Quadratic Equations — Multiple Choice Questions

The roots of the equation px² + qx + r = 0, where p ≠ 0, are given by:

1. $x = \dfrac{-p \pm \sqrt{q^{2} - 4pr}}{2p}$

2. $x = \dfrac{-q \pm \sqrt{q^{2} - 2pr}}{4p}$

3. $x = \dfrac{-q \pm \sqrt{q^{2} - 4pr}}{2p}$

4. $x = \dfrac{-q \pm \sqrt{q^{2} - 4pr}}{2q}$

Answer

Given,

⇒ px² + qx + r = 0

Comparing equation px² + qx + r = 0 with ax² + bx + c = 0, we get :

a = p, b = q and c = r.

By formula,

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

Substituting values we get :

x = $\dfrac{-q \pm \sqrt{q^2 - 4pr}}{2p}$.

Hence, option 3 is the correct option.

The discriminant of the quadratic equation ax² + bx + c = 0, a ≠ 0 is given by:

1. b² - 2ac

2. b² - ac

3. b² - 4ac

4. none of these

Answer

We know that,

Discriminant (D) = b² - 4ac.

Hence, option 3 is the correct option.

For real roots of a quadratic equation, the discriminant must be:

1. greater than or equal to zero

2. greater than zero

3. less than or equal to zero

4. less than zero

Answer

If the Discriminant is greater than 0, then the roots are real and distinct.

If the Discriminant is equal to 0, then the roots are real and equal.

Thus, for real roots of a quadratic equation, the discriminant must be greater than or equal to zero.

Hence, option 1 is the correct option.

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

1. p² = 4qr

2. q² = 4pr

3. –q² = 4pr

4. 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 the roots of the quadratic equation, ax² + bx + c = 0, a ≠ 0 are real and equal, then each root is equal to:

1. $\dfrac{-a}{2b}$

2. $\dfrac{-b}{2a}$

3. $\dfrac{-2a}{b}$

4. $\dfrac{-c}{2a}$

Answer

Let us consider the quadratic equation ax² + bx + c = 0, a ≠ 0 and a, b, c are real numbers.

Let r₁ and r₂ be the roots of this equation.

r₁ = $\dfrac{-b + \sqrt{D}}{2a}$

r₂ = $\dfrac{-b - \sqrt{D}}{2a}$

if the roots are real and equal then D = 0,

r₁ = $\dfrac{-b + \sqrt{0}}{2a}$

r₁ = $\dfrac{-b}{2a}$

r₂ = $\dfrac{-b - \sqrt{0}}{2a}$

r₂ = $\dfrac{-b}{2a}$

r₁ = r₂ = $\dfrac{-b}{2a}$

Hence, each root can be given by $\dfrac{-b}{2a}$.

Hence, option 2 is the correct option.

If the discriminant of the quadratic equation, ax² + bx + c = 0, a ≠ 0 is greater than zero and a perfect square and a, b, c are rational, then the roots are:

1. rational and equal

2. irrational and unequal

3. irrational and equal

4. rational and unequal

Answer

We know that,

If the Discriminant is greater than 0 and a perfect square and a, b, c are rational, then the roots are rational and unequal.

Hence, option 4 is the correct option.

If the discriminant of a quadratic equation, ax² + bx + c = 0, is greater than zero and a perfect square and b is irrational, then the roots are:

1. irrational and unequal

2. irrational and equal

3. rational and unequal

4. rational and equal

Answer

We know that,

If the Discriminant is greater than 0 and a perfect square and b is irrational, then the roots are irrational and unequal.

Hence, option 1 is the correct option.

Which of the following is a quadratic equation?

1. x² - $2\sqrt{x}$ + 7 = 0

2. 2x² - 5x = (x - 1)²

3. $x - \dfrac{1}{x} = 2x^{2}$

4. x² + $\dfrac{1}{x^{2}}$ = 2

Answer

Solving,

⇒ 2x² - 5x = (x - 1)²

⇒ 2x² - 5x = [(x)² + (1)² - 2 × x × 1]

⇒ 2x² - 5x = x² + 1 - 2x

⇒ 2x² - 5x - x² - 1 + 2x = 0

⇒ 2x² - x² - 5x + 2x - 1 = 0

⇒ x² - 3x - 1 = 0

The highest power of equation is 2 and it's equivalent to ax² + bx + c = 0 . Hence the equation is a quadratic equation.

