and b = -10

Substituting value of a and b in 12x2 = ax - b,

⇒ 12x2 = (-7)x - (-10)

⇒ 12x2 = -7x + 10

⇒ 12x2 + 7x - 10 = 0

⇒ 12x2 + 15x - 8x - 10 = 0

⇒ 3x(4x + 5) - 2(4x + 5) = 0

⇒ (3x - 2)(4x + 5) = 0

⇒ 3x - 2 = 0 or 4x + 5 = 0      [Using Zero-product rule]

⇒ 3x = 2 or 4x = - 5

⇒ x = 23 or x=−54\dfrac{2}{3} \text{ or } x = -\dfrac{5}{4}.

Hence, x = 23 or −54\dfrac{2}{3} \text{ or } -\dfrac{5}{4}.

Question 13

Use the substitution 2x + 3 = y to solve for x, if 4(2x + 3)2 - (2x + 3) - 14 = 0.

Answer

Substituting, 2x + 3 = y in 4(2x + 3)2 - (2x + 3) - 14 = 0 we get,

⇒ 4y2 - y - 14 = 0

⇒ 4y2 - 8y + 7y - 14 = 0

⇒ 4y(y - 2) + 7(y - 2) = 0

⇒ (4y + 7)(y - 2) = 0

⇒ 4y + 7 = 0 or y - 2 = 0

⇒ 4y = -7 or y = 2

⇒ y = −74-\dfrac{7}{4} or y = 2.

∴ 2x + 3 = −74-\dfrac{7}{4} or 2x + 3 = 2

⇒ 2x = −74−3-\dfrac{7}{4} - 3 or 2x = 2 - 3

⇒ 2x = −7−124\dfrac{-7 - 12}{4} or 2x = -1

⇒ 2x = −194 or x=−12-\dfrac{19}{4} \text{ or } x = -\dfrac{1}{2}

⇒ x = −198 or x=−12-\dfrac{19}{8} \text{ or } x = -\dfrac{1}{2}

Hence, x = −198 or x=−12-\dfrac{19}{8} \text{ or } x = -\dfrac{1}{2}.

Question 14

If x ≠ 0 and a ≠ 0, solve :

xa−a+bx=b(a+b)ax\dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax}

Answer

Given,

⇒xa−a+bx=b(a+b)ax⇒x2−a(a+b)ax=ab+b2ax⇒x2−a2−abax×ax=ab+b2⇒x2−a2−ab=ab+b2⇒x2=a2+ab+ab+b2⇒x2=a2+2ab+b2⇒x2=(a+b)2⇒x2−(a+b)2=0⇒(x−(a+b))(x+(a+b))=0⇒x=a+b or x=−(a+b).\Rightarrow \dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax} \\[1em] \Rightarrow \dfrac{x^2 - a(a + b)}{ax} = \dfrac{ab + b^2}{ax} \\[1em] \Rightarrow \dfrac{x^2 - a^2 - ab}{ax} \times ax = ab + b^2 \\[1em] \Rightarrow x^2 - a^2 - ab = ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + ab + ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + 2ab + b^2 \\[1em] \Rightarrow x^2 = (a + b)^2 \\[1em] \Rightarrow x^2 - (a + b)^2 = 0 \\[1em] \Rightarrow (x - (a + b))(x + (a + b)) = 0 \\[1em] \Rightarrow x = a + b \text{ or } x = -(a + b).

Hence, x = a + b or -(a + b).

Question 15

Solve :

(1200x+2)(x−10)−1200=60.\Big(\dfrac{1200}{x} + 2\Big)(x - 10) - 1200 = 60.

Answer

Given,

⇒(1200x+2)(x−10)−1200=60⇒(1200+2x)(x−10)x=60+1200⇒1200x−12000+2x2−20xx=1260⇒2x2+1180x−12000=1260x⇒2x2+1180x−1260x−12000=0⇒2x2−80x−12000=0⇒2(x2−40x−6000)=0⇒x2−40x−6000=0⇒x2−100x+60x−6000=0⇒x(x−100)+60(x−100)=0⇒(x+60)(x−100)=0⇒(x+60)=0 or (x−100)=0⇒x=−60 or x=100.\Rightarrow \Big(\dfrac{1200}{x} + 2\Big)(x - 10) - 1200 = 60 \\[1em] \Rightarrow \dfrac{(1200 + 2x)(x - 10)}{x} = 60 + 1200 \\[1em] \Rightarrow \dfrac{1200x - 12000 + 2x^2 - 20x }{x} = 1260 \\[1em] \Rightarrow 2x^2 + 1180x - 12000 = 1260x \\[1em] \Rightarrow 2x^2 + 1180x -1260x - 12000 = 0 \\[1em] \Rightarrow 2x^2 - 80x - 12000 = 0 \\[1em] \Rightarrow 2(x^2 - 40x - 6000) = 0 \\[1em] \Rightarrow x^2 - 40x - 6000 = 0 \\[1em] \Rightarrow x^2 - 100x + 60x - 6000 = 0 \\[1em] \Rightarrow x(x - 100) + 60(x - 100) = 0 \\[1em] \Rightarrow (x + 60)(x - 100) = 0 \\[1em] \Rightarrow (x + 60) = 0 \text{ or } (x - 100) = 0 \\[1em] \Rightarrow x = -60 \text{ or } x = 100.

Hence, x = -60 or 100.

Exercise 5(C)

Question 1(a)

If x2 - 3x + 2 = 0, values of x correct to one decimal place are :

  1. 2.0 and 1.0

  2. 2.0 or 1.0

  3. 3.0 and 2.0

  4. 3.0 or 2.0

Answer

Comparing equation x2 - 3x + 2 = 0 with ax2 + bx + c = 0, we get :

a = 1, b = -3 and c = 2.

By formula,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get :

x=−(−3)±(−3)2−4×1×22×1=3±9−82=3±12=3±12=3+12 or 3−12=42 or 22=2.0 or 1.0x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 2}}{2 \times 1} \\[1em] = \dfrac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \dfrac{3 \pm \sqrt{1}}{2} \\[1em] = \dfrac{3 \pm 1}{2} \\[1em] = \dfrac{3 + 1}{2} \text{ or } \dfrac{3 - 1}{2} \\[1em] = \dfrac{4}{2} \text{ or } \dfrac{2}{2} \\[1em] = 2.0 \text{ or } 1.0

Hence, Option 2 is the correct option.

Question 1(b)

If x2 - 4x - 5 = 0, values of x correct to two decimal places are :

  1. 5.00 or -1.00

  2. 5 or 1

  3. 5.0 and -1.0

  4. -5.00 and -1.00

Answer

Comparing equation x2 - 4x - 5 = 0 with ax2 + bx + c = 0, we get :

a = 1, b = -4 and c = -5.

By formula,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get :

x=−(−4)±(−4)2−4×1×−52×1=4±16+202=4±362=4±62=4+62 or 4−62=102 or −22=5.00 or −1.00x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times -5}}{2 \times 1} \\[1em] = \dfrac{4 \pm \sqrt{16 + 20}}{2} \\[1em] = \dfrac{4 \pm \sqrt{36}}{2} \\[1em] = \dfrac{4 \pm 6}{2} \\[1em] = \dfrac{4 + 6}{2} \text{ or } \dfrac{4 - 6}{2} \\[1em] = \dfrac{10}{2} \text{ or } \dfrac{-2}{2} \\[1em] = 5.00 \text{ or } -1.00

Hence, Option 1 is the correct option.

Question 1(c)

If x2 - 8x - 9 = 0; values of x correct to one significant figure are :

  1. 9 and -1

  2. 9 or -1

  3. 9.0 or -1.0

  4. 9.00 or -1.00

Answer

Comparing equation x2 - 8x - 9 = 0 with ax2 + bx + c = 0, we get :

a = 1, b = -8 and c = -9.

By formula,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get :

x=−(−8)±(−8)2−4×1×−92×1=8±64+362=8±1002=8±102=8+102 or 8−102=182 or −22=9 or −1x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times -9}}{2 \times 1} \\[1em] = \dfrac{8 \pm \sqrt{64 + 36}}{2} \\[1em] = \dfrac{8 \pm \sqrt{100}}{2} \\[1em] = \dfrac{8 \pm 10}{2} \\[1em] = \dfrac{8 + 10}{2} \text{ or } \dfrac{8 - 10}{2} \\[1em] = \dfrac{18}{2} \text{ or } \dfrac{-2}{2} \\[1em] = 9 \text{ or } -1

Hence, Option 2 is the correct option.

Question 1(d)

If x2 - 2x - 3 = 0; values of x correct to two significant figures are :

  1. 3.0 and 1.0

  2. -1.0 and 3.0

  3. 3.0 or -1.0

  4. 3.00 or -1.00

Answer

Given,

⇒ x2 - 2x - 3 = 0

⇒ x2 - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ x + 1 = 0 or x - 3 = 0

⇒ x = -1 or x = 3.

Rounding off to two significant figures, we get :

⇒ x = -1.0 or x = 3.0

Hence, Option 3 is the correct option.

Question 1(e)

The value (values) of x satisfying the equation x2 - 6x - 16 = 0 is/are :

  1. 8 or -2

  2. -8 or 2

  3. 8 and -2

  4. -8 or 2

Answer

Comparing equation x2 - 6x - 16 = 0 with ax2 + bx + c = 0, we get :

a = 1, b = -6 and c = -16.

