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

 Let’s solve this step by step.

Take a positive number R, when it is multiplied by 8 it becomes 8R, squaring it becomes 64R², when it is divided by 4 it becomes 16R², then it is given that

√16R² = 4R = Q

Given:

ax2 + bx + c = 0 is a quadratic equation 

Roots are in the ratio c : 1

i.e α : β = c : 1 or β : α = c : 1

Concept Used:

Sum of roots (α + β) = -b/a

Product of roots (α.β) = c/a

Calculation:

Given α : β = c : 1     ----(i)

According to the question 

(α + β) = -b/a      ----(ii)

(α.β) = c/a      ----(iii)

On squaring the (i) equation, we get 

α² + β² + 2α.β = \(\frac{b^2}{a^2}\)

Dividing the above equation by α.β, we get 

⇒ \(\frac{\alpha^2}{\alpha .\beta} + \frac{\beta^2}{\alpha .\beta} + 2 = \frac{\frac{b^2}{a^2}}{\frac{c}{a}}\)

⇒ \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2 = \frac{b^2}{ac}\)

Now, from (i)

⇒ \(c + \frac{1}{c} + 2 = \frac{b^2}{ac}\) 

Multiplying the above equation by ac both side, we get 

⇒ ac² + a + 2ac = b²

⇒ a (c2 + 1 + 2c) = b2

⇒ a (c + 1)2 = b2

∴ The correct relation is b2 = a(c + 1)2.

Given:

Divisors are 104, 78, and 260

Concept used:

Least Common Multiple (LCM) and quadratic equations.

Solution:

The least number exactly divisible by 104, 78, and 260 is the LCM of these three numbers.

The LCM of 104, 78, and 260 is 1560.

Therefore, P = 1560

The problem states that P + 40 = Q². Substituting P into this equation gives 1560 + 40 = Q², so Q² = 1600

Solving for Q gives Q = 40 (we only take the positive root because Q represents a real number in this context).

Therefore, the positive value of Q is 40.

Given:

(1 + t2)x2 +2tcx + (c2 − a2) = 0

Concept used: 

If an equation has equal roots,

D = 0

Calculation:

In the given equation,

a = (1 + t²)

b = 2tc

c = (c2 − a2)

⇒ b² - 4ac = 0

⇒ 4t²c² - 4(1 + t2)(c2 − a2) = 0

⇒ t²c² - t²c² + a² - c² + a²t² = 0

⇒ a²(1 + t²) = c² 

∴ a2(1 + t2) = c2 is true.

Given:

The given equation = \({x^2} + 3√ 2x + 6 = 0\)

Concept used:

When quadratic equation in the form ax² + bx + c

Then, sum of roots (α + β) = -b/a

Product of roots (αβ) = c/a

(a + b)² = a² + b² + 2ab

Calculation:

According to above concept,

Sum of roots = -b/a = -3/√2

Product of roots = c/a = 6

Now,

⇒ (α + β)² = α² + β² + 2αβ 

⇒ (-3√2)² = α2 + β2 + 2 × 6

⇒ 18 = α2 + β2 + 12

⇒ α2 + β2 = 6

Now,

⇒  α2 + β2 = 6 = -b/a

⇒  (αβ)² = (6)² = 36 = c/a

So, the required equation = ax² + bx + c = x2 - 6x + 36 

∴ The required equation is x2 - 6x + 36 = 0.

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:

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:

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:

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.

Given:

One root of the equation is \(5 - 2\sqrt 5 \)

Concept: 

If one root of the quadratic equation is in this form \(\left( {a + \sqrt b }\right)\) then the other roots must be conjugate \(\left( {a - \sqrt b }\right)\) and vice-versa.

Quadratic equation: x² - (sum of root) + (product of root) = 0

Calculation: 

Let α = \(5 - 2\sqrt 5 \) and β = \(5 + 2\sqrt 5 \) 

sum of root = α + β = \(5 - 2\sqrt 5 + 5 + 2\sqrt 5 = 10\)  

Product of root = α β = \(\left( {5 - 2\sqrt 5 } \right)\left( {5 + 2\sqrt 5 }\right)\) = 25 - 20 = 5

Quadratic equation = x² - (α + β)x + α β = 0

Now, Quadratic equation = x² - 10x + 5 = 0 

Hence, required quadratic equation is x² - 10x + 5 = 0