Hence, option 2 is the correct option.

Which of the following is a quadratic equation?

1. x² + 1 = (2 - x)² + 3

2. 2x² + 3 = (5 + x)(2x - 3)

3. x³ - x² = (x - 1)³

4. none of these

Answer

Solving,

⇒ x³ - x² = (x - 1)³

⇒ x³ - x² = [(x)³ - (1)³ - 3x × 1 × (x - 1)]

⇒ x³ - x² = x³ - 1 - 3x(x - 1)

⇒ x³ - x² = x³ - 1 - 3x² + 3x

⇒ x³ - x² - x³ + 1 + 3x² - 3x = 0

⇒ x³ - x³ - x² + 3x² - 3x + 1 = 0

⇒ 2x² - 3x + 1 = 0

The highest power of equation is 2 and it's equivalent to ax² + bx + c = 0 . Hence the equation is a quadratic equation.

Hence, option 3 is the correct option.

Which of the following is not a quadratic equation?

1. 3x - x² = x² + 5

2. (x + 2)² = 2(x² - 5)

3. $(\sqrt{2}x + 3)^{2} = 2x^{2} + 6$

4. (x - 1)² = 3x² + x - 2

Answer

Solving,

$\Rightarrow (\sqrt{2}x + 3)^{2} = 2x^{2} + 6 \\[1em] \Rightarrow 2x^2 + 6\sqrt{2}x + 9 = 2x^{2} + 6 \\[1em] \Rightarrow 2x^2 - 2x^2 + 6\sqrt{2}x + 9 - 6 = 0 \\[1em] \Rightarrow 6\sqrt{2}x + 3 = 0 \\[1em]$

The highest power of equation is 1 and it's not equivalent to ax² + bx + c = 0 . Thus it is a linear equation.

Hence, option 3 is the correct option.

The roots of the quadratic equation 2x² - x - 6 = 0 are:

1. -2, $\dfrac{3}{2}$

2. 2, $\dfrac{-3}{2}$

3. -2, $\dfrac{-3}{2}$

4. 2, $\dfrac{3}{2}$

Answer

Given,

⇒ 2x² - x - 6 = 0

⇒ 2x² - 4x + 3x - 6 = 0

⇒ 2x(x - 2) + 3(x - 2) = 0

⇒ (2x + 3)(x - 2) = 0

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

⇒ 2x = -3 or x = 2

⇒ x = $-\dfrac{3}{2}$ or x = 2

Hence, option 2 is the correct option.

Which of the following quadratic equations has 2 and 3 as its roots?

1. x² - 5x + 6 = 0

2. x² + 5x + 6 = 0

3. x² - 5x - 6 = 0

4. x² + 5x - 6 = 0

Answer

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 - 5x + 6 = 0.

Hence, option 1 is the correct option.

Which of the following is a root of the quadratic equation, 3x² + 13x + 14 = 0 ?

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

2. $-\dfrac{3}{2}$

3. $-\dfrac{5}{3}$

4. $-\dfrac{7}{3}$

Answer

Given,

⇒ 3x² + 13x + 14 = 0

⇒ 3x² + 6x + 7x + 14 = 0

⇒ 3x(x + 2) + 7(x + 2) = 0

⇒ (3x + 7)(x + 2) = 0

⇒ 3x + 7 = 0 or x + 2 = 0      [Using Zero-product rule]

⇒ 3x = -7 or x = -2

⇒ x = $-\dfrac{7}{3}$ or x = -2

Hence, option 4 is the correct option.

If x = $\dfrac{-1}{2}$ is a solution of the quadratic equation 3x² + 2kx - 3 = 0, then the value of k is:

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

2. $-\dfrac{5}{4}$

3. $-\dfrac{9}{4}$

4. $-\dfrac{4}{5}$

Answer

Given,

x = $\dfrac{-1}{2}$

Equation : 3x² + 2kx - 3 = 0

Substituting value of x in equation:

$\Rightarrow 3x^2 + 2kx - 3 = 0 \\[1em] \Rightarrow 3\Big(\dfrac{-1}{2}\Big)^2 + 2k\Big(\dfrac{-1}{2}\Big) - 3 = 0 \\[1em] \Rightarrow 3\Big(\dfrac{1}{4}\Big) - k - 3 = 0 \\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big) - k = 3 \\[1em] \Rightarrow k = \dfrac{3}{4} - 3 \\[1em] \Rightarrow k = \dfrac{3 - 12}{4} \\[1em] \Rightarrow k = -\dfrac{9}{4}.$

Hence, option 3 is the correct option.

