Quadratic Equation MCQ Quiz - Objective Question with Answer for Quadratic Equation - Download Free PDF

Equation I: x² − 35x + 294 = 0

⇒ Discriminant = 35² − 4×294 = 1225 − 1176 = 49

⇒ Roots: x = [35 ± 7] ÷ 2 ⇒ x = 21 or 14

Equation II: y² − 38y + 345 = 0

⇒ Discriminant = 38² − 4×345 = 1444 − 1380 = 64

⇒ Roots: y = [38 ± 8] ÷ 2 ⇒ y = 23 or 15

Possible values:

x ∈ {14, 21}

y ∈ {15, 23}

Since sometimes x < y (e.g. 14 < 15) and sometimes x > y (e.g. 21 > 15), the relationship between x and y cannot be uniquely determined.

Thus, the correct answer is x = y or relationship cannot be established.

Given:

The quadratic equation is 4m² + 6m + 2 = 0

Formula used:

To find the roots of a quadratic equation ax² + bx + c = 0, use the quadratic formula:

m = \(\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Where:

a = coefficient of m², b = coefficient of m, c = constant term

Calculation:

Here, a = 4, b = 6, c = 2

⇒ m = \(\dfrac{-6 \pm \sqrt{6^2 - 4 \times 4 \times 2}}{2 \times 4}\)

⇒ m = \(\dfrac{-6 \pm \sqrt{36 - 32}}{8}\)

⇒ m = \(\dfrac{-6 \pm \sqrt{4}}{8}\)

⇒ m = \(\dfrac{-6 \pm 2}{8}\)

Case 1: m = \(\dfrac{-6 + 2}{8} = \dfrac{-4}{8} = -\dfrac{1}{2}\)

Case 2: m = \(\dfrac{-6 - 2}{8} = \dfrac{-8}{8} = -1\)

∴ The roots of the equation are -1/2 and -1. The correct answer is option (3).

Given:

Sum of roots = 4 - 3√(2)

Product of roots = -28

Formula Used:

The quadratic equation based on sum (S) and product (P) of roots is:

x² - (Sum of roots) × x + Product of roots = 0

Calculation:

Substitute the values of Sum of roots = 4 - 3√(2) and Product of roots = 28 :

⇒ x² - (4 - 3√2) × x + (-28) = 0

⇒ x² - (4x - 3√2x) - 28 = 0

The quadratic equation is x² - (4 - 3√2)x - 28 = 0.

Given:

(2x + 3)² - (x + 1)²

Formula used:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Calculation:

(2x + 3)² - (x + 1)²

⇒ [(2x)² + 2 × 2x × 3 + 3²] - [(x)² + 2 × x × 1 + 1²]

⇒ [4x² + 12x + 9] - [x² + 2x + 1]

⇒ 4x² + 12x + 9 - x² - 2x - 1

⇒ (4x² - x²) + (12x - 2x) + (9 - 1)

⇒ 3x² + 10x + 8

∴ The correct answer is option (4).

Given:

The quadratic equation is x² - 2x + 13 = 0

Formula used:

The discriminant (D) of a quadratic equation ax² + bx + c = 0 is given by:

D = b² - 4ac

Where: a = coefficient of x², b = coefficient of x, and c = constant term.

Calculation:

Here, a = 1, b = -2, c = 13

⇒ D = (-2)² - 4 × 1 × 13

⇒ D = 4 - 52

⇒ D = -48

Since the discriminant (D) is negative, the quadratic equation has no real roots.

∴ The correct answer is option (2).

Let the fraction be x.

Reciprocal = 1/x

Then,

x + 3/x = 73/20

⇒ x² + 3 = 73x/20

⇒ 20x² – 73x + 60 = 0

⇒ x = {- (-73) + √ (5329 – 4800)}/40     or        x = {- (-73) - √ (5329 – 4800)}/40

⇒ x = 96/40 = 12/5                                or        x = 50/40 = 5/4

Given:

3x² – ax + 6 = ax² + 2x + 2

⇒ 3x² – ax² – ax – 2x + 6 – 2 = 0

⇒ (3 – a)x² – (a + 2)x + 4 = 0

Concept Used:

If a quadratic equation (ax2 + bx + c=0) has equal roots, then discriminant should be zero i.e. b2 – 4ac = 0

Calculation:

⇒ D = B² – 4AC = 0

⇒ (a + 2)² – 4(3 – a)4 = 0

⇒ a² + 4a + 4 – 48 + 16a = 0

⇒ a² + 20a – 44 = 0

⇒ a² + 22a – 2a – 44 = 0

⇒ a(a + 22) – 2(a + 22) = 0

⇒ a = 2, -22

Given:

x2 – x – 1 = 0

Formula used:

If the given equation is ax² + bx + c = 0

Then Sum of roots = -b/a

And Product of roots = c/a

Calculation:

As α and β are roots of x² – x – 1 = 0, then

⇒ α + β = -(-1) = 1

⇒ αβ = -1

Now, if (α/β) and (β/α) are roots then,

⇒ Sum of roots = (α/β) + (β/α)

⇒ Sum of roots = (α² + β²)/αβ

⇒ Sum of roots = [(α + β)² – 2αβ]/αβ

⇒ Sum of roots = (1)² – 2(-1)]/(-1) = -3

⇒ Product of roots = (α/β) × (β/α) = 1

Now, then the equation is,

⇒ x² – (Sum of roots)x + Product of roots = 0

⇒ x² – (-3)x + (1) = 0

Given:

Two roots are 2 + √5 and 2 - √5.

