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

Concept -

If α and β are the roots of the polynomial f(x) = ax2 + bx + c then

sum of roots = -b/a

product of roots = c/a

Explanation -

Now we have -

If α and β are the roots of the polynomial f(x) = x2 + x + 1

then α + β = -1 .....(i)

and α.β = 1...... (ii)

Now we want to find the value of  \(\rm \frac{1}{\alpha}+\frac{1}{\beta}\)

= \(\frac{\alpha+\beta}{\alpha.\beta}\)

Now put the value of equation (i) and (ii) we get -

= -1/1 = -1

Hence option (3) is true.

Given:

α and β are the roots of the quadratic equation.

α + β = 12

α - β = 4

Formula used:

For a quadratic equation with roots α and β, the equation is:  x² - (α + β)x + (αβ) = 0

Calculations:

From the given information:

α + β = 12 (sum of roots)

α - β = 4 (difference of roots)

We can find α and β by solving these two equations:

Adding (α + β) and (α - β):

α + β + α - β = 12 + 4

⇒ 2α = 16

⇒ α = 8

Substitute α = 8 in α + β = 12:

8 + β = 12

⇒ β = 4

Now, substitute α = 8 and β = 4 in the quadratic equation:

The quadratic equation is:

x² - (α + β)x + (αβ) = 0

⇒ x² - (12)x + (8 × 4) = 0

⇒ x² - 12x + 32 = 0

∴ The required quadratic equation is: x² - 12x + 32 = 0.

Given:

Quadratic equation: 3x² - 5x + 3 = 0

Formula Used:

Discriminant (D) = b² - 4ac

Where a, b, and c are coefficients of the quadratic equation ax² + bx + c = 0.

Calculation:

Here, a = 3, b = -5, c = 3

⇒ Discriminant (D) = b² - 4ac

⇒ D = (-5)² - 4 × 3 × 3

⇒ D = 25 - 36

⇒ D = -11

The discriminant of the quadratic equation is -11.

Calculation

So, 2x2 + 7x – 30 = 0  

⇒ 2x2 + 12x − 5x – 30 = 0

⇒ 2x(x + 6) − 5(x + 6) =0

⇒ (x + 6)(2x − 5) = 0

So, x = - 6 and 5/2

3y2 - 6y - 24 = 0

3y2 + 12y − 6y − 24=0

⇒ 3y (y + 4) − 6(y + 4) =0

⇒ (y + 4) (3y − 6) =0

So, y = −4 and 2

So, the relationship cannot be established between x and y

Option (5) is the correct Option.

 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:

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:

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:

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.

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