If the equation, x² - ax + 1 = 0 has two distinct and real roots, then:

1. |a| ≥ 2

2. |a| ≤ 2

3. |a| > 2

4. |a| < 2

Answer

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

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

We know that,

Discriminant (D) = b² - 4ac

= (-a)² - 4 × 1 × 1

= a² - 4

Since equations has distinct real roots,

⇒ D > 0

⇒ a² - 4 > 0

⇒ a² > 4

⇒ |a| > $\sqrt{4}$

⇒ |a| > 2.

Hence, option 3 is the correct option.

The positive value of k for which the equations x² + kx + 64 = 0 and x² - 8x + k = 0 will both have real roots, is

1. 16

2. 8

3. 12

4. 4

Answer

Given,

x² + kx + 64 = 0     ....(1)

x² - 8x + k = 0     ....(2)

Comparing x² + kx + 64 = 0 with ax² + bx + c = 0 we get,

a = 1, b = k and c = 64.

We know that,

Discriminant (D) = b² - 4ac

= (k)² - 4 × (1) × (64)

= k² - 256

Since equations has real roots,

⇒ D ≥ 0

⇒ k² - 256 ≥ 0

⇒ k² ≥ 256

⇒ |k| ≥ $\sqrt{256}$

⇒ |k| ≥ 16

⇒ 16 ≤ k ≤ -16

Taking only positive value,

⇒ k ≥ 16 .........(3)

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

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

We know that,

Discriminant (D) = b² - 4ac

= (-8)² - 4 × (1) × (k)

= (64) - 4k

Since equations has real roots,

⇒ D ≥ 0

⇒ 64 - 4k ≥ 0

⇒ 4k ≤ 64

⇒ k ≤ $\dfrac{64}{4}$

⇒ k ≤ 16 .........(4)

From (3) and (4), we get :

⇒ k = 16.

Hence, option 1 is the correct option.

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

1. 4

2. 3

3. 5

4. 2

Answer

Given,

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

Substituting value of x = 3 in equation, we get :

⇒ (3)² - p(3) + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 12 = 3p

⇒ p = $\dfrac{12}{3}$

⇒ p = 4.

Hence, option 1 is the correct option.

The solution set for the quadratic equation 2x² - x + $\dfrac{1}{8}$ = 0 is:

1. $\Big\lbrace\dfrac{1}{4}, \dfrac{1}{4}\Big\rbrace$

2. $\Big\lbrace-\dfrac{1}{4}, \dfrac{1}{4}\Big\rbrace$

3. $\Big\lbrace-\dfrac{1}{2}, \dfrac{1}{4}\Big\rbrace$

4. {4, 4}

Answer

Given,

⇒ 2x² - x + $\dfrac{1}{8}$ = 0

Multiply the equation with 16, we get:

⇒ 16(2x² - x + $\dfrac{1}{8}$ = 0)

⇒ 32x² - 16x + $\dfrac{16}{8}$ = 0

⇒ 32x² - 16x + 2 = 0

⇒ 2(16x² - 8x + 1) = 0

⇒ 16x² - 8x + 1 = 0

⇒ 16x² - 4x - 4x + 1 = 0

⇒ 4x(4x - 1) - 1(4x - 1) = 0

⇒ (4x - 1)(4x - 1) = 0

⇒ (4x - 1) = 0 or (4x - 1) = 0      [Using Zero-product rule]

⇒ 4x = 1 or 4x = 1

⇒ x = $\dfrac{1}{4}$ or x = $\dfrac{1}{4}$

Hence, option 1 is the correct option.

