Quadratic Equations — Exercise 5(D)

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.

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.

If the roots of equation x² - 6x + k = 0 are real and distinct, the value of k is :

1. > -9

2. > -6

3. < 6

4. < 9

Answer

Given,

Roots of equation x² - 6x + k = 0 are real and distinct.

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-6)² - 4 × 1 × k > 0

⇒ 36 - 4k > 0

⇒ 4k < 36

⇒ k < $\dfrac{36}{4}$

⇒ k < 9.

Hence, Option 4 is the correct option.

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.

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.

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

1. 7

2. -7

3. $\dfrac{4}{3}$

4. $-\dfrac{4}{3}$

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.

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.

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.

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.

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

x² + $2\sqrt{3}x - 9$ = 0

Answer

Comparing x² + $2\sqrt{3}$x - 9 = 0 with ax² + bx + c = 0 we get,

a = 1, b = $2\sqrt{3}$ and c = -9.

We know that,

Discriminant = D = b² - 4ac = $(2\sqrt{3})$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.

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 = $\dfrac{204}{12}$

⇒ n = 17

Hence, n = 17.

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.

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.

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.

Use quadratic formula to solve:

$\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$

Answer

Given,

$\Rightarrow \dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{x - 7 - x - 4}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{-11}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow -11 \times 30 = 11(x + 4)(x - 7) \\[1em] \Rightarrow \dfrac{-11 \times 30}{11} = x^2 - 7x + 4x - 28 \\[1em] \Rightarrow -30 = x^2 - 3x - 28 \\[1em] \Rightarrow x^2 - 3x + 2 = 0$

Now we have,

x² - 3x + 2 = 0

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

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

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \dfrac{3 \pm \sqrt{1}}{2} \\[1em] = \dfrac{3 \pm 1}{2} \\[1em] = \dfrac{3 + 1}{2} \text{ or } \dfrac{3 - 1}{2} \\[1em] = \dfrac{4}{2} \text{ or } \dfrac{2}{2} \\[1em] = 2 \text{ or } 1.$

Hence, x = 1 or 2.

Use quadratic formula to solve:

x² = 4x

Answer

Given,

x² = 4x

x² - 4x = 0

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

a = 1, b = -4 and c = 0.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$\Rightarrow x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(0)}}{2(1)} \\[1em] = \dfrac{4 \pm \sqrt{16 - 0}}{2} \\[1em] = \dfrac{4 \pm 4}{2} \\[1em] = \dfrac{4 + 4}{2} \text{ or } \dfrac{4 - 4}{2} \\[1em] = \dfrac{8}{2} \text{ or } \dfrac{0}{2} \\[1em] = 4 \text{ or } 0.$

Hence, x = 0 or 4.

Use quadratic formula to solve:

3y + $\dfrac{5}{16y}$ = 2

Answer

Given,

3y + $\dfrac{5}{16y}$ = 2

$\dfrac{48y^2 + 5}{16y}$ = 2

48y² + 5 = 32y

48y² - 32y + 5 = 0

Comparing 48y² - 32y + 5 = 0 with ax² + bx + c = 0 we get,

a = 48, b = -32 and c = 5.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$\Rightarrow x = \dfrac{-(-32) \pm \sqrt{(-32)^2 - 4(48)(5)}}{2(48)} \\[1em] = \dfrac{32 \pm \sqrt{1024 - 960}}{96} \\[1em] = \dfrac{32 \pm \sqrt{64}}{96} \\[1em] = \dfrac{32 \pm 8}{96} \\[1em] = \dfrac{32 + 8}{96} \text{ or } \dfrac{32 - 8}{96} \\[1em] = \dfrac{40}{96} \text{ or } \dfrac{24}{96} \\[1em] = \dfrac{5}{12} \text{ or } \dfrac{1}{4}.$

Hence, x = $\dfrac{5}{12} \text{ or } \dfrac{1}{4}$.

From each of the following equations, find the value of constant 'k' so that each equation has real and equal roots.

(i) kx(x - 2) + 6 = 0

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Answer

(i) kx(x - 2) + 6 = 0

Given,

kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

Comparing kx² - 2kx + 6 = 0 with ax² + bx + c = 0 we get,

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

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (-2k)² - 4(k)(6) = 0

⇒ 4k² - 24k = 0

⇒ 4k(k - 6) = 0

⇒ 4k = 0 or (k - 6) = 0

⇒ k = 0 or k = 6

If k = 0, substituting in L.H.S. of kx² - 2kx + 6 = 0 we get,

= (0)x² - 2(0)x + 6

= 6

L.H.S. ≠ R.H.S.

Thus, k ≠ 0.

Substituting k = 0 in kx² - 2kx + 6 = 0 we get,

⇒ (6)x² - 2(6)x + 6 = 0

⇒ 6x² - 12x + 6 = 0

⇒ 6x² - 6x - 6x + 6 = 0

⇒ 6x(x - 1) - 6(x - 1) = 0

⇒ (6x - 6)(x - 1) = 0

⇒ (6x - 6) = 0 or (x - 1) = 0

⇒ x = $\dfrac{6}{6}$ or x = 1

⇒ x = 1 equation has real and equal roots

∴ k = 6

Hence, k = 6.

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Given,

(k + 4)x² + (k + 1)x + 1 = 0

Comparing (k + 4)x² + (k + 1)x + 1 = 0 with ax² + bx + c = 0 we get,

a = k + 4, b = k + 1 and c = 1.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (k + 1)² - 4(k + 4)(1) = 0

⇒ k² + 2k + 1 - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k + 3)(k - 5) = 0

⇒ k + 3 = 0 or k - 5 = 0

⇒ k = -3 or k = 5.

If k = 5, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (5 + 4)x² + (5 + 1)x + 1 = 0

⇒ 9x² + 6x + 1 = 0

⇒ (3x + 1)² = 0

⇒ 3x + 1 = 0

⇒ x = $\dfrac{-1}{3}$

Both roots are the same.

If k = -3, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (-3 + 4)x² + (-3 + 1)x + 1 = 0

⇒ x² - 2x + 1 = 0

⇒ (x - 1)² = 0

⇒ x - 1 = 0

⇒ x = 1

Both roots are the same.

Hence, k = -3 or 5.