By formula,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get :

x=−(−6)±(−6)2−4×1×−162×1=6±36+642=6±1002=6±102=6+102 or 6−102=162 or −42=8 or −2x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times -16}}{2 \times 1} \\[1em] = \dfrac{6 \pm \sqrt{36 + 64}}{2} \\[1em] = \dfrac{6 \pm \sqrt{100}}{2} \\[1em] = \dfrac{6 \pm 10}{2} \\[1em] = \dfrac{6 + 10}{2} \text{ or } \dfrac{6 - 10}{2} \\[1em] = \dfrac{16}{2} \text{ or } \dfrac{-4}{2} \\[1em] = 8 \text{ or } -2

Hence, Option 1 is the correct option.

Question 2(i)

Solve the following equation for x and give your answer correct to one decimal place :

x2 - 8x + 5 = 0

Answer

Comparing x2 - 8x + 5 = 0 with ax2 + bx + c = 0 we get,

a = 1, b = -8 and c = 5.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−8)±(−8)2−4.(1).(5)2(1)=8±64−202=8±442=8±2112=4±11=4+11 and 4−11=4+3.3 and 4−3.3=7.3 and 0.7\Rightarrow x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \dfrac{8 \pm \sqrt{64 - 20}}{2} \\[1em] = \dfrac{8 \pm \sqrt{44}}{2} \\[1em] = \dfrac{8 \pm 2\sqrt{11}}{2} \\[1em] = 4 \pm \sqrt{11} \\[1em] = 4 + \sqrt{11} \text{ and } 4 - \sqrt{11} \\[1em] = 4 + 3.3 \text{ and } 4 - 3.3 \\[1em] = 7.3 \text{ and } 0.7

Hence, x = 7.3 and 0.7

Question 2(ii)

Solve the following equation for x and give your answer correct to one decimal place :

5x2 + 10x - 3 = 0

Answer

Comparing 5x2 + 10x - 3 = 0 with ax2 + bx + c = 0 we get,

a = 5, b = 10 and c = -3.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(10)±(10)2−4.(5).(−3)2(5)=−10±100+6010=−10±16010=−10±41010=−10±12.810=−10+12.810 and −10−12.810=2.810 and −22.810=0.28 and −2.28≈0.3 and −2.3\Rightarrow x = \dfrac{-(10) \pm \sqrt{(10)^2 - 4.(5).(-3)}}{2(5)} \\[1em] = \dfrac{-10 \pm \sqrt{100 + 60}}{10} \\[1em] = \dfrac{-10 \pm \sqrt{160}}{10} \\[1em] = \dfrac{-10 \pm 4\sqrt{10}}{10} \\[1em] = \dfrac{-10 \pm 12.8}{10} \\[1em] = \dfrac{-10 + 12.8}{10} \text{ and } \dfrac{-10 - 12.8}{10} \\[1em] = \dfrac{2.8}{10} \text{ and } \dfrac{-22.8}{10} \\[1em] = 0.28 \text{ and } -2.28 \\[1em] \approx 0.3 \text{ and } -2.3

Hence, x = 0.3 and -2.3

Question 3(i)

Solve the following equation for x and give your answer correct to two decimal places :

2x2 - 10x + 5 = 0

Answer

Comparing 2x2 - 10x + 5 = 0 with ax2 + bx + c = 0 we get,

a = 2, b = -10 and c = 5.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−10)±(−10)2−4.(2).(5)2(2)=10±100−404=10±604=10±2154=10±7.744=10+7.744 and 10−7.744=17.744 and 2.264=4.44 and 0.56\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4.(2).(5)}}{2(2)} \\[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \dfrac{10 \pm \sqrt{60}}{4} \\[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \\[1em] = \dfrac{10 \pm 7.74}{4} \\[1em] = \dfrac{10 + 7.74}{4} \text{ and } \dfrac{10 - 7.74}{4} \\[1em] = \dfrac{17.74}{4} \text{ and } \dfrac{2.26}{4} \\[1em] = 4.44 \text{ and } 0.56

Hence, x = 4.44 and 0.56

Question 3(ii)

Solve the following equation for x and give your answer correct to two decimal places :

4x + 6x\dfrac{6}{x} + 13 = 0

Answer

Given,

⇒4x+6x+13=0⇒4x2+6+13xx=0⇒4x2+13x+6=0\Rightarrow 4x + \dfrac{6}{x} + 13 = 0 \\[1em] \Rightarrow \dfrac{4x^2 + 6 + 13x}{x} = 0 \\[1em] \Rightarrow 4x^2 + 13x + 6 = 0

Comparing 4x2 + 13x + 6 = 0 with ax2 + bx + c = 0 we get,

a = 4, b = 13 and c = 6.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(13)±(−13)2−4.(4).(6)2(4)=−13±169−968=−13±738=−13±8.548=−13+8.548 and −13−8.548=−4.468 and −21.548=−0.56 and −2.69\Rightarrow x = \dfrac{-(13) \pm \sqrt{(-13)^2 - 4.(4).(6)}}{2(4)} \\[1em] = \dfrac{-13 \pm \sqrt{169 - 96}}{8} \\[1em] = \dfrac{-13 \pm \sqrt{73}}{8} \\[1em] = \dfrac{-13 \pm 8.54}{8} \\[1em] = \dfrac{-13 + 8.54}{8} \text{ and } \dfrac{-13 - 8.54}{8} \\[1em] = \dfrac{-4.46}{8} \text{ and } \dfrac{-21.54}{8} \\[1em] = -0.56 \text{ and } -2.69

Hence, x = -0.56 and -2.69

Question 3(iii)

Solve the following equation for x and give your answer correct to two decimal places :

4x2 - 5x - 3 = 0

Answer

Comparing 4x2 - 5x - 3 = 0 with ax2 + bx + c = 0 we get,

a = 4, b = -5 and c = -3.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−5)±(−5)2−4.(4).(−3)2(4)=5±25+488=5±738=5±8.548=5+8.548 and 5−8.548=13.548 and −3.548=1.69 and −0.44\Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4.(4).(-3)}}{2(4)} \\[1em] = \dfrac{5 \pm \sqrt{25 + 48}}{8} \\[1em] = \dfrac{5 \pm \sqrt{73}}{8} \\[1em] = \dfrac{5 \pm 8.54}{8} \\[1em] = \dfrac{5 + 8.54}{8} \text{ and } \dfrac{5 - 8.54}{8} \\[1em] = \dfrac{13.54}{8} \text{ and } \dfrac{-3.54}{8} \\[1em] = 1.69 \text{ and } -0.44

Hence, x = 1.69 and -0.44

Question 4(i)

Solve the following equation for x, giving your answer correct to 3 decimal places :

3x2 - 12x - 1 = 0

Answer

Comparing 3x2 - 12x - 1 = 0 with ax2 + bx + c = 0 we get,

a = 3, b = -12 and c = -1.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−12)±(−12)2−4.(3).(−1)2(3)=12±144+126=12±1566=12±12.496=12+12.496 or 12−12.496=24.496 or −0.496=4.082 or −0.082\Rightarrow x = \dfrac{-(-12) \pm \sqrt{(-12)^2 - 4.(3).(-1)}}{2(3)} \\[1em] = \dfrac{12 \pm \sqrt{144 + 12}}{6} \\[1em] = \dfrac{12 \pm \sqrt{156}}{6} \\[1em] = \dfrac{12 \pm 12.49}{6} \\[1em] = \dfrac{12 + 12.49}{6} \text{ or } \dfrac{12 - 12.49}{6} \\[1em] = \dfrac{24.49}{6} \text{ or } \dfrac{-0.49}{6} \\[1em] = 4.082 \text{ or } -0.082

Hence, x = 4.082 or -0.082

Question 4(ii)

Solve the following equation for x, giving your answer correct to 3 decimal places :

x2 - 16x + 6 = 0

Answer

Comparing x2 - 16x + 6 = 0 with ax2 + bx + c = 0 we get,

a = 1, b = -16 and c = 6.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−16)±(−16)2−4.(1).(6)2(1)=16±256−242=16±2322=16±15.2322=16+15.2322 or 16−15.2322=31.2322 or 0.7682=15.616 or 0.384\Rightarrow x = \dfrac{-(-16) \pm \sqrt{(-16)^2 - 4.(1).(6)}}{2(1)} \\[1em] = \dfrac{16 \pm \sqrt{256 - 24}}{2} \\[1em] = \dfrac{16 \pm \sqrt{232}}{2} \\[1em] = \dfrac{16 \pm 15.232}{2} \\[1em] = \dfrac{16 + 15.232}{2} \text{ or } \dfrac{16 - 15.232}{2} \\[1em] = \dfrac{31.232}{2} \text{ or } \dfrac{0.768}{2} \\[1em] = 15.616 \text{ or } 0.384

Hence, x = 15.616 or 0.384

Question 4(iii)

Solve the following equation for x, giving your answer correct to 3 decimal places :

2x2 + 11x + 4 = 0

Answer

Comparing 2x2 + 11x + 4 = 0 with ax2 + bx + c = 0 we get,

a = 2, b = 11 and c = 4.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(11)±(11)2−4.(2).(4)2(2)=−11±121−324=−11±894=−11±9.4344=−11+9.4344 or −11−9.4344=−1.5664 or −20.4344=−0.392 or −5.109\Rightarrow x = \dfrac{-(11) \pm \sqrt{(11)^2 - 4.(2).(4)}}{2(2)} \\[1em] = \dfrac{-11 \pm \sqrt{121 - 32}}{4} \\[1em] = \dfrac{-11 \pm \sqrt{89}}{4} \\[1em] = \dfrac{-11 \pm 9.434}{4} \\[1em] = \dfrac{-11 + 9.434}{4} \text{ or } \dfrac{-11 - 9.434}{4} \\[1em] = \dfrac{-1.566}{4} \text{ or } \dfrac{-20.434}{4} \\[1em] = -0.392 \text{ or } -5.109

Hence, x = -0.392 or -5.109

Question 5

Solve the following equation and give your answer correct to 3 significant figures :

5x2 - 3x - 4 = 0

Answer

Comparing 5x2 - 3x - 4 = 0 with ax2 + bx + c = 0 we get,

a = 5, b = -3 and c = -4.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−3)±(−3)2−4.(5).(−4)2(5)=3±9+8010=3±8910=3±9.43410=3+9.43410 or 3−9.43410=12.43410 or −6.43410=1.2434 or −0.6434≈1.243 or −0.643\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4.(5).(-4)}}{2(5)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 80}}{10} \\[1em] = \dfrac{3 \pm \sqrt{89}}{10} \\[1em] = \dfrac{3 \pm 9.434}{10} \\[1em] = \dfrac{3 + 9.434}{10} \text{ or } \dfrac{3 - 9.434}{10} \\[1em] = \dfrac{12.434}{10} \text{ or } \dfrac{-6.434}{10} \\[1em] = 1.2434 \text{ or } -0.6434 \\[1em] \approx 1.243 \text{ or } -0.643

Hence, x = 1.243 or -0.643

Question 6

Solve for x using the quadratic formula. Write your answer correct to two significant figures.

