Quadratic Equations — Multiple Choice Questions

Which of the following is not a quadratic equation?

1. (x + 2)² = 2(x + 3)
2. x² + 3x = (-1)(1 - 3x)
3. (x + 2)(x - 1) = x² - 2x - 3
4. x³ - x² + 2x + 1 = (x + 1)³

Answer

Option 1:

$(x + 2)^2 = 2(x + 3) \\[0.5em] \Rightarrow x^2 + 4 + 4x = 2x + 6 \\[0.5em] \Rightarrow x^2 + 4x - 2x + 4 - 6 = 0 \\[0.5em] \Rightarrow x^2 + 2x - 2 = 0$

It is a quadratic equation as highest power of x is 2.

Option 2:

$x^2 + 3x = (-1)(1 - 3x) \\[0.5em] \Rightarrow x^2 + 3x = -1 + 3x \\[0.5em] \Rightarrow x^2 + 3x - 3x + 1 = 0 \\[0.5em] x^2 + 1 = 0$

It is a quadratic equation as the highest power of x is 2.

Option 3:

$(x + 2)(x - 1) = x^2 - 2x - 3 \\[0.5em] \Rightarrow x^2 - x + 2x - 2 = x^2 - 2x - 3 \\[0.5em] \Rightarrow x^2 - x^2 - x + 2x + 2x - 2 + 3 = 0 \\[0.5em] 3x + 1 = 0$

It is not a quadratic equation as highest power of x is not 2.

Option 4:

$x^3 - x^2 + 2x + 1 = (x + 1)^3 \\[0.5em] \Rightarrow x^3 - x^2 + 2x + 1 = x^3 + 1 + 3x(x + 1) \\[0.5em] \Rightarrow x^3 - x^2 + 2x + 1 = x^3 + 1 + 3x^2 + 3x \\[0.5em] \Rightarrow x^3 - x^3 - x^2 - 3x^2 + 2x - 3x + 1 - 1 = 0 \\[0.5em] \Rightarrow -4x^2 - x = 0 \\[0.5em] 4x^2 + x = 0$

It is a quadratic equation as highest power of x is 2.

∴ Option 3 is the correct option.

If 3 is a root of the quadratic equation x² - px + 3 = 0 then the value of p is :

1. 4

2. 3

3. 5

4. 2

Answer

Given,

3 is a root of the quadratic equation x² - px + 3 = 0

∴ 3² - 3p + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 3p = 12

⇒ p = $\dfrac{12}{3}$ = 4.

Hence, Option 1 is the correct option.

The roots of the equation x² - 3x - 10 = 0 are

1. 2, -5
2. -2, 5
3. 2, 5
4. -2, -5

Answer

Given,

$x^2 - 3x - 10 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 2x - 10 = 0 \\[0.5em] \Rightarrow x(x - 5) + 2(x - 5) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 5) = 0 \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 5 = 0 \\[0.5em] x = -2 \text{ or } x = 5 \\[0.5em]$

∴ Option 2 is the correct option

If one root of a quadratic equation with rational coefficients is $\dfrac{3 - \sqrt{5}}{2}$, then the other root is

1. $\dfrac{-3 - \sqrt{5}}{2}$

2. $\dfrac{-3 + \sqrt{5}}{2}$

3. $\dfrac{3 + \sqrt{5}}{2}$

4. $\dfrac{\sqrt{3} + 5}{2}$

Answer

Irrational roots occur in conjugate pair .

Hence if one root is $\dfrac{3 - \sqrt{5}}{2}$ then the other root is =

$\dfrac{3 + \sqrt{5}}{2}$

∴ Option 3 is the correct option

If the equation 2x² - 5x + (k + 3) = 0 has equal roots then the value of k is

1. $\dfrac{9}{8}$

2. $-\dfrac{9}{8}$

3. $\dfrac{1}{8}$

4. $-\dfrac{1}{8}$

Answer

Given ,

2x² - 5x + (k + 3) = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -5 , c = k + 3

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-5)^2 - 4 \times 2 \times (k + 3) = 0 \\[0.5em] \Rightarrow 25 - 8(k + 3) = 0 \\[0.5em] \Rightarrow 25 - 8k - 24 = 0 \\[0.5em] \Rightarrow 1 - 8k = 0 \\[0.5em] \Rightarrow 8k = 1 \\[0.5em] k = \dfrac{1}{8}$

∴ Option 3 is the correct option.

The value(s) of k for which the quadratic equation 2x² - kx + k = 0 has equal roots is (are)

1. 0 only
2. 4
3. 8 only
4. 0, 8

Answer

Given ,

2x² - kx + k = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -k , c = k

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-k)^2 - 4 \times 2 \times k = 0 \\[0.5em] \Rightarrow k^2 - 8k = 0 \\[0.5em] \Rightarrow k(k - 8) = 0 \\[0.5em] \Rightarrow k = 0 \text{ or } k - 8 = 0 \\[0.5em] k = 0 \text{ or } k = 8 \\[0.5em]$

∴ Option 4 is the correct option.

