Practice the quadratic equations problems using the quadratic formula. In this article, you can find the worksheet on quadratic formula for free. We can solve the quadratic equation using different methods. One of the easiest methods for solving a quadratic equation is the quadratic formula. These Quadratic Equations questions will help you to score maximum marks in the exams. To find the quadratic equation problems you have to write the given question in the standard form.
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Quadratic Equation Formula
The standard form of the quadratic equation is ax² + bx + c = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
where a, b are coefficients of x², x and c are constant.
Quadratic Formula Questions with Answers
Example 1.
Solve the equation x² – 9 = 0
Solution:
Given,
x² – 9 = 0
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 1
b = 0
c = -9
x = [-0 ± √(0² – 4.1.(-9))]/2.1
x = ±6i/2
x = +3i
x = -3i
Example 2.
Solve the equation 2x² + 5x = 9
Solution:
Given,
2x² + 5x = 9
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
2x² + 5x – 9 = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 2
b = 5
c = -9
x = [-5 ± √(5² – 4.2.(-9))]/2.2
x = [-5 ± √(25 + 72)]/4
x = [-5 ± √97]/4
x = [-5 + √97]/4
x = [-5 – √97]/4
Example 3.
Solve the equation 6x² – 11 = 0
Solution:
Given,
6x² – 11 = 0
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 6
b = 0
c = -11
x = [-0 ± √(0² – 4.6.(-11))]/2.6
x = ±√264/12
x = +√264/12 = 1.35401
x = -√264/12 = -1.35401
Example 4.
Solve the equation n² + 10n + 21 = 0
Solution:
Given,
n² + 10n + 21 = 0
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 1
b = 10
c = 21
x = [-10 ± √(10² – 4.1.(21))]/2.1
x = [-10 ± √(100 – 84)]/2
x = [-10 ± √(16)]/2
x = [-10 ± 4]/2
x = [-10 + 4]/2
x = -6/2
x = -3
x = [-10 – 4]/2
x = -14/2
x = -7
Example 5.
Solve the equation 6x² + 12x + 25 = 0
Solution:
Given,
6x² + 12x + 25 = 0
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 6
b = 12
c = 25
x = [-12 ± √(12² – 4.6.(25))]/2.6
x = [-12 ± √(144 – 600)]/12
x = [-12 ± √(-456)]/12
x = [-12 ±2√(114)i]/12
x = -1 + 1.779i
x = -1 – 1.779i
Example 6.
Solve the equation x² – 3x + 10x + 7 = 0
Solution:
Given,
x² – 3x + 10x + 7 = 0
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
x² – 3x + 10x + 7 = 0
x² + 7x + 7 = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 1
b = 7
c = 7
x = [-7 ± √(7² – 4.1.(7))]/2.1
x = [-7 ± √(21)]/2
x = [-7 + √(21)]/2
x = [-7 – √(21)]/2
Example 7.
Solve the equation 6x² + 2x = -3
Solution:
Given,
6x² + 2x = -3
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
6x² + 2x + 3 = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 6
b = 2
c = 3
x = [-2 ± √(2² – 4.6.(3))]/2.6
x = [-2 ± √(4 – 72)]/12
x = [-7 ± √(-68)]/12
x = [-7 ± √(21)]/12
x = [-7 + √(21)]/12
x = [-7 – √(21)]/12
Example 8.
Solve the equation 3x² – 5x = 8
Solution:
Given,
3x² – 5x = 8
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
3x² – 5x – 8 = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 3
b = -5
c = -8
x = [-(-5) ± √(5² – 4.3.(-8))]/2.3
x = [5 ± √(25 – (-96))]/6
x = [5 ± √(-68)]/6
x = [5 ± √(121)]/6
x = [5 ± 11]/6
x = [5 + 11]/6
x = 16/6 = 8/3
x = [5 – 11]/6
x = -6/6
x = -1
Example 9.
Solve the equation x² + 2x + 1 = 0
Solution:
Given,
x² + 2x + 1 = 0
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
x² + 2x + 1 = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 1
b = 2
c = 1
x = [-2 ± √(2² – 4.1.(1))]/2.1
x = [-2 ± √(4 – 4)]/2
x = [-2 ± √(0)]/2
x = [-2± 0]/2
x = -2/2
x = -1
Example 10.
Solve the equation 3x² – 23 = 6x
Solution:
Given,
3x² – 23 = 6x
Convert the given equation into a standard form of the quadratic equation.
The standard form of the quadratic equation is ax² + bx + c = 0
3x² – 6x – 23 = 0
(α, β) = [-b ± √(b² – 4ac)]/2a
a = 3
b = -6
c = -23
x = [-(-6) ± √((-6)² – 4.3.(-23))]/2.3
x = [6 ± √(36 – (-276))]/6
x = [6 ± 2√(78)]/6
x = [6 + 2√(78)]/6
x = [6 – 2√(78)]/6