(x - 1)2 - 3x + 4 = 0

Answer

Given,

⇒ (x - 1)2 - 3x + 4 = 0

⇒ x2 + 1 - 2x - 3x + 4 = 0

⇒ x2 - 5x + 5 = 0

Comparing x2 - 5x + 5 = 0 with ax2 + bx + c = 0 we get,

a = 1, b = -5 and c = 5.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values of a, b and c in above equation we get,

⇒x=−(−5)±(−5)2−4.(1).(5)2(1)=5±25−202=5±52=5±2.22=5+2.22 or 5−2.22=7.22 or 2.82=3.6 or 1.4\Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \dfrac{5 \pm \sqrt{25 - 20}}{2} \\[1em] = \dfrac{5 \pm \sqrt{5}}{2} \\[1em] = \dfrac{5 \pm 2.2}{2} \\[1em] = \dfrac{5 + 2.2}{2} \text{ or } \dfrac{5 - 2.2}{2} \\[1em] = \dfrac{7.2}{2} \text{ or } \dfrac{2.8}{2} \\[1em] = 3.6 \text{ or } 1.4

Hence, x = 3.6 or 1.4

Exercise 5(D)

Question 1(a)

Equation 2x2 - 3x + 1 = 0 has :

  1. distinct and real roots

  2. no real roots

  3. equal roots

  4. imaginary roots

Answer

Comparing equation 2x2 - 3x + 1 = 0, with ax2 + bx + c = 0, we get :

a = 2, b = -3 and c = 1.

By formula,

D = b2 - 4ac

= (-3)2 - 4 × 2 × 1

= 9 - 8

= 1; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b2 - 4ac > 0

∴ Roots are real and unequal.

Hence, Option 1 is the correct option.

Question 1(b)

Which of the following equations has two real and distinct roots ?

  1. x2 - 5x + 6 = 0

  2. x2 - 3x + 6 = 0

  3. x2 - 2x + 5 = 0

  4. x2 - 4x + 6 = 0

Answer

Comparing equation x2 - 5x + 6 = 0, with ax2 + bx + c = 0, we get :

a = 1, b = -5 and c = 6.

By formula,

D = b2 - 4ac

= (-5)2 - 4 × 1 × 6

= 25 - 24

= 1; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b2 - 4ac > 0

∴ Roots are real and distinct.

Hence, Option 1 is the correct option.

Question 1(c)

If the roots of x2 - px + 4 = 0 are equal, the value (values) of p is/are :

  1. 4 and -4

  2. 4

  3. -4

  4. 4 or -4

Answer

Comparing equation x2 - px + 4 = 0, with ax2 + bx + c = 0, we get :

a = 1, b = -p and c = 4.

Since, roots are equal.

∴ D = 0

∴ b2 - 4ac = 0

⇒ (-p)2 - 4 × 1 × 4 = 0

⇒ p2 - 16 = 0

⇒ p2 - 42 = 0

⇒ (p - 4)(p + 4) = 0

⇒ (p - 4) = 0 or (p + 4) = 0

⇒ p = 4 or p = -4.

Hence, Option 4 is the correct option.

Question 1(d)

Which of the following equations has imaginary roots ?

  1. x2 + 10x - 3 = 0

  2. 2x2 - 5x + 9 = 0

  3. x2 + 5x + 4 = 0

  4. 5x2 - 8x - 1 = 0

Answer

Comparing equation x2 + 10x - 3 = 0, with ax2 + bx + c = 0, we get :

a = 1, b = 10 and c = -3.

By formula,

D = b2 - 4ac

= (10)2 - 4 × 1 × -3

= 100 + 12

= 112; which is positive.

Comparing equation 2x2 - 5x + 9 = 0, with ax2 + bx + c = 0, we get :

a = 2, b = -5 and c = 9.

By formula,

D = b2 - 4ac

= (-5)2 - 4 × 2 × 9

= 25 - 72

= -47; which is negative.

∴ Roots are imaginary.

Hence, Option 2 is the correct option.

Question 1(e)

One root of equation 3x2 - mx + 4 = 0 is 1, the value of m is :

  1. 7

  2. -7

  3. 43\dfrac{4}{3}

  4. −43-\dfrac{4}{3}

Answer

Since, 1 is the root of equation 3x2 - mx + 4 = 0.

∴ x = 1, will satisfy the equation 3x2 - mx + 4 = 0.

⇒ 3(1)2 - m(1) + 4 = 0

⇒ 3 - m + 4 = 0

⇒ 7 - m = 0

⇒ m = 7.

Hence, Option 1 is the correct option.

Question 2(i)

Without solving, comment upon the nature of roots of the following equation :

7x2 - 9x + 2 = 0

Answer

Comparing 7x2 - 9x + 2 = 0 with ax2 + bx + c = 0 we get,

a = 7, b = -9 and c = 2.

We know that,

Discriminant = D = b2 - 4ac = (-9)2 - 4(7)(2)

= 81 - 56 = 25; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b2 - 4ac > 0

∴ The roots are real and unequal.

Question 2(ii)

Without solving, comment upon the nature of roots of the following equation :

6x2 - 13x + 4 = 0

Answer

Comparing 6x2 - 13x + 4 = 0 with ax2 + bx + c = 0 we get,

a = 6, b = -13 and c = 4.

We know that,

Discriminant = D = b2 - 4ac = (-13)2 - 4(6)(4)

= 169 - 96 = 73; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b2 - 4ac > 0

∴ The roots are real and unequal.

Question 2(iii)

Without solving, comment upon the nature of roots of the following equation :

25x2 - 10x + 1 = 0

Answer

Comparing 25x2 - 10x + 1 = 0 with ax2 + bx + c = 0 we get,

a = 25, b = -10 and c = 1.

We know that,

Discriminant = D = b2 - 4ac = (-10)2 - 4(25)(1)

= 100 - 100 = 0;

Since, a, b and c are real numbers; a ≠ 0 and b2 - 4ac = 0

∴ The roots are real and equal.

Question 2(iv)

Without solving, comment upon the nature of roots of the following equation :

x2 + 23x−92\sqrt{3}x - 9 = 0

Answer

Comparing x2 + 232\sqrt{3}x - 9 = 0 with ax2 + bx + c = 0 we get,

a = 1, b = 232\sqrt{3} and c = -9.

We know that,

Discriminant = D = b2 - 4ac = (23)(2\sqrt{3})2 - 4(1)(-9)

= 12 + 36 = 48; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b2 - 4ac > 0

∴ The roots are real and unequal.

Question 3

The equation 3x2 - 12x + (n - 5) = 0 has equal roots. Find the value of n.

Answer

Comparing 3x2 - 12x + (n - 5) = 0 with ax2 + bx + c = 0 we get,

a = 3, b = -12 and c = (n - 5).

Since equations have equal roots,

∴ D = 0

⇒ (-12)2 - 4.(3).(n - 5) = 0

⇒ 144 - 12(n - 5) = 0

⇒ 144 - 12n + 60 = 0

⇒ 204 - 12n = 0

⇒ 12n = 204

⇒ n = 20412\dfrac{204}{12}

⇒ n = 17

Hence, n = 17.

Question 4

Find the values of 'm', if the following equation has equal roots :

(m - 2)x2 - (5 + m)x + 16 = 0

Answer

Comparing (m - 2)x2 - (5 + m)x + 16 = 0 with ax2 + bx + c = 0 we get,

a = (m - 2), b = -(5 + m) and c = 16.

Since equations have equal roots,

∴ D = 0

⇒ (-(5 + m))2 - 4.(m - 2).(16) = 0

⇒ 25 + m2 + 10m - 64(m - 2) = 0

⇒ 25 + m2 + 10m - 64m + 128 = 0

⇒ m2 - 54m + 153 = 0

⇒ m2 - 51m - 3m + 153 = 0

⇒ m(m - 51) - 3(m - 51) = 0

⇒ (m - 3)(m - 51) = 0

⇒ (m - 3) = 0 or (m - 51) = 0

⇒ m = 3 or m = 51.

Hence, m = 3 or 51.

Question 5

Find the value of k for which the equation 3x2 - 6x + k = 0 has distinct and real roots.

