Quadratic Equations — Test Yourself

If (k + 2)x² - 2x + 1 = 0 has real roots then greater value of k(∈ Z) is :

1. 1

2. 3

3. -1

4. none of these

Answer

Given,

Equation : (k + 2)x² - 2x + 1 = 0

Comparing (k + 2)x² - 2x + 1 = 0 with ax² + bx + c = 0 we get,

a = (k + 2), b = -2 and c = 1.

Since, roots are real,

∴ D ≥ 0

⇒ b² - 4ac ≥ 0

⇒ (-2)² - 4(k + 2)(1) ≥ 0

⇒ 4 - 4(k + 2) ≥ 0

⇒ 4 - 4k - 8 ≥ 0

⇒ -4k - 4 ≥ 0

⇒ 4k ≤ -4

⇒ k ≤ -1.

Thus, greatest value of k = -1.

Hence, option 3 is the correct option.

Find the greatest value of k ∈ N for which the equation x² - 4x + k = 0 has distinct real roots.

1. -4

2. 3

3. 1

4. 4

Answer

Given,

x² - 4x + k = 0

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

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

Since roots are distinct and real,

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-4)² - 4(1)(k) > 0

⇒ 16 - 4k > 0

⇒ 16 > 4k

⇒ $\dfrac{16}{4}$ > k

⇒ k < 4

Since k ∈ N,

Possible natural numbers less than 4 are: 1, 2, 3

The greatest value is: 3

Hence, Option 2 is the correct option.

If the quadratic equation kx² + kx + 1 = 0 has real and equal 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 equal roots.

∴ D = 0

∴ b² - 4ac = 0

⇒ k² - 4 × k × 1 = 0

⇒ k² - 4k = 0

⇒ k(k - 4) = 0

⇒ k = 0 or (k - 4) = 0.

⇒ k = 0 or k = 4.

Hence, Option 4 is the correct option.

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.

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.

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

1. 1

2. -1

3. 2

4. -2

Answer

Given,

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

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

Hence, Option 1 is the correct option.

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

Assertion (A): x = 3.

Reason (R): $\sqrt{15 - 2x} = x$

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

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,

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

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

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

Hence, option 1 is correct.

A quadratic equation 2x² + 5x - 3 = 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.

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

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

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 + $\sqrt{2}$. 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 - $\sqrt{2}$ 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 + $\sqrt{2}$ is a surd, the other root must be 3 - $\sqrt{2}$.

So, statement (2) is true.

Hence, option 1 is correct.

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 = -$\dfrac{5}{2}$ or x = -5.

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

Solve :

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

Answer

Given,

$\Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{x + p + q} \\[1em] \Big(\dfrac{1}{p} + \dfrac{1}{q}\Big) + \Big(\dfrac{1}{x} - \dfrac{1}{x + p + q}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{x + p + q - x}{x(x + p + q)}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{p + q}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{1}{pq} + \dfrac{1}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{x(x + p + q) + pq}{xpq(x + p + q)} \Big) = 0 \\[1em] (p + q)\Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] \therefore \Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] x^2 + x(p + q) + pq = 0 \\[1em]$

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

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

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

Substituting value in above equation we get,

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

Hence, x = -q or -p.

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 = $\dfrac{1}{4}$ or x = $\dfrac{3}{2}$.

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

Hence, no solution.

(ii) Here, x ∈ Q

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

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

Substituting values we get,

$\Rightarrow \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-(a + 2)(a + 1))}}{2(1)} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a + 2)(a + 1)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + a + 2a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 8 + 1}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{-1 \pm 2a + 3}{2} \\[1em] = \dfrac{-1 + (2a + 3)}{2} \text{ or } \dfrac{-1 - (2a + 3)}{2} \\[1em] = \dfrac{2a + 2}{2} \text{ or } \dfrac{-2a - 4}{2} \\[1em] = (a + 1) \text{ or } -(a + 2).$

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

If m and n are roots of the equation :

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

Answer

Solving,

$\Rightarrow \dfrac{1}{x} - \dfrac{1}{x - 2} = 3 \\[1em] \dfrac{x - 2 - x}{x(x - 2)} = 3 \\[1em] \dfrac{-2}{x(x - 2)} = 3 \\[1em] -2 = 3x(x - 2) \\[1em] -2 = 3x^2 - 6x \\[1em] 3x^2 - 6x + 2 = 0 \\[1em]$

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

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

We know that,

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

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \\[1em] x = \dfrac{6 \pm \sqrt{36 - 24}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{12}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{4 \times 3}}{6} \\[1em] x = \dfrac{6 \pm 2\sqrt{3}}{6} \\[1em] x = \dfrac{2(3 \pm \sqrt{3})}{6} \\[1em] x = \dfrac{3 + \sqrt{3}}{3}, \dfrac{3 - \sqrt{3}}{3} \\[1em] \therefore m = \dfrac{3 + \sqrt{3}}{3} \text{ and } n = \dfrac{3 - \sqrt{3}}{3}.$

Solving m × n,

$m × n = \Big(\dfrac{3 + \sqrt{3}}{3}\Big)\Big(\dfrac{3 - \sqrt{3}}{3}\Big) \\[1em] = \dfrac{(3)^2 - (\sqrt{3})^2}{9} \\[1em] = \dfrac{9 - 3}{9} \\[1em] = \dfrac{6}{9} \\[1em] = \dfrac{2}{3}.$

Hence, m × n = $\dfrac{2}{3}$.

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

Answer

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

$\therefore 8\Big(\dfrac{3}{4}\Big)^2 + m \times \dfrac{3}{4} + 15 = 0 \\[1em] \Rightarrow 8 \times \dfrac{9}{16} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{9}{2} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{3m}{4} + \dfrac{30 + 9}{2} = 0 \\[1em] \Rightarrow \dfrac{3m}{4} = -\dfrac{39}{2} \\[1em] \Rightarrow m = -\dfrac{39}{2} \times \dfrac{4}{3} \\[1em] \Rightarrow m = -26.$

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 = $\dfrac{3}{4}$ or x = $\dfrac{5}{2}$.

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

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

$= \dfrac{-(3 - 2a) \pm \sqrt{(3 - 2a)^2 - 4(1)(-6a)}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{9 + 4a^2 - 12a + 24a}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{(2a - 3) + (2a + 3)}{2} \text{ or } \dfrac{(2a - 3) - (2a + 3)}{2} \\[1em] = \dfrac{4a}{2} \text{ or } \dfrac{-6}{2} \\[1em] = 2a \text{ or } -3.$

Hence, the other root is 2a.

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 = $\dfrac{5}{2}$ or x = -5.

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

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 = $-\dfrac{2}{a + b} \text{ or } \dfrac{3}{a + b}$.

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

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.

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.

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

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

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