Quadratic formula questions are given here to help students of Classes 10 and 11 in solving quadratic equations, whose roots are not real numbers. As we know, finding the roots of quadratic equations is simple using the factorisation technique, but there are some cases where we cannot use the approach. In such cases, we must use a technique that helps us find the roots of quadratics effortlessly. That is the quadratic formula.

What is the Quadratic formula?

In mathematics, the quadratic formula is one of the effective techniques for solving quadratic equations, and finding the roots of quadratic equations. The quadratic formula for calculating the roots of a quadratic equation ax² + bx + c = 0 is given by:

x = [-b ± √(b² – 4ac)]/2a

Here, D = b² – 4ac is called the discriminant of the quadratic equation, and based on this, we can define the nature of the roots of a quadratic equation.

• If D < 0, the roots are not real, i.e., imaginary.
• If D = 0, the roots are real and equal.
• If D > 0, the roots are real and unequal.

Learn more about the Quadratic formula and its derivation here.

Also, check other methods of finding the roots of quadratic equations given below.

• Factorisation
• Completing the square

However, the quadratic formula is used to find the roots of a quadratic equation when the above two methods are not sufficient, i.e., to find the imaginary roots.

Quadratic Formula Questions and Answers

1. Using the quadratic formula, find the roots of the quadratic equation 2x² – 7x + 6 = 0.

Solution:

Given,

2x² – 7x + 6 = 0

When compared with the standard form, we have;

a = 2, b = -7 and c = 6

D = b² – 4ac

= (-7)² – 4(2)(6)

= 49 – 48

= 1 > 0

That means the roots are real and unequal.

Using the quadratic formula,

x = [-b ± √(b² – 4ac)]/2a

= [-(-7) ± √1]/ 2(2)

= (7 ± 1)/4

x = (7 + 1)/4, x = (7 – 1)/4

x = 8/4, x = 6/4

x = 2, x = 3/2

Therefore, 2 and 3/2 are the roots of the given equation.

2. Find whether the equation 5x² – 2x – 10 = 0 has real roots. If real roots exist, find them.

Solution:

Given,

5x² – 2x – 10 = 0

Here, a = 5, b = -2, and c = -10

Let us calculate the discriminant value.

D = b² – 4ac

= (-2)² – 4(5)(-10)

= 4 + 200

= 204 > 0

That means the given quadratic equation has two real and unequal roots.

Now, using the quadratic formula, we have;

x = [-b ± √(b² – 4ac)]/2a

= [-(-2) ± √204]/ 2(5)

= (2 ± √(4 × 51))/10

x = (2 + 2√51)/10, x = (2 – 2√51)/10

x = 2(1 + √51)/10, x = 2(1 – √51)/10

x = (1 + √51)/5 , x = (1 – √51)/5

Therefore, the real roots of the given quadratic equation are (1 + √51)/5 and (1 – √51)/5.

3. Calculate the roots of the quadratic equation x – 1/x = 3, x ≠ 0, using the quadratic formula.

Solution:

Given the quadratic equation is:

x – 1/x = 3

(x² – 1)/x = 3

x² – 1 = 3x

x² – 3x – 1 = 0

Here, a = 1, b = -3 and c = -1

D = b² – 4ac

= (-3)² – 4(1)(-1)

= 9 + 4

= 13 > 0

Thus, the roots of the given quadratic equation are real and unequal.

Using the quadratic formula,

x = [-b ± √(b² – 4ac)]/2a

= [-(-3) ± √13]/ 2(1)

= (3 ± √13)/2

x = (3 + √13)/2, x = (3 – √13)/2

Hence, (3 + √13)/2 and (3 – √13)/2 are the roots of the given quadratic equation.

4. What are the roots of the quadratic equation 2x² – √5x + 1 = 0?

Solution:

Given,

2x² – √5x + 1 = 0

Here, a = 2, b = -√5 and c = 1

D = b² – 4ac

= (-√5)² – 4(2)(1)

= 5 – 8

= -3 < 0

That means the roots are imaginary.

Using the quadratic formula, we have;

x = [-b ± √(b² – 4ac)]/2a

= [-(-√5) ± √(-3)]/ 2(2)

= (√5 ± i√3)/4

x = (√5 + i√3)/4, (√5 – i√3)/4

Hence, (√5 + i√3)/4 and (√5 – i√3)/4 are the roots of the quadratic equation 2x² – √5x + 1 = 0.

5. Find the roots of the quadratic equation 4x² + 4√3x + 3 = 0.

Solution:

Given,

4x² + 4√3x + 3 = 0

Here, a = 4, b = 4√3 and c = 3

b² – 4ac = (4√3)² – 4(4)(3)

= 48 – 48

= 0

That means the roots of the given quadratic equation are real and equal.