Answer

Comparing 3x2 - 6x + k = 0 with ax2 + bx + c = 0 we get,

a = 3, b = -6 and c = k.

Since equations have distinct and real roots,

∴ D > 0

⇒ (-6)2 - 4.(3).(k) > 0

⇒ 36 - 12k > 0

⇒ 12k < 36

⇒ k < 3.

Hence, k < 3.

Question 6

Given that 2 is a root of the equation 3x2 - p(x + 1) = 0 and that the equation px2 - qx + 9 = 0 has equal roots, find the values of p and q.

Answer

Since, 2 is the root hence, it satisfies the equation 3x2 - p(x + 1) = 0.

⇒ 3(2)2 - p(2 + 1) = 0

⇒ 3(4) - 3p = 0

⇒ 3p = 12

⇒ p = 4.

Substituting value of p in px2 - qx + 9 = 0

⇒ 4x2 - qx + 9 = 0

Comparing 4x2 - qx + 9 = 0 with ax2 + bx + c = 0 we get,

a = 4, b = -q and c = 9.

Since equation has equal roots,

∴ D = 0

⇒ (-q)2 - 4.(4).(9) = 0

⇒ q2 - 144 = 0

⇒ q2 = 144

⇒ q = 12 or -12.

Hence, p = 4 and q = 12 or -12.

Exercise 5(E)

Question 1(a)

If x4 - 5x2 + 4 = 0; the values of x are :

  1. 1 or 2

  2. ± 1 or ± 2

  3. -1 and 2

  4. -1 and -2

Answer

Given,

⇒ x4 - 5x2 + 4 = 0

⇒ x4 - 4x2 - x2 + 4 = 0

⇒ x2(x2 - 4) - 1(x2 - 4) = 0

⇒ (x2 - 1)(x2 - 4) = 0

⇒ x2 - 1 = 0 or x2 - 4 = 0

⇒ x2 = 1 or x2 = 4

⇒ x = 1\sqrt{1} or x = 4\sqrt{4}

⇒ x = ± 1 or x = ± 2.

Hence, Option 2 is the correct option.

Question 1(b)

For equation 1x+1x−5=310\dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10}; one value of x is :

  1. −53-\dfrac{5}{3}

  2. 10

  3. -10

  4. 5

Answer

Given,

⇒1x+1x−5=310⇒x−5+xx(x−5)=310⇒2x−5x2−5x=310⇒10(2x−5)=3(x2−5x)⇒20x−50=3x2−15x⇒3x2−15x−20x+50=0⇒3x2−35x+50=0⇒3x2−30x−5x+50=0⇒3x(x−10)−5(x−10)=0⇒(3x−5)(x−10)=0⇒3x−5=0 or x−10=0⇒3x=5 or x=10⇒x=53 or x=10.\Rightarrow \dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{x - 5 + x}{x(x - 5)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x - 5}{x^2 - 5x} = \dfrac{3}{10} \\[1em] \Rightarrow 10(2x - 5) = 3(x^2 - 5x) \\[1em] \Rightarrow 20x - 50 = 3x^2 - 15x \\[1em] \Rightarrow 3x^2 - 15x - 20x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 35x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 30x - 5x + 50 = 0 \\[1em] \Rightarrow 3x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (3x - 5)(x - 10) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow 3x = 5 \text{ or } x = 10 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = 10.

Hence, Option 2 is the correct option.

Question 1(c)

Which of the following is correct for the equation 1x−3−1x+5\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = 1 ?

  1. x ≠ 3 and x = -5

  2. x = 3 and x ≠ -5

  3. x ≠ 3 and x ≠ -5

  4. x > 3 and x < 5

Answer

Given,

1x−3−1x+5\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = 1

So, in above equation.

x - 3 and x + 5 cannot be equal to zero.

⇒ x - 3 ≠ 0

⇒ x ≠ 3

⇒ x + 5 ≠ 0

⇒ x ≠ -5.

Hence, Option 3 is the correct option.

Question 1(d)

The product of two integers is -18; the integers are :

  1. -2 and 9 or -9 and 2

  2. -6 and -3 or 6 and 3

  3. -9 and -2 or 9 and 2

  4. 1 and 18 or 18 and 1.

Answer

In 1st option :

-2 × 9 = 18 and -9 × 2 = -18.

Hence, Option 1 is the correct option.

Question 2

Solve :

2x4 - 5x2 + 3 = 0

Answer

Let x2 = y,

⇒ 2x4 - 5x2 + 3 = 0

⇒ 2y2 - 5y + 3 = 0

⇒ 2y2 - 2y - 3y + 3 = 0

⇒ 2y(y - 1) - 3(y - 1) = 0

⇒ (2y - 3)(y - 1) = 0

⇒ 2y - 3 = 0 or y - 1 = 0      [Zero product rule]

⇒ 2y = 3 or y = 1

⇒ y = 32\dfrac{3}{2} or y = 1.

∴ x2 = 32\dfrac{3}{2} or x2 = 1

⇒ x = ±32=±1.22\pm \sqrt{\dfrac{3}{2}} = \pm 1.22 or x = 1=±1\sqrt{1} = \pm 1

Hence, x = +1.22, -1.22, +1, -1.

Question 3

Solve :

x4 - 2x2 - 3 = 0

Answer

Let x2 = y,

⇒ x4 - 2x2 - 3 = 0

⇒ y2 - 2y - 3 = 0

⇒ y2 - 3y + y - 3 = 0

⇒ y(y - 3) + 1(y - 3) = 0

⇒ (y + 1)(y - 3) = 0

⇒ y + 1 = 0 or y - 3 = 0      [Zero product rule]

⇒ y = -1 or y = 3

∴ x2 = -1 or x2 = 3

Since, square of a number cannot be negative,

∴ x2 = 3

⇒ x = 3=±1.73\sqrt{3} = \pm 1.73

Hence, x = +1.73, -1.73.

Question 4(i)

Solve :

(x2 - x)2 + 5(x2 - x) + 4 = 0

Answer

Let x2 - x = a

Substituting value in (x2 - x)2 + 5(x2 - x) + 4 = 0 we get,

⇒ a2 + 5a + 4 = 0

⇒ a2 + 4a + a + 4 = 0

⇒ a(a + 4) + 1(a + 4) = 0

⇒ (a + 1)(a + 4) = 0

⇒ a + 1 = 0 or a + 4 = 0      [Zero product rule]

⇒ a = -1 or a = -4

∴ x2 - x = -1 and x2 - x = -4

Solving, x2 - x = -1

⇒ x2 - x = -1

⇒ x2 - x + 1 = 0

Comparing above equation with ax2 + bx + x = 0 we get,

a = 1, b = -1, c = 1

Discriminant = D = b2 - 4ac = (-1)2 - 4(1)(1) = 1 - 4 = -3.

-3 < 0

∴ No real solution.

Solving, x2 - x = -4

⇒ x2 - x + 4 = 0

Comparing above equation with ax2 + bx + x = 0 we get,

a = 1, b = -1, c = 4

Discriminant = D = b2 - 4ac = (-1)2 - 4(1)(4) = 1 - 16 = -15.

-15 < 0

∴ No real solution.

Hence, there is no real solution.

Question 4(ii)

Solve :

(x2 - 3x)2 - 16(x2 - 3x) - 36 = 0

Answer

Let x2 - 3x = a

Substituting value in (x2 - 3x)2 - 16(x2 - 3x) - 36 = 0 we get,

⇒ a2 - 16a - 36 = 0

⇒ a2 - 18a + 2a - 36 = 0

⇒ a(a - 18) + 2(a - 18) = 0

⇒ (a + 2)(a - 18) = 0

⇒ a + 2 = 0 or a - 18 = 0      [Zero product rule]

⇒ a = -2 or a = 18

∴ x2 - 3x = -2 and x2 - 3x = 18

Solving, x2 - 3x = -2

⇒ x2 - 3x = -2

⇒ x2 - 3x + 2 = 0

⇒ x2 - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x - 1 = 0 or x - 2 = 0      [Zero product rule]

⇒ x = 1 or x = 2.

Solving, x2 - 3x = 18

⇒ x2 - 3x = 18

⇒ x2 - 3x - 18 = 0

⇒ x2 - 6x + 3x - 18 = 0

⇒ x(x - 6) + 3(x - 6) = 0

⇒ (x + 3)(x - 6) = 0

⇒ x + 3 = 0 or x - 6 = 0      [Zero product rule]

⇒ x = -3 or x = 6.

Hence, x = 1, 2, -3, 6.

Question 5(i)

Solve :

xx−3+x−3x=52\sqrt{\dfrac{x}{x - 3}} + \sqrt{\dfrac{x - 3}{x}} = \dfrac{5}{2}

Answer

Let xx−3=\sqrt{\dfrac{x}{x - 3}} = a .......(i)

⇒xx−3+x−3x=52⇒a+1a=52⇒a2+1a=52⇒2(a2+1)=5a⇒2a2+2=5a⇒2a2−5a+2=0⇒2a2−4a−a+2=0⇒2a(a−2)−1(a−2)=0⇒(2a−1)(a−2)=0⇒2a−1=0 or a−2=0⇒a=12 or a=2.\Rightarrow \sqrt{\dfrac{x}{x - 3}} + \sqrt{\dfrac{x - 3}{x}} = \dfrac{5}{2} \\[1em] \Rightarrow a + \dfrac{1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow 2(a^2 + 1) = 5a \\[1em] \Rightarrow 2a^2 + 2 = 5a \\[1em] \Rightarrow 2a^2 - 5a + 2 = 0 \\[1em] \Rightarrow 2a^2 - 4a - a + 2 = 0 \\[1em] \Rightarrow 2a(a - 2) - 1(a - 2) = 0 \\[1em] \Rightarrow (2a - 1)(a - 2) = 0 \\[1em] \Rightarrow 2a - 1 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.

