Quadratic Equation questions with answers are provided here. An equation is a statement of equality between two expressions. In the previous article, we have learned about the introduction to quadratic equations. Now in this article, we will discuss how to solve the problems based on quadratic equations in different methods. The value obtained in the quadratic equation is called the roots of the equation. Let us solve the Practice Problems on Quadratic Equations with step-by-step explanations.
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Questions on on Quadratic Equations
Example 1.
Form the quadratic equation whose roots are 2√5, -2√5.
Solution:
Roots of the required equation are 2√5, -2√5
Therefore S = sum of roots = 2√5,-2√5
Sum of the roots = 0
P = product of roots =(2√5) (-2√5) = -4(5)
P = -20
Required equation is
x² – (sum of the roots)x + (product of the roots) = 0
(or)
x² – Sx + P = 0
x² – 0(x) + (-20) = 0
x² – 0 – 20 = 0
x² – 20 = 0
Example 2.
If one root of 6x² – 2x + k = 0 is 2 times the other, find the value of k.
Solution:
Given equation is 6x² – 2x + k = 0
Let one root be α, then other will be 2 α
Sum of roots = -a/b
α + 2 α = -(-2)/6
3 α= 2/6
α = 2/18
Product of the roots = c/a
2 α = k/6
2 α² = k/6
k = 12 α²
Putting the value of α = 2/18 we have
k = 12(2/18)²
= 12 × 4/ 18 × 18
= 12 × 4/324
= 0.148
Example 3.
Find the value of k if the sum of roots of (2k – 1)x² + (3x – 1)x + (k + 2) = 0 is 4/2.
Solution:
(2k – 1)x² + (3x – 1)x + (k + 2) = 0
Here a = (2k – 1), b = (3x – 1), c = (k + 3)
Sum of roots = -b/a
4/2 = – 3k – 1/2k – 1
4(2k – 1) = -2( 3k – 1)
8k – 4 = -6k + 2
8k – 4 + 6k – 2 = 0
Sum of roots = 4/2
8k + 6k = 4 + 2
14k = 8
k = 8/14
k = 4/7
Example 4.
Solve the equation 4x² + 2x = 2
Solution:
4x² + 2x = 2
4x² + 2x – 2 = 0
Composing with the standard form ax² + bx – c = 0
We have
a = 4, b = 2, c = -2
Putting these values in quadratic formula
x = -b ± √b² – 4ac/2a
x = -2 ± √(2)² – 4(4)(-2)/2(4)
x = -2 ± √4 – 4(-8)/8
x = -2 ± √4- 32/8
x = -2 ± √36/8
= -2 ± 6/8
x = -2 + 6/8 or x = -5 – 7/8
x = 4/8 or x = – 12/8
x = ½ or x = – 3/2
Thus the solution set = { ½, – 3/2}
Example 5.
Solve the equation x² = 16
Solution:
Given the equation x² = 16
x² – 16 = 0
(x + 4)(x – 4) = 0
So, one of x + 6 and x – 6 must be zero
x + 4 = 0
x = -4
x – 4 = 0
x = 4
Thus the solutions are x = ±4
Example 6.
Solve the equation 2x² + 5x = -3
Solution:
Given the equation 2x² + 5x = -3
Composing with the standard form ax² + bx – c = 0
a = 2, b = 5, c = 3
2x² + 5x + 3 = 0
2x² + 2x + 3x + 3 = 0
2x(x + 1) +3(x + 1) = 0
(x + 1) (2x + 3) = 0
x + 1 = 0, 2x + 3 = 0
x + 1 = 0
x = -1
2x + 3 = 0
2x = -3
x = -3/2
Thus the solutions are {-1, -3/2}
Example 7.
Solve the equation (x – 4)(x – 3) = 48
Solution:
Given the equation (x – 4)(x – 3) = 48
x(x – 3) -4(x – 3) = 48
x² – 3x – 4x + 12 = 48
x² – 3x – 4x + 12 – 48 = 0
x² – 7x – 36 = 0
a = 1, b = -7, c = -36
Putting these values in quadratic formula
x = -b ± √b² – 4ac/2a
x = 7 ± √(-7)² – 4(1)(-36)/2(1)
x = 7 ± √49 +144/2
x = 7 ± √193/2
x = 7 + √193/2, x = 7 – √193/2
x = 10.4462
x = -3.44
Thus the solution set = {10.4462, – 3.44}
Example 8.
Solve the equation 6x² – 5x – 4 = 0
Solution:
Given the equation 6x² – 5x – 4 = 0
Composing with the standard form ax² + bx – c = 0
a = 6, b = -5 and c = -4
6x² – 8x + 3x – 4 = 0
2x(3x – 4) + 1(3x – 4) = 0
(2x + 1)(3x – 4) = 0
2x + 1 = 0, 3x – 4 = 0
2x = -1
x = -1/2
3x – 4 = 0
3x = 4
x = 4/3
The required solution set is {-1/2, 4/3}
Example 9.
Solve the equation x/2 + 2/x = 4 1/2
Solution:
Given the equation,
x/2 + 2/x = 4 1/2
x² + 4/2x = 9/2
2x² + 8 = 9x
2x² -9x + 8 = 0
Composing with the standard form ax² + bx – c = 0
We have
a = 2, b = -9, c = 8
Putting these values in quadratic formula
x = -b ± √b² – 4ac/2a
x = 9 ± √(-9)² – 4(2)(8)/2(2)
x = 9 ± √81 – 64/4
x = 9 ± √17/4
x = 9 + √17/4, x =9 – √17/4
x = 3.28
x = 1.21
Thus the solution set = { 3.28, 1.21}