The solution set for the quadratic equation 2x² = 288, is:

1. {12, 12}

2. {-12, -12}

3. {-12, 18}

4. {-12, 12}

Answer

Given,

2x² = 288

$\Rightarrow 2x^2 = 288 \\[1em] \Rightarrow x^2 = \dfrac{288}{2} \\[1em] \Rightarrow x^2 = 144 \\[1em] \Rightarrow x = \sqrt{144} \\[1em] \Rightarrow x = \pm 12 \\[1em] \Rightarrow x = 12 \text{ or } x = -12.$

The solution set = {-12, 12}

Hence, option 4 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, 0

3. 8 only

4. 0, 8

Answer

Given,

Equation : 2x² - kx + k = 0

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

For equal roots,

⇒ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (-k)² - 4 × 2 × k = 0

⇒ k² - 8k = 0

⇒ k(k - 8) = 0

⇒ k = 0 or k - 8 = 0

⇒ k = 0 or k = 8.

Hence, Option 4 is the correct option.

In solving a quadratic equation, one of the values of the variables x is 233.356. The solution rounded to two significant figures is :

1. 233.36

2. 233.35

3. 233.3

4. 230

Answer

x = 233.356

On rounding off to two significant figures

x = 230.

Hence, Option 4 is the correct option.

If the roots of the quadratic equation, px(x - 2) + 6 = 0 are equal, then the value of p is:

1. 0

2. 4

3. 6

4. none of these

Answer

Given,

⇒ px(x - 2) + 6 = 0

⇒ px² - 2px + 6 = 0

Comparing px² - 2px + 6 = 0 with ax² + bx + c = 0 we get,

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

We know that,

Since equations has equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ (-2p)² - 4(p)(6) = 0

⇒ 4p² - 24p = 0

⇒ 4p(p - 6) = 0

⇒ 4p = 0 or p - 6 = 0      [Using Zero-product rule]

⇒ p = 0 or p = 6.

Since p = 0 would make the equation no longer quadratic, we take:

p = 6

Hence, option 3 is the correct option.

If the quadratic equation, px² - $2\sqrt{5}$ px + 15 = 0 has two equal roots, then the value of p is:

1. 0

2. 3

3. 6

4. both 0 and 3

Answer

Comparing px² - $2\sqrt{5}$ px + 15 = 0 with ax² + bx + c = 0 we get,

a = p, b = $-2\sqrt{5}$ p and c = 15.

We know that,

Since equations has equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ ($-2\sqrt{5}$ p)² - 4(p)(15) = 0

⇒ 20p² - 60p = 0

⇒ 20p(p - 3) = 0

⇒ 20p = 0 or (p - 3) = 0      [Using Zero-product rule]

⇒ p = 0 or p = 3.

Since p = 0 would make the equation no longer quadratic, we take:

p = 3

Hence, option 2 is the correct option.

If 1 is a root of the quadratic equation, ky² + ky + 3 = 0, then the value of k is:

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

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

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

4. $-\dfrac{3}{2}$

Answer

Given,

1 is a root of the quadratic equation, ky² + ky + 3 = 0.

Substituting value of y = 1 in equation :

⇒ k(1)² + k(1) + 3 = 0

⇒ k + k + 3 = 0

⇒ 2k + 3 = 0

⇒ 2k = -3

⇒ k = $-\dfrac{3}{2}$.

Hence, option 4 is the correct option.

If the equation x² + 5kx + 16 = 0 has no real roots, then :

1. k > $\dfrac{8}{5}$

2. k < $-\dfrac{8}{5}$

3. $-\dfrac{8}{5} \lt k \lt \dfrac{8}{5}$

4. none of these

Answer

Comparing x² + 5kx + 16 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 5k and c = 16.

We know that,

Since equations has no real roots,

⇒ D < 0

⇒ b² - 4ac < 0

⇒ (5k)² - 4(1)(16) < 0

⇒ 25k² - 64 < 0

⇒ 25k² < 64

$\Rightarrow k^2 \lt \dfrac{64}{25} \\[1em] \Rightarrow |k| \lt \sqrt{\dfrac{64}{25}} \\[1em] \Rightarrow |k| \lt \dfrac{8}{5} \\[1em] \Rightarrow -\dfrac{8}{5} \lt k \lt \dfrac{8}{5}.$

Hence, option 3 is the correct option.