Substituting value of a = 12\dfrac{1}{2} in (i) we get,

⇒xx−3=12\Rightarrow \sqrt{\dfrac{x}{x - 3}} = \dfrac{1}{2}

Squaring both sides we get,

⇒xx−3=14⇒4x=x−3⇒4x−x=−3⇒3x=−3⇒x=−1.\Rightarrow \dfrac{x}{x - 3} = \dfrac{1}{4} \\[1em] \Rightarrow 4x = x - 3 \\[1em] \Rightarrow 4x - x = -3 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1.

Substituting value of a = 2 in (i) we get,

⇒xx−3=2\Rightarrow \sqrt{\dfrac{x}{x - 3}} = 2

Squaring both sides we get,

⇒xx−3=4⇒x=4(x−3)⇒x=4x−12⇒4x−x=12⇒3x=12⇒x=4.\Rightarrow \dfrac{x}{x - 3} = 4 \\[1em] \Rightarrow x = 4(x - 3) \\[1em] \Rightarrow x = 4x - 12 \\[1em] \Rightarrow 4x - x = 12 \\[1em] \Rightarrow 3x = 12 \\[1em] \Rightarrow x = 4.

Hence, x = -1, 4.

Question 5(ii)

Solve :

(2x−3x−1)−4(x−12x−3)=3\Big(\dfrac{2x - 3}{x - 1}\Big) - 4\Big(\dfrac{x - 1}{2x -3}\Big) = 3

Answer

Let 2x−3x−1=\dfrac{2x - 3}{x - 1} = a .......(i)

⇒a−4a=3⇒a2−4a=3⇒a2−4=3a⇒a2−3a−4=0⇒a2−4a+a−4=0⇒a(a−4)+1(a−4)=0⇒(a+1)(a−4)=0⇒a=−1 or a=4.\Rightarrow a - \dfrac{4}{a} = 3 \\[1em] \Rightarrow \dfrac{a^2 - 4}{a} = 3 \\[1em] \Rightarrow a^2 - 4 = 3a \\[1em] \Rightarrow a^2 - 3a - 4 = 0 \\[1em] \Rightarrow a^2 - 4a + a - 4 = 0 \\[1em] \Rightarrow a(a - 4) + 1(a - 4) = 0 \\[1em] \Rightarrow (a + 1)(a - 4) = 0 \\[1em] \Rightarrow a = -1 \text{ or } a = 4.

Substituting value of a = -1 in (i) we get,

⇒2x−3x−1=−1⇒2x−3=−1(x−1)⇒2x−3=−x+1⇒2x+x=1+3⇒3x=4⇒x=43⇒x=113\Rightarrow \dfrac{2x - 3}{x - 1} = -1 \\[1em] \Rightarrow 2x - 3 = -1(x - 1) \\[1em] \Rightarrow 2x - 3 = -x + 1 \\[1em] \Rightarrow 2x + x = 1 + 3 \\[1em] \Rightarrow 3x = 4 \\[1em] \Rightarrow x = \dfrac{4}{3} \\[1em] \Rightarrow x = 1\dfrac{1}{3}

Substituting value of a = 4 in (i) we get,

⇒2x−3x−1=4⇒2x−3=4(x−1)⇒2x−3=4x−4⇒4x−2x=−3+4⇒2x=1⇒x=12.\Rightarrow \dfrac{2x - 3}{x - 1} = 4 \\[1em] \Rightarrow 2x - 3 = 4(x - 1) \\[1em] \Rightarrow 2x - 3 = 4x - 4 \\[1em] \Rightarrow 4x - 2x = -3 + 4 \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.

Hence, x = 113,121\dfrac{1}{3}, \dfrac{1}{2}.

Question 5(iii)

Solve :

(3x+1x+1)+(x+13x+1)=52\Big(\dfrac{3x + 1}{x + 1}\Big) + \Big(\dfrac{x + 1}{3x + 1}\Big) = \dfrac{5}{2}

Answer

Let (3x+1x+1)=a\Big(\dfrac{3x + 1}{x + 1}\Big) = a .......(i)

⇒(3x+1x+1)+(x+13x+1)=52⇒a+1a=52⇒a2+1a=52⇒2(a2+1)=5a⇒2a2+2−5a=0⇒2a2−5a+2=0⇒2a2−4a−a+2=0⇒2a(a−2)−1(a−2)=0⇒(2a−1)(a−2)=0⇒2a−1=0 or a−2=0⇒2a=1 or a=2⇒a=12 or a=2.\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) + \Big(\dfrac{x + 1}{3x + 1}\Big) = \dfrac{5}{2} \\[1em] \Rightarrow a + \dfrac{1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow 2(a^2 + 1) = 5a \\[1em] \Rightarrow 2a^2 + 2 - 5a = 0 \\[1em] \Rightarrow 2a^2 - 5a + 2 = 0 \\[1em] \Rightarrow 2a^2 - 4a - a + 2 = 0 \\[1em] \Rightarrow 2a(a - 2) - 1(a - 2) = 0 \\[1em] \Rightarrow (2a - 1)(a - 2) = 0 \\[1em] \Rightarrow 2a - 1 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow 2a = 1 \text{ or } a = 2 \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.

Substituting value of a = 12\dfrac{1}{2} in (i) we get,

⇒(3x+1x+1)=12⇒2(3x+1)=x+1⇒6x+2=x+1⇒6x−x=1−2⇒5x=−1⇒x=−15\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = \dfrac{1}{2} \\[1em] \Rightarrow 2(3x + 1) = x + 1 \\[1em] \Rightarrow 6x + 2 = x + 1 \\[1em] \Rightarrow 6x - x = 1 - 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\dfrac{1}{5}

Substituting value of a = 2 in (i) we get,

⇒(3x+1x+1)=2⇒3x+1=2(x+1)⇒3x+1=2x+2⇒3x−2x=2−1⇒x=1.\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = 2 \\[1em] \Rightarrow 3x + 1 = 2(x + 1) \\[1em] \Rightarrow 3x + 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 - 1 \\[1em] \Rightarrow x = 1.

Hence, x = 1, −15-\dfrac{1}{5}.

Question 6

Solve :

9(x2+1x2)−9(x+1x)−52=09(x^2 + \dfrac{1}{x^2}) - 9(x + \dfrac{1}{x}) - 52 = 0

Answer

Let x+1x=ax + \dfrac{1}{x} = a ........(i)

Squaring, both sides we get,

⇒x2+1x2+2=a2⇒x2+1x2=a2−2.......(ii)\Rightarrow x^2 + \dfrac{1}{x^2} + 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 - 2 .......(ii)

Substituting the values from equations (i) and (ii) we get,

⇒ 9(a2 - 2) - 9a - 52 = 0

⇒ 9a2 - 18 - 9a - 52 = 0

⇒ 9a2 - 9a - 70 = 0

⇒ 9a2 - 30a + 21a - 70 = 0

⇒ 3a(3a - 10) + 7(3a - 10) = 0

⇒ (3a + 7)(3a - 10) = 0

⇒ (3a + 7) = 0 or 3a - 10 = 0      [Zero product rule]

⇒ 3a = -7 or 3a = 10

⇒ a=−73 or a=103a = -\dfrac{7}{3} \text{ or } a = \dfrac{10}{3}

Considering a = 103\dfrac{10}{3} we get,

⇒x+1x=103⇒x2+1x=103⇒3(x2+1)=10x⇒3x2+3=10x⇒3x2−10x+3=0⇒3x2−9x−x+3=0⇒3x(x−3)−1(x−3)=0⇒(3x−1)(x−3)=0⇒3x−1=0 or x−3=0⇒3x=1 or x=3⇒x=13 or x=3.\Rightarrow x + \dfrac{1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow 3(x^2 + 1) = 10x \\[1em] \Rightarrow 3x^2 + 3 = 10x \\[1em] \Rightarrow 3x^2 - 10x + 3 = 0 \\[1em] \Rightarrow 3x^2 - 9x - x + 3 = 0 \\[1em] \Rightarrow 3x(x - 3) - 1(x - 3) = 0 \\[1em] \Rightarrow (3x - 1)(x - 3) = 0 \\[1em] \Rightarrow 3x - 1 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow 3x = 1 \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{1}{3} \text{ or } x = 3.

Considering a = −73-\dfrac{7}{3} we get,

⇒x+1x=−73⇒x2+1x=−73⇒3(x2+1)=−7x⇒3x2+3=−7x⇒3x2+7x+3=0\Rightarrow x + \dfrac{1}{x} = -\dfrac{7}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = -\dfrac{7}{3} \\[1em] \Rightarrow 3(x^2 + 1) = -7x \\[1em] \Rightarrow 3x^2 + 3 = -7x \\[1em] \Rightarrow 3x^2 + 7x + 3 = 0

Comparing 3x2 + 7x + 3 = 0 with ax2 + bx + c = 0 we get,

a = 3, b = 7 and c = 3.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

⇒x=−7±(7)2−4(3)(3)2(3)=−7±49−366=−7±136.\Rightarrow x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)} \\[1em] = \dfrac{-7 \pm \sqrt{49 - 36}}{6} \\[1em] = \dfrac{-7 \pm \sqrt{13}}{6}.

Hence, x = 3, 13,−7±136.\dfrac{1}{3}, \dfrac{-7 \pm \sqrt{13}}{6}.