If the equation 3x² - kx + 2k = 0 has equal roots, then the value(s) of k is (are)

1. 6
2. 0 only
3. 24 only
4. 0 or 24

Answer

Given ,

3x² - kx + 2k = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 3 , b = -k , c = 2k

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-k)^2 - 4 \times 3 \times 2k = 0 \\[0.5em] \Rightarrow k^2 - 24k = 0 \\[0.5em] \Rightarrow k(k - 24) = 0 \\[0.5em] \Rightarrow k = 0 \text{ or } k - 24 = 0 \\[0.5em] k = 0 \text{ or } k = 24 \\[0.5em]$

∴ Option 4 is the correct option.

If the equation (k + 1)x² - 2(k - 1)x + 1 = 0 has equal roots, then the values of k are

1. 1, 3

2. 0, 3

3. 0, 1

4. 0, $\dfrac{3}{4}$

Answer

Given ,

(k + 1)x² - 2(k - 1)x + 1 = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= (k + 1) , b = -2(k - 1) , c = 1

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-2(k - 1))^2 - 4 \times (k + 1) \times 1 = 0 \\[0.5em] \Rightarrow 4(k^2 + 1 - 2k) - 4(k + 1) = 0 \\[0.5em] \Rightarrow 4k^2 + 4 - 8k - 4k - 4 = 0 \\[0.5em] \Rightarrow 4k^2 - 12k = 0 \\[0.5em] \Rightarrow 4k(k - 3) = 0 \\[0.5em] \Rightarrow 4k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3.$

∴ Option 2 is the correct option.

If the equation 2x² - 6x + p = 0 has real and different roots, then the values of p are given by

1. $p \lt \dfrac{9}{2}$

2. $p \le \dfrac{9}{2}$

3. $p \gt \dfrac{9}{2}$

4. $p \ge \dfrac{9}{2}$

Answer

Given ,

2x² - 6x + p = 0 has real and different roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -6 , c = p

Since, equation has real and different roots

∴ b² - 4ac > 0

$\Rightarrow (-6)^2 - 4 \times 2 \times p \gt 0 \\[0.5em] \Rightarrow (-6)^2 - 8p \gt 0 \\[0.5em] \Rightarrow 36 - 8p \gt 0 \\[0.5em] \Rightarrow 8p \lt 36 \\[0.5em] \Rightarrow p \lt \dfrac{36}{8} \\[0.5em] p \lt \dfrac{9}{2}$

∴ Option 1 is the correct option.

The quadratic equation $2x^2 - \sqrt{5}x + 1 = 0$ has

1. two distinct real roots
2. two equal real roots
3. no real roots
4. more than two real roots

Answer

In order to find nature of roots we need to find the value of, b² - 4ac

Given,

$2x^2 - \sqrt{5}x + 1 = 0$

Comparing equation with ax² + bx + c = 0
a= 2 , b = -$\sqrt{5}$ , c = 1

Putting values in b² - 4ac

$(-\sqrt{5})^2 - 4 \times 2 \times 1 \\[0.5em] \Rightarrow 5 - 8 \\[0.5em] \Rightarrow -3$

Since, b² - 4ac = -3 < 0 , hence there are no real roots

∴ Option 3 is the correct option.

If the roots of equation x² - 6x + k = 0 are real and distinct, then 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.

The roots of the quadratic equation px² - qx + r = 0 are real and equal if :

(a) p² = 4qr

(b) q² = 4pr

(c) –q² = 4pr

(d) p² > 4qr

Answer

By formula,

D = b² - 4ac

For equation, px² - qx + r = 0

D = (-q)² - 4 × p × r

We know that,

Roots of a quadratic equation are real and equal if discriminant = 0.

⇒ q² - 4pr = 0

⇒ q² = 4pr.

Hence, Option 2 is the correct option.

If x² + kx + 6 = (x - 2)(x - 3) for all values of x, then the value of k is :

1. -5

2. -3

3. -2

4. 5

Answer

Given,

⇒ x² + kx + 6 = (x - 2)(x - 3)

⇒ x² + kx + 6 = x² - 3x - 2x + 6

⇒ x² + kx + 6 = x² - 5x + 6

⇒ x² - x² + kx + 6 - 6 = -5x

⇒ kx = -5x

⇒ k = -5.

Hence, Option 1 is the correct option.

The roots of quadratic equation x² - 1 = 0 are :

1. 0

2. 1

3. -1

4. ±1

Answer

Solving,

⇒ x² - 1 = 0

⇒ (x + 1)(x - 1) = 0

⇒ x + 1 = 0 or x - 1 = 0

⇒ x = -1 or x = 1.

Hence, Option 4 is the correct option.