The roots of the quadratic equation 3x² = 6x is:

1. 0

2. 2

3. 0 and 2

4. 0 and 6

Answer

⇒ 3x² = 6x

⇒ 3x² - 6x = 0

⇒ 3x(x - 2) = 0      [Using Zero-product rule]

⇒ 3x = 0 or (x - 2) = 0

⇒ x = 0 or x = 2.

Hence, option 3 is the correct option.

The solution set for the quadratic equation 2x² + kx - k² = 0 is:

1. {k, k}

2. {-k, k}

3. $\Big\lbrace-k, \dfrac{k}{2}\Big\rbrace$

4. $\Big\lbrace\dfrac{-k}{2}, k\Big\rbrace$

Answer

⇒ 2x² + kx - k² = 0

⇒ 2x² + 2kx - kx - k² = 0

⇒ 2x(x + k) - k(x + k) = 0

⇒ (2x - k)(x + k) = 0

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

⇒ 2x = k or x = -k

⇒ x = $\dfrac{k}{2}$ or x = -k

The solution set: $\Big\lbrace-k, \dfrac{k}{2}\Big\rbrace$.

Hence, option 3 is the correct option.

The solution set for the equation, 25x(x + 1) = -4, is:

1. $\Big\lbrace\dfrac{1}{5}, \dfrac{4}{5}\Big\rbrace$

2. $\Big\lbrace-\dfrac{1}{5}, \dfrac{4}{5}\Big\rbrace$

3. $\Big\lbrace-\dfrac{4}{5}, -\dfrac{1}{5}\Big\rbrace$

4. $\Big\lbrace-\dfrac{4}{5}, \dfrac{1}{5}\Big\rbrace$

Answer

⇒ 25x(x + 1) = -4

⇒ 25x² + 25x + 4 = 0

⇒ 25x² + 5x + 20x + 4 = 0

⇒ 5x(5x + 1) + 4(5x + 1) = 0

⇒ (5x + 4)(5x + 1) = 0

⇒ (5x + 4) = 0 or (5x + 1) = 0      [Using Zero-product rule]

⇒ 5x = -4 or 5x = -1

⇒ x = $\dfrac{-4}{5}$ or x = $\dfrac{-1}{5}$.

The solution set: $\Big\lbrace-\dfrac{4}{5}, -\dfrac{1}{5}\Big\rbrace$.

Hence, option 3 is the correct option.

The discriminant of the equation, 3x² - 2x + $\dfrac{1}{3}$ = 0 is :

1. 0

2. 1

3. 2

4. 4

Answer

Comparing 3x² - 2x + $\dfrac{1}{3}$ = 0 with ax² + bx + c = 0 we get,

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

We know that,

⇒ D = b² - 4ac

⇒ D = (-2)² - 4(3) $\Big(\dfrac{1}{3}\Big)$

⇒ D = 4 - 4

⇒ D = 0.

Hence, option 1 is the correct option.

The value of the discriminant of the equation, $\sqrt{3}x^2 + 10x + 7\sqrt{3}$ = 0 is:

1. 4

2. 16

3. -16

4. -12

Answer

Comparing $\sqrt{3}x^2 + 10x + 7\sqrt{3}$ = 0 with ax² + bx + c = 0 we get,

a = $\sqrt{3}$, b = 10 and c = $7\sqrt{3}$.

We know that,

⇒ D = b² - 4ac

⇒ D = (10)² - $4 \times (\sqrt{3}) \times (7\sqrt{3})$

⇒ D = 100 - 84

⇒ D = 16.

Hence, option 2 is the correct option.

The value of the discriminant of the equation, x² - ($\sqrt{2}$ + 1)x + $\sqrt{2}$ = 0 is:

1. 3 + $2\sqrt{2}$

2. 1 - $2\sqrt{2}$

3. 3 - $2\sqrt{2}$

4. 2 - $\sqrt{2}$

Answer

Comparing x² - ($\sqrt{2}$ + 1)x + $\sqrt{2}$ = 0 with ax² + bx + c = 0 we get,

a = 1, b = -($\sqrt{2}$ + 1) and c = $\sqrt{2}$.