Question 7

Solve :

(x2+1x2)−3(x−1x)−2=0(x^2 + \dfrac{1}{x^2}) - 3(x - \dfrac{1}{x}) - 2 = 0

Answer

Let x−1x=ax - \dfrac{1}{x} = a ........(i)

Squaring, both sides we get,

⇒x2+1x2−2=a2⇒x2+1x2=a2+2.......(ii)\Rightarrow x^2 + \dfrac{1}{x^2} - 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 + 2 .......(ii)

Substituting the values from equations (i) and (ii) we get,

(x2+1x2)−3(x−1x)−2=0(x^2 + \dfrac{1}{x^2}) - 3(x - \dfrac{1}{x}) - 2 = 0

⇒ a2 + 2 - 3a - 2 = 0

⇒ a2 - 3a = 0

⇒ a(a - 3) = 0

⇒ a = 0 or a - 3 = 0

⇒ a = 0 or a = 3.

Considering a = 0 we get,

⇒x−1x=0⇒x2−1x=0⇒x2−1=0⇒(x−1)(x+1)=0⇒x−1=0 or x+1=0⇒x=1 or x=−1.\Rightarrow x - \dfrac{1}{x} = 0 \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 0 \\[1em] \Rightarrow x^2 - 1 = 0 \\[1em] \Rightarrow (x - 1)(x + 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -1.

Considering a = 3 we get,

⇒x−1x=3⇒x2−1x=3⇒x2−1=3x⇒x2−3x−1=0\Rightarrow x - \dfrac{1}{x} = 3 \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 3 \\[1em] \Rightarrow x^2 - 1 = 3x \\[1em] \Rightarrow x^2 - 3x - 1 = 0

Comparing x2 - 3x - 1 = 0 with ax2 + bx + c = 0 we get,

a = 1, b = -3 and c = -1.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

⇒x=−(−3)±(−3)2−4(1)(−1)2(1)=3±9+42=3±132.\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 4}}{2} \\[1em] = \dfrac{3 \pm \sqrt{13}}{2}.

Hence, x = 1, -1, 3±132\dfrac{3 \pm \sqrt{13}}{2}.

Question 8

Solve :

(x2 + 5x + 4)(x2 + 5x + 6) = 120

Answer

Let x2 + 5x = a we get,

⇒ (a + 4)(a + 6) = 120

⇒ a2 + 6a + 4a + 24 = 120

⇒ a2 + 10a + 24 - 120 = 0

⇒ a2 + 10a - 96 = 0

⇒ a2 + 16a - 6a - 96 = 0

⇒ a(a + 16) - 6(a + 16) = 0

⇒ (a - 6)(a + 16) = 0

⇒ (a - 6) = 0 or (a + 16) = 0

⇒ a = 6 or a = -16.

Considering a = 6 we get,

⇒ x2 + 5x = 6

⇒ x2 + 5x - 6 = 0

⇒ x2 + 6x - x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x - 1)(x + 6) = 0

⇒ (x - 1) = 0 or (x + 6) = 0

⇒ x = 1 or x = -6.

Considering a = -16 we get,

⇒ x2 + 5x = -16

⇒ x2 + 5x + 16 = 0

In this case,

D = b2 - 4ac = 52 - 4(1)(16) = 25 - 64 = -39 < 0.

It means roots are imaginary in this case.

Hence, x = 1, -6.

Question 9

Solve :

2(2x−1x+3)−3(x+32x−1)=52\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5

x ≠ -3, 12\dfrac{1}{2}

Answer

Substituting, 2x−1x+3\dfrac{2x - 1}{x + 3} = a in above equation

⇒2a−31a=5⇒2a−3a=5⇒2a2−3a=5⇒2a2−3=5a⇒2a2−5a−3=0⇒2a2−6a+a−3=0⇒2a(a−3)+1(a−3)=0⇒(2a+1)(a−3)=0⇒2a+1=0 or a−3=0⇒a=−12 or a=3.\Rightarrow 2a - 3\dfrac{1}{a} = 5 \\[1em] \Rightarrow 2a - \dfrac{3}{a} = 5 \\[1em] \Rightarrow \dfrac{2a^2 - 3}{a} = 5 \\[1em] \Rightarrow 2a^2 - 3 = 5a \\[1em] \Rightarrow 2a^2 - 5a - 3 = 0 \\[1em] \Rightarrow 2a^2 - 6a + a - 3 = 0 \\[1em] \Rightarrow 2a(a - 3) + 1(a - 3) = 0 \\[1em] \Rightarrow (2a + 1)(a - 3) = 0 \\[1em] \Rightarrow 2a + 1 = 0 \text{ or } a - 3 = 0 \\[1em] \Rightarrow a = -\dfrac{1}{2} \text{ or } a = 3.

Considering, a = 3

⇒2x−1x+3=3⇒2x−1=3(x+3)⇒2x−1=3x+9⇒3x−2x=−1−9⇒x=−10.\Rightarrow \dfrac{2x - 1}{x + 3} = 3 \\[1em] \Rightarrow 2x - 1 = 3(x + 3) \\[1em] \Rightarrow 2x - 1 = 3x + 9 \\[1em] \Rightarrow 3x - 2x = -1 - 9 \\[1em] \Rightarrow x = -10.

Considering, a = -12\dfrac{1}{2}

⇒2x−1x+3=−12⇒2(2x−1)=−1(x+3)⇒4x−2=−x−3⇒4x+x=−3+2⇒5x=−1⇒x=−15.\Rightarrow \dfrac{2x - 1}{x + 3} = -\dfrac{1}{2} \\[1em] \Rightarrow 2(2x - 1) = -1(x + 3) \\[1em] \Rightarrow 4x - 2 = -x - 3 \\[1em] \Rightarrow 4x + x = -3 + 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\dfrac{1}{5}.

Hence, x = −15,−10.-\dfrac{1}{5}, -10.

Question 10

Solve: 5x+1 + 52-x = 53 + 1

Answer

Given,

5x+1 + 52-x = 53 + 1

⇒ 5x.51 + 52.5-x = 125 + 1

⇒ 5x.51 + 525x\dfrac{5^2}{5^x} = 126

Let 5x be y.

⇒ 5y + 25y\dfrac{25}{\text{y}} = 126

⇒ 5y2 + 25 = 126y

⇒ 5y2 - 126y + 25 = 0

⇒ 5y2 - 125y - y + 25 = 0

⇒ 5y(y - 25) - (y - 25) = 0

⇒ (y - 25)(5y - 1) = 0

⇒ (y - 25) = 0 or (5y - 1) = 0

⇒ y = 25 or y = 15\dfrac{1}{5}

⇒ y = 52 or y = 5-1

Substituting the value of y,

⇒ 5x = 52 or 5x = 5-1

⇒ x = 2 or -1

Hence, the value of x = 2 or -1.

Test Yourself

Question 1(a)

If 3x2 - kx - 15 = (3x - 5)(x + 3), the value of k is :

  1. 4

  2. -4

  3. 3

  4. -3

Answer

Given,

⇒ 3x2 - kx - 15 = 3x2 + 9x - 5x - 15

⇒ 3x2 - kx - 15 = 3x2 + 4x - 15

From above,

⇒ -k = 4

⇒ k = -4.

Hence, Option 2 is the correct option.

Question 1(b)

If the quadratic equation kx2 + kx + 1 = 0 has real and distinct roots, the value of k is :

  1. 0

  2. 4

  3. 0 and 4

  4. 0 or 4

Answer

Comparing equation kx2 + kx + 1 = 0, with ax2 + bx + c = 0 , we get :

a = k, b = k and c = 1.

Since, quadratic equation has real and distinct roots.

∴ D > 0

∴ b2 - 4ac > 0

⇒ k2 - 4 × k × 1 > 0

⇒ k2 - 4k > 0

⇒ k(k - 4) > 0

⇒ k > 0 or k > 4.

Considering k > 0,

k = 4 can give real and distinct roots for kx2 + kx + 1 = 0.

Hence, Option 2 is the correct option.

Question 1(c)

If x2 - 4x = 5, the value of x is :

  1. 5

  2. -1

  3. 5 or -1

  4. 5 and -1

Answer

Given,

⇒ x2 - 4x = 5

⇒ x2 - 4x - 5 = 0

⇒ x2 - 5x + x - 5 = 0

⇒ x(x - 5) + 1(x - 5) = 0

⇒ (x + 1)(x - 5) = 0

⇒ (x + 1) = 0 or (x - 5) = 0

⇒ x = -1 or x = 5.

Hence, Option 3 is the correct option.

Question 1(d)

If x2 - 7x = 0, the value of x is :

  1. 7

  2. 0

  3. 0 and 7

  4. 0 or 7

Answer

Given,

⇒ x2 - 7x = 0

⇒ x(x - 7) = 0

⇒ x = 0 or x - 7 = 0

⇒ x = 0 or x = 7.

Hence, Option 4 is the correct option.

Question 1(e)

If x = 1 is a root of the equation x+kx−2\sqrt{x} + kx - 2 = 0; the value of k is :

  1. 1

  2. -1

  3. 2

  4. -2

Answer

Given,

x = 1 is a root of the equation x+kx−2\sqrt{x} + kx - 2 = 0.

⇒1+k(1)−2=0⇒1+k−2=0⇒k−1=0⇒k=1.\Rightarrow \sqrt{1} + k(1) - 2 = 0 \\[1em] \Rightarrow 1 + k - 2 = 0 \\[1em] \Rightarrow k - 1 = 0 \\[1em] \Rightarrow k = 1.