By using the quadratic formula, we get,

x = [-b ± √(b² – 4ac)]/2a

x = (-4√3 ± √0)/8

x = (-4√3)/8

x = -√3/2

Therefore, -√3/2 and -√3/2 are the roots of the given equation.

6. What will be the roots of the quadratic equation 10x² − 9x + 6 = 0?

Solution:

Given the quadratic equation is:

10x² − 9x + 6 = 0

Here, a = 10, b = -9 and c = 6.

b² – 4ac = (-9)² – 4(10)(6)

= 81 – 240

= -159

Using the quadratic formula, we have;

x = [-(-9) ± √(b² – 4ac)]/2a

= [9 ± √(-159)]/ 2(10)

= (9 ± i√159)/20

x = (9 + i√159)/20, (9 – i√159)/20

Hence, (9 + i√159)/20 and (9 – i√159)/20 are the roots of the quadratic equation 10x² − 9x + 6 = 0.

7. Solve the following equation using the quadratic formula.

4x² – 1 = -8x

Solution:

Given,

4x² – 1 = -8x

4x² + 8x – 1 = 0

Here, a = 4, b = 8 and c = -1

Now, b² – 4ac = (8)² – 4(4)(-1)

= 64 + 16

= 80 > 0

That means the given quadratic equation contains two real and unequal roots.

Now, by using the quadratic formula, we have;

x = [-b ± √(b² – 4ac)]/2a

= [-8 ± √80]/ 2(4)

= (-8 ± √(16 × 5))/8

x = (-8 + 4√5)/8, x = (-8 – 4√5)/8

x = 4(-2 + √5)/8, x = 4(-2 – √5)/8

x = (-2 + √5)/2, x = (-2 – √5)/2

Therefore, the real roots of the given quadratic equation are (-2 + √5)/2 and x = (-2 – √5)/2.

8. What is the nature of the roots of 5x² + 3x + 1 = 0? Find them.

Solution:

Given,

5x² + 3x + 1 = 0

Here, a = 5, b = 3 and c = 1.

Let us calculate the discriminant to determine the nature of the roots of 5x² + 3x + 1 = 0.

D = b² – 4ac

= (3)² – 4(5)(1)

= 9 – 20

= -11 < 0

Thus, the given quadratic equation contains two imaginary roots.

Using the quadratic formula, we have;

x = [-b ± √(b² – 4ac)]/2a

= [-3 ± √(-11)]/ 2(5)

= (-3 ± i√11)/10

x = (-3 + i√11)/10, (-3 – i√11)/10

Hence, (-3 + i√11)/10 and (-3 – i√11)/10 are the roots of the given quadratic equation.

9. Solve: m² − 31 − 2m = −6 − 3m² − 4m

Solution:

Given,

m² − 31 − 2m = −6 − 3m² − 4m

m² – 31 – 2m + 6 + 3m² + 4m = 0

4m² + 2m – 25 = 0

Comparing with the standard form of quadratic equation ax² + bx + c = 0, we get;

a = 4, b = 2 and c = -25

Now, b² – 4ac = (2)² – 4(4)(-25)

= 4 + 400

= 404 > 0

That means the given equation has two real and unequal roots.

Using the quadratic formula,

x = [-b ± √(b² – 4ac)]/2a

m = [-2 ± √404]/ 2(4)

= [-2 ± √(4 × 101)]/8

= (-2 ± 2√101)/8

Therefore, m = (-2 + 2√101)/8 and m = (-2 – 2√101)/8.

10. Write the roots of the quadratic equation 3x² +3x = -3 using the quadratic formula.

Solution:

Given,

3x² +3x = -3

3x² +3x + 3 = 0

3(x² + x + 1) = 0

x² + x + 1 = 0

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

Using the quadratic formula,

x = [-b ± √(b² – 4ac)]/2a

= [-1 ± √{1² – 4(1)(1)}]/ 2(1)

= [-1 ± √(1 – 4)]/ 2

= (-1 ± √-3)/2

= (-1 ± i√3)/2

x = (-1 + i√3)/2, x = (-1 – i√3)/2

Therefore, the roots of 3x² + 3x = -3 are (-1 + i√3)/2 and (-1 – i√3)/2.

Practice Questions on Quadratic Formula

1. Find the roots of the quadratic equation 6x² – 7x + 2 = 0.
2. Find the roots of the quadratic equation x² – (√5/2)x – 1 = 0 using the quadratic formula.
3. Determine the nature of the roots of 3x² – 4√3x + 4 = 0 quadratic equations. If the real roots exist, find them.
4. What are the roots of the quadratic equation 6x² + 3 = -2x?
5. Solve for x: 8x² + 7x − 8 = 0