Concept used:

The quadratic equation is:

x2 - (Sum of roots)x + Product of roots = 0

Calculation:

Let the roots of the equation be A and B.

A = 2 + √5 and B = 2 - √5

⇒ A + B = 2 + √5 + 2 - √5 = 4

⇒ A × B = (2 + √5)(2 - √5) = 4 - 5 = -1

Then equation is

∴ x2 - 4x - 1 = 0

For a quadratic equation, ax² + bx + c = 0,

Sum of the roots = (-b/a) = 4/1

Product of the roots = c/a = -1/1

Then, b = -4

So, the sign of coefficient of x is negative. 

Given our polynomial is (3x2 + ax + 4) and it is perfectly divisible by (x - 5), the remainder is (0) when (x = 5).

So, let's substitute (x = 5) into (3x2 + ax + 4) and set it equal to (0):

[3(5)2 + a(5) + 4 = 0]

[3(25) + 5a + 4 = 0]

[75 + 5a + 4 = 0]

[79 + 5a = 0]

Solving for (a), we get:

[5a = -79]

[a = -79/5

[a = -15.8]

∴ The value of (a) is (-15.8).

Alternate Method3x² + ax + 4 is perfectly divisible by x – 5,

⇒ 3 × 25 + 5a + 4 = 0

⇒ 5a = -79

Given:

The quadratic equation kx (x - 2) + 6 = 0 

Formula used:

b² = 4ac

Calculation:

kx(x – 2) + 6 = 0

⇒ kx² – 2kx + 6 = 0

Since the roots are equal

⇒ b² = 4ac

⇒ (-2k)² = 4 × k × 6

⇒ 4k² = 4k(6)

⇒ k = 6

∴ The value of k is 6.

Given:

5x2 + 2x + Q = 2

Given α = 1/β ⇒ α.β = 1 ----(i)

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Let α and β be the two roots of the above quadratic equation.

The sum of the roots is given by:

α + β = − b/a = −(coefficient of x/coefficient of x²)

The product of the roots is given by:

α × β = c/a = (constant term /coefficient of x²)

Calculation:

Let the roots of 5x² + 2x + Q -2 = 0 are α and β

According to the question,

α = 1/β 

⇒  α.β = 1 

Compare with general equation ax² + bx + c = 0

a = 5, b = 2, c = Q - 2

⇒  (Q – 2)/5 = 1

⇒ Q - 2 = 5

⇒ Q = 7

Hence, Q² = 7² = 49.

Given:

6x2 + 3x2 – 5x + 1

Calculation:

6x² + 3x² – 5x + 1

⇒ 9x² – 5x + 1

Let 'a' and 'b' be two roots of the equations

As we know,

Sum of roots (α + β) = (-b)/a = 5/9

Product of roots (αβ) = c/a = 1/9

According to the question

⇒ 1/α + 1/β

⇒ (α + β)/αβ

Given:

The given equation is ax² + x + b = 0

Concept used:

General form of the quadratic equation is ax2 + x + b = 0

Condition for roots,

For equal and real roots, b2 – 4ac = 0 

For unequal and real roots, b2 – 4ac > 0

For imaginary roots, b2 – 4ac < 0  

Calculation:

For equal and real roots, b2 – 4ac = 0 

⇒ b² = 4ac

After comparing with the general form of the quadratic equation we'll get

b = 1, a = a and c = b

Then, b2 = 4ac

⇒ 1 = 4ab

⇒ ab = 1/4

∴ The correct relation is ab = 1/4

Given:

x⁴ + y⁴ + z⁴ = 3(14 + 9.8xyz);

P = x² + y² - z²; Q = - x² + y² + z²; R = x² - y² + z²

Calculation:

Put y = z = 0

x⁴ = 42

⇒ P = x²

⇒ Q = - x²

⇒ R = x²

Now,

(P - Q + R)2 - (P2 + Q2 + R2)

⇒ (x² - (-x²) + x²)2 - [(x²)2 + (-x²)2 + (x²)²]

⇒ (x2 + x2 + x2)2 - [x4 + x⁴ + x⁴]

⇒ (3x²)² - (3x⁴)

⇒ 9x⁴ - 3x⁴

⇒ 6x⁴ = 6 × 42 = 252

∴ The correct answer is 252.