Hence, Option 1 is the correct option.

Question 1(f)

The equation 15−2x=x\sqrt{15 - 2x} = x.

Assertion (A): x = 3.

Reason (R): 15−2x=x\sqrt{15 - 2x} = x

⇒15−2x=x2⇒x2−2x−15=0⇒x=−5 or 3\Rightarrow 15 - 2x = x^2 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x = -5 \text{ or } 3

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true and R is correct reason for A.

  4. Both A and R are true and R is incorrect reason for A.

Answer

A is true, R is false.

Reason

Given,

⇒15−2x=x⇒15−2x=x2⇒x2+2x−15=0⇒x2+5x−3x−15=0⇒x(x+5)−3(x+5)=0⇒(x+5)(x−3)=0⇒(x+5)=0 or (x−3)=0⇒x=−5 or x=3\Rightarrow\sqrt{15 - 2x} = x\\[1em] \Rightarrow 15 - 2x = x^2\\[1em] \Rightarrow x^2 + 2x - 15 = 0\\[1em] \Rightarrow x^2 + 5x - 3x - 15 = 0\\[1em] \Rightarrow x(x + 5) - 3(x + 5) = 0\\[1em] \Rightarrow (x + 5)(x - 3) = 0\\[1em] \Rightarrow (x + 5) = 0 \text{ or } (x - 3) = 0\\[1em] \Rightarrow x = -5 \text{ or } x = 3

The quadratic equation mentioned in the reason is x2−2x−15=0x^2 - 2x - 15 = 0 whereas we see that the correct quadratic equation is x2+2x−15=0x^2 + 2x - 15 = 0
∴ Reason (R) is false.

Our solution shows that one of the roots is 3.
∴ Assertion (A) is true.

Hence, option 1 is correct.

Question 1(g)

A quadratic equation ax2 + bx + c = 0 ; where a, b and c are real numbers and a ≠ 0.

Assertion (A): The roots of equation 2x2 + 5x - 3 = 0 are real and unequal.

Reason (R): For the equation ax2 + bx + c = 0, the roots are real and unequal if b2 - 4ac > 0.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true and R is correct reason for A.

  4. Both A and R are true and R is incorrect reason for A.

Answer

Both A and R are true and R is correct reason for A.

Reason

Given, 2x2 + 5x - 3 = 0

As we know that the roots of equation ax2 + bx + c = 0 are real and unequal if b2 - 4ac > 0.

⇒ b2 - 4ac = 52 - 4 x 2 x (-3)

= 25 + 24 = 49 > 0

So, Assertion (A) is true.

And, Reason (R) is also true and it clearly explain assertion as a positive discriminant (b2 - 4ac > 0) guarantees that the roots are real and unequal

Hence, option 3 is correct.

Question 1(h)

One root of a quadratic equation is 3 + 2\sqrt{2}.

Statement (1): The other root of the given quadratic equation is 3 - 2\sqrt{2}.

Statement (2): If one root of the given quadratic equation is in the form of a surd, the other root is its conjugate.

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Both the statements are true.

Reason

We are given that one root of the quadratic equation is 3 + 2\sqrt{2}. If the quadratic equation has real coefficients, the conjugate of a root that involves a surd (i.e., a square root or irrational number) must also be a root of the quadratic equation.

Therefore, 3 - 2\sqrt{2} is other root of the given quadratic equation.

So, statement (1) is true.

The property of conjugate roots holds for quadratic equations with real coefficients, meaning if one root involves a surd, the other root will be its conjugate. In this case, since 3 + 2\sqrt{2} is a surd, the other root must be 3 - 2\sqrt{2}.

So, statement (2) is true.

Hence, option 1 is correct.

Question 1(i)

The quadratic equation 2x2 - 9x + 12 = 0.

Statement (1): Sum of the roots of the equation = 4124\dfrac{1}{2}.

Statement (2): In an quadratic equation ax2 + bx + c = 0, sum of the roots = −ba-\dfrac{\text{b}}{\text{a}}.

Options

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Both the statements are true.

Reason

The given quadratic equation 2x2 - 9x + 12 = 0

Here, a = 2

b = -9

c = 12

Sum of roots = −ba=−−92=92=412-\dfrac{\text{b}}{\text{a}} = -\dfrac{-9}{2} = \dfrac{9}{2} = 4\dfrac{1}{2}

So, Statement (1) is true.

Statement (2) is a well-known property of quadratic equations. The sum of the roots of a quadratic equation ax2 + bx + c = 0 is indeed −ba-\dfrac{\text{b}}{\text{a}}

So, Statement (2) is true.

Hence, option 1 is correct.

Question 2

If p - 15 = 0 and 2x2 + px + 25 = 0; find the values of x.

Answer

Given,

⇒ p - 15 = 0

⇒ p = 15.

Substituting value of p in 2x2 + px + 25 = 0 we get,

⇒ 2x2 + 15x + 25 = 0

⇒ 2x2 + 10x + 5x + 25 = 0

⇒ 2x(x + 5) + 5(x + 5) = 0

⇒ (2x + 5)(x + 5) = 0

⇒ 2x + 5 = 0 or x + 5 = 0

⇒ 2x = -5 or x = -5

⇒ x = -52\dfrac{5}{2} or x = -5.

Hence, x = -5 or −52-\dfrac{5}{2}.

Question 3

Solve :

1p+1q+1x=1x+p+q\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{x + p + q}

Answer

Given,

⇒1p+1q+1x=1x+p+q(1p+1q)+(1x−1x+p+q)=0(q+ppq)+(x+p+q−xx(x+p+q))=0(q+ppq)+(p+qx(x+p+q))=0(p+q)(1pq+1x(x+p+q))=0(p+q)(x(x+p+q)+pqxpq(x+p+q))=0(p+q)(x2+x(p+q)+pqxpq(x+p+q))=0∴(x2+x(p+q)+pqxpq(x+p+q))=0x2+x(p+q)+pq=0\Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{x + p + q} \\[1em] \Big(\dfrac{1}{p} + \dfrac{1}{q}\Big) + \Big(\dfrac{1}{x} - \dfrac{1}{x + p + q}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{x + p + q - x}{x(x + p + q)}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{p + q}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{1}{pq} + \dfrac{1}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{x(x + p + q) + pq}{xpq(x + p + q)} \Big) = 0 \\[1em] (p + q)\Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] \therefore \Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] x^2 + x(p + q) + pq = 0 \\[1em]

Comparing above equation with ax2 + bx + c = 0 we get,

a = 1, b = (p + q), c = pq

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting value in above equation we get,

⇒x=−(p+q)±(p+q)2−4.(1).pq2(1)=−(p+q)±p2+q2+2pq−4pq2=−(p+q)±(p−q)22=−(p+q)±(p−q)2=−(p+q)+(p−q)2 or −(p+q)−(p−q)2=−p−q+p−q2 or −p−q−p+q2=−2q2 or −2p2=−q or −p.\Rightarrow x = \dfrac{-(p + q) \pm \sqrt{(p + q)^2 - 4.(1).pq}}{2(1)} \\[1em] = \dfrac{-(p + q) \pm \sqrt{p^2 + q^2 + 2pq - 4pq}}{2} \\[1em] = \dfrac{-(p + q) \pm \sqrt{(p - q)^2}}{2} \\[1em] = \dfrac{-(p + q) \pm (p - q)}{2} \\[1em] = \dfrac{-(p + q) + (p - q)}{2} \text{ or } \dfrac{-(p + q) - (p - q)}{2} \\[1em] = \dfrac{-p - q + p - q}{2} \text{ or } \dfrac{-p - q - p + q}{2} \\[1em] = \dfrac{-2q}{2} \text{ or } \dfrac{-2p}{2} \\[1em] = -q \text{ or } -p.

Hence, x = -q or -p.

Question 4

Solve the quadratic equation 8x2 - 14x + 3 = 0

(i) When x ∈ I (integers)

(ii) When x ∈ Q (rational numbers)

Answer

Given,

⇒ 8x2 - 14x + 3 = 0

⇒ 8x2 - 12x - 2x + 3 = 0

⇒ 4x(2x - 3) - 1(2x - 3) = 0

⇒ (4x - 1)(2x - 3) = 0

⇒ 4x - 1 = 0 or 2x - 3 = 0

⇒ 4x = 1 or 2x = 3

⇒ x = 14\dfrac{1}{4} or x = 32\dfrac{3}{2}.

(i) Since, there is no integer in the solution,

Hence, no solution.

(ii) Here, x ∈ Q

Hence, x = 14,32\dfrac{1}{4}, \dfrac{3}{2}.

Question 5

Solve, using formula :

x2 + x - (a + 2)(a + 1) = 0

Answer

Comparing above equation with ax2 + bx + c = 0 we get,

a = 1, b = 1, c = -(a + 2)(a + 1)

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get,

⇒−1±(1)2−4(1)(−(a+2)(a+1))2(1)=−1±1+4(a+2)(a+1)2=−1±1+4(a2+a+2a+2)2=−1±1+4(a2+3a+2)2=−1±4a2+12a+8+12=−1±4a2+12a+92=−1±(2a+3)22=−1±2a+32=−1+(2a+3)2 or −1−(2a+3)2=2a+22 or −2a−42=(a+1) or −(a+2).\Rightarrow \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-(a + 2)(a + 1))}}{2(1)} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a + 2)(a + 1)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + a + 2a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 8 + 1}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{-1 \pm 2a + 3}{2} \\[1em] = \dfrac{-1 + (2a + 3)}{2} \text{ or } \dfrac{-1 - (2a + 3)}{2} \\[1em] = \dfrac{2a + 2}{2} \text{ or } \dfrac{-2a - 4}{2} \\[1em] = (a + 1) \text{ or } -(a + 2).