We know that,

$\Rightarrow D = b^2 - 4ac \\[1em] = [-(\sqrt{2} + 1)]^2 - 4 \times 1 \times (\sqrt{2}) \\[1em] = (\sqrt2)^2 + 2 \times \sqrt2 \times 1 + 1^2 - 4(\sqrt{2}) \\[1em] = 2 + 2\sqrt2 + 1 - 4\sqrt{2} \\[1em] = 3 - 2\sqrt2.$

Hence, option 3 is the correct option.

The value of the discriminant of the equation 2x² - 3x + 5 = 0, is:

1. 31

2. $\sqrt{-31}$

3. $\sqrt{31}$

4. -31

Answer

Comparing 2x² - 3x + 5 = 0 with ax² + bx + c = 0 we get,

a = 2, b = -3 and c = 5.

We know that,

⇒ D = b² - 4ac

⇒ D = (-3)² - 4(2)(5)

⇒ D = 9 - 40

⇒ D = -31.

Hence, option 4 is the correct option.

If the equation, ax² + 2x + a = 0 has two real and equal roots, then:

1. a = 0, 1

2. a = 1, 1

3. a = 0, -1

4. a = -1, 1

Answer

Comparing ax² + 2x + a = 0 with ax² + bx + c = 0 we get,

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

We know that,

Since equations has real and equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ (2)² - 4(a)(a) = 0

⇒ 4 - 4a² = 0

⇒ 4 = 4a²

⇒ a² = $\dfrac{4}{4}$

⇒ a² = 1

⇒ a = $\sqrt{1}$

⇒ a = ± 1

⇒ a = -1, 1.

Hence, option 4 is the correct option.

The given quadratic equation 3x² + $\sqrt7x$ + 2 = 0 has :

1. two equal real roots.

2. two distinct real roots.

3. more than two real roots.

4. no real roots.

Answer

Comparing the given quadratic equation with ax2 + bx + c = 0, we get :

a = 3, b = $\sqrt7$ and c = 2

Discriminant = b2 - 4ac

= $(\sqrt7)^2$ - 4 x 3 x 2

= 7 - 24

= -17.

Since, D < 0, there are no real roots.

Hence, option 4 is the correct option.

What is the nature of the roots of the equation, 2x² - 6x + 3 = 0?

1. rational and unequal

2. irrational and unequal

3. real and equal

4. imaginary and unequal

Answer

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

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

We know that,

⇒ D = b² - 4ac

⇒ D = (-6)² - 4(2)(3)

⇒ D = 36 - 24

⇒ D = 12

⇒ D > 0 roots are real and unequal.

⇒ D = 12 is not a perfect square, roots are irrational.

Thus, roots are irrational and unequal.

Hence, option 2 is the correct option.

The nature of the roots of the equation, 3x² - $4\sqrt{3}x$ + 4 = 0 is:

1. real and equal

2. irrational and unequal

3. rational and unequal

4. imaginary and unequal

Answer

Comparing 3x² - $4\sqrt{3}x$ + 4 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -$4\sqrt{3}$ and c = 4.

We know that,

⇒ D = b² - 4ac

⇒ D = (-$4\sqrt{3}$)² - 4(3)(4)

⇒ D = 16(3) - 48

⇒ D = 48 - 48

⇒ D = 0.

Thus, roots are real and equal.

Hence, option 1 is the correct option.

If -5 is a root of the quadratic equation 2x² + px - 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then the value of k is:

1. $\dfrac{7}{4}$

2. $\dfrac{5}{4}$

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

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

Answer

Given,

-5 is a root of the quadratic equation 2x² + px - 15 = 0.

Substituting value of x = -5 in 2x² + px - 15 = 0, we get:

⇒ 2(-5)² + p(-5) - 15 = 0

⇒ 2 × 25 - 5p - 15 = 0

⇒ 50 - 5p - 15 = 0

⇒ 35 - 5p = 0

⇒ 5p = 35

⇒ p = $\dfrac{35}{5}$

⇒ p = 7.

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

⇒ 7(x² + x) + k = 0

⇒ 7x² + 7x + k = 0

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

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

Since equation has equal roots,

⇒ Discriminant = 0

⇒ b² - 4ac = 0

⇒ (7)² - 4(7)(k) = 0

⇒ 49 - 28k = 0

⇒ 28k = 49

⇒ k = $\dfrac{49}{28}$

⇒ k = $\dfrac{7}{4}$.

Hence, option 1 is the correct option.