Hence, x = (a + 1) or -(a + 2).

Question 6

If m and n are roots of the equation :

1x−1x−2=3\dfrac{1}{x} - \dfrac{1}{x - 2} = 3; where x ≠ 0 and x ≠ 2; find m × n.

Answer

Solving,

⇒1x−1x−2=3x−2−xx(x−2)=3−2x(x−2)=3−2=3x(x−2)−2=3x2−6x3x2−6x+2=0\Rightarrow \dfrac{1}{x} - \dfrac{1}{x - 2} = 3 \\[1em] \dfrac{x - 2 - x}{x(x - 2)} = 3 \\[1em] \dfrac{-2}{x(x - 2)} = 3 \\[1em] -2 = 3x(x - 2) \\[1em] -2 = 3x^2 - 6x \\[1em] 3x^2 - 6x + 2 = 0 \\[1em]

Comparing 3x2 - 6x + 2 = 0 with ax2 + bx + c = 0 we get,

a = 3, b = -6 and c = 2.

We know that,

x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=−(−6)±(−6)2−4(3)(2)2(3)x=6±36−246x=6±126x=6±4×36x=6±236x=2(3±3)6x=3+33,3−33∴m=3+33 and n=3−33.x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \\[1em] x = \dfrac{6 \pm \sqrt{36 - 24}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{12}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{4 \times 3}}{6} \\[1em] x = \dfrac{6 \pm 2\sqrt{3}}{6} \\[1em] x = \dfrac{2(3 \pm \sqrt{3})}{6} \\[1em] x = \dfrac{3 + \sqrt{3}}{3}, \dfrac{3 - \sqrt{3}}{3} \\[1em] \therefore m = \dfrac{3 + \sqrt{3}}{3} \text{ and } n = \dfrac{3 - \sqrt{3}}{3}.

Solving m × n,

m×n=(3+33)(3−33)=(3)2−(3)29=9−39=69=23.m × n = \Big(\dfrac{3 + \sqrt{3}}{3}\Big)\Big(\dfrac{3 - \sqrt{3}}{3}\Big) \\[1em] = \dfrac{(3)^2 - (\sqrt{3})^2}{9} \\[1em] = \dfrac{9 - 3}{9} \\[1em] = \dfrac{6}{9} \\[1em] = \dfrac{2}{3}.

Hence, m × n = 23\dfrac{2}{3}.

Question 7

One root of the quadratic equation 8x2 + mx + 15 = 0 is 34\dfrac{3}{4}. Find the value of m. Also, find other root of equation.

Answer

Given, 34\dfrac{3}{4} is root of 8x2 + mx + 15 = 0

∴8(34)2+m×34+15=0⇒8×916+3m4+15=0⇒92+3m4+15=0⇒3m4+30+92=0⇒3m4=−392⇒m=−392×43⇒m=−26.\therefore 8\Big(\dfrac{3}{4}\Big)^2 + m \times \dfrac{3}{4} + 15 = 0 \\[1em] \Rightarrow 8 \times \dfrac{9}{16} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{9}{2} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{3m}{4} + \dfrac{30 + 9}{2} = 0 \\[1em] \Rightarrow \dfrac{3m}{4} = -\dfrac{39}{2} \\[1em] \Rightarrow m = -\dfrac{39}{2} \times \dfrac{4}{3} \\[1em] \Rightarrow m = -26.

Substituting value of m in equation,

⇒ 8x2 + mx + 15 = 0

⇒ 8x2 - 26x + 15 = 0

⇒ 8x2 - 20x - 6x + 15 = 0

⇒ 4x(2x - 5) - 3(2x - 5) = 0

⇒ (4x - 3)(2x - 5) = 0

⇒ 4x - 3 = 0 or 2x - 5 = 0

⇒ 4x = 3 or 2x = 5

⇒ x = 34\dfrac{3}{4} or x = 52\dfrac{5}{2}.

Hence, m = -26 and other root = 52\dfrac{5}{2}.

Question 8

Show that one root of the quadratic equation x2 + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

Answer

Substituting x = -3 in x2 + (3 - 2a)x - 6a = 0,

⇒ (-3)2 + (3 - 2a)(-3) - 6a = 0

⇒ 9 - 9 + 6a - 6a = 0

⇒ 0 = 0.

Hence, -3 is one root of the quadratic equation.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

=−(3−2a)±(3−2a)2−4(1)(−6a)2=2a−3±9+4a2−12a+24a2=2a−3±4a2+12a+92=2a−3±(2a+3)22=(2a−3)+(2a+3)2 or (2a−3)−(2a+3)2=4a2 or −62=2a or −3.= \dfrac{-(3 - 2a) \pm \sqrt{(3 - 2a)^2 - 4(1)(-6a)}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{9 + 4a^2 - 12a + 24a}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{(2a - 3) + (2a + 3)}{2} \text{ or } \dfrac{(2a - 3) - (2a + 3)}{2} \\[1em] = \dfrac{4a}{2} \text{ or } \dfrac{-6}{2} \\[1em] = 2a \text{ or } -3.

Hence, the other root is 2a.

Question 9

Find the solution of the quadratic equation 2x2 - mx - 25n = 0; if m + 5 = 0 and n - 1 = 0.

Answer

Given, m + 5 = 0 and n - 1 = 0

∴ m = -5 and n = 1.

Substituting values of m and n in 2x2 - mx - 25n = 0 we get,

⇒ 2x2 - (-5)x - 25(1) = 0

⇒ 2x2 + 5x - 25 = 0

⇒ 2x2 + 10x - 5x - 25 = 0

⇒ 2x(x + 5) - 5(x + 5) = 0

⇒ (2x - 5)(x + 5) = 0

⇒ (2x - 5) = 0 or x + 5 = 0

⇒ x = 52\dfrac{5}{2} or x = -5.

Hence, x = 52\dfrac{5}{2} or -5.

Question 10

Solve : (a + b)2x2 - (a + b)x - 6 = 0; a + b ≠ 0.

Answer

Let (a + b)x = y

⇒ (a + b)2x2 - (a + b)x - 6 = 0

⇒ y2 - y - 6 = 0

⇒ y2 - 3y + 2y - 6 = 0

⇒ y(y - 3) + 2(y - 3) = 0

⇒ (y + 2)(y - 3) = 0

⇒ (y + 2) = 0 or y - 3 = 0

⇒ y = -2 or y = 3.

∴ (a + b)x = -2 or (a + b)x = 3

⇒ x = −2a+b or 3a+b-\dfrac{2}{a + b} \text{ or } \dfrac{3}{a + b}.

Hence, x = −2a+b or 3a+b.-\dfrac{2}{a + b} \text{ or } \dfrac{3}{a + b}.

Question 11

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.

x2 + 2(m - 1)x + (m + 5) = 0

Answer

Since, equation has equal roots, D = 0.

∴ b2 - 4ac = 0

⇒ (2(m - 1))2 - 4(1)(m + 5) = 0

⇒ (2m - 2)2 - (4m + 20) = 0

⇒ 4m2 + 4 - 8m - 4m - 20 = 0

⇒ 4m2 - 12m - 16 = 0

⇒ 4(m2 - 3m - 4) = 0

⇒ m2 - 3m - 4 = 0

⇒ m2 - 4m + m - 4 = 0

⇒ m(m - 4) + 1(m - 4) = 0

⇒ (m + 1)(m - 4) = 0

⇒ m + 1 = 0 or m - 4 = 0

⇒ m = -1 or m = 4.

Hence, m = -1, 4.

Question 12

Find the value of k for which equation 4x2 + 8x - k = 0 has real roots.

Answer

Since equations has real roots, D ≥ 0

∴ b2 - 4ac ≥ 0

⇒ 82 - 4(4)(-k) ≥ 0

⇒ 64 + 16k ≥ 0

⇒ 16k ≥ -64

Dividing both sides by 16 we get,

⇒ k ≥ -4

Hence, k ≥ -4.

Question 13

If -2 is a root of the equation 3x2 + 7x + p = 1, find the value of p. Now find the value of k so that the roots of the equation x2 + k(4x + k - 1) + p = 0 are equal.

Answer

Since, -2 is a root of the equation 3x2 + 7x + p = 1,

∴ 3(-2)2 + 7(-2) + p = 1

⇒ 3(4) - 14 + p = 1

⇒ 12 - 14 + p = 1

⇒ -2 + p = 1

⇒ p = 3.

Substituting value of p in x2 + k(4x + k - 1) + p = 0 we get,

⇒ x2 + k(4x + k - 1) + 3 = 0

⇒ x2 + 4kx + k2 - k + 3 = 0

Since, roots are equal, D = 0.

∴ b2 - 4ac = 0

⇒ (4k)2 - 4(1)(k2 - k + 3) = 0

⇒ 16k2 - 4k2 + 4k - 12 = 0

⇒ 12k2 + 4k - 12 = 0

⇒ 4(3k2 + k - 3) = 0

⇒ 3k2 + k - 3 = 0

⇒k=−1±(1)2−4(3)(−3)2(3)=−1±1+366=−1±376.\Rightarrow k = \dfrac{-1 \pm \sqrt{(1)^2 - 4(3)(-3)}}{2(3)} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 36}}{6} \\[1em] = \dfrac{-1 \pm \sqrt{37}}{6}.

Hence, p = 3 and k = −1±376\dfrac{-1 \pm \sqrt{37}}{6}.

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