SOLVE QUADRATIC EQUATION BY COMPLETING THE SQUARE
There are three ways to solve a quadratic equation.
ii) Quadratic formula
iii) Completing the square.
Here we are going to see, how to solve quadratic equation using completing the square method.
If the quadratic equation is in the form of ax2 + bx + c = 0
Step 1 :
Move the constant to the other side of the equal sign.
Step 2 :
Check if the leading coefficient of x2 is 1. If it is not 1, then divide the quadratic equation by the leading coefficient.
Step 3 :
Write the coefficient of x as multiple of 2.
Step 4 :
So far, the equation will be in the form of a2 + 2ab (or) a2 - 2ab
Step 5 :
Add b2 on both sides, and complete the formula for (a+b)2 or (a-b)2
Step 6 :
Using square root property, solve for the variable x.
Solve for exact values of x by completing the square:
Problem 1 :
x2 + 4x + 1 = 0
Solution:
x2 + 4x + 1 = 0
x2 + 4x = -1
Half of the coefficient of x(4/2) is 2.
x2 + 4x + 22 = -1 + 22
(x + 2)2 = 3
Taking square roots on both sides, we get
(x + 2) = ± √3
x = -2 ± √3
Problem 2 :
x2 - 4x + 1 = 0
Solution:
x2 - 4x + 1 = 0
x2 - 4x = -1
Half of the coefficient of x(4/2) is 2.
x2 - 4x + 22 = -1 + 22
(x - 2)2 = 3
Taking square roots on both sides, we get
(x - 2) = ± √3
x = 2 ± √3
Problem 3 :
x2 + 6x + 2 = 0
Solution:
x2 + 6x + 2 = 0
x2 + 6x = -2
Half of the coefficient of x(6/2) is 3.
x2 + 6x + 32 = -2 + 32
(x + 3)2 = 7
Taking square roots on both sides, we get
(x + 3) = ± √7
x = -3 ± √7
Problem 4 :
x2 - 14x + 46 = 0
Solution:
x2 - 14x + 46 = 0
x2 - 14x = -46
Half of the coefficient of x(14/2) is 7.
x2 - 14x + 72 = -46 + 72
(x - 7)2 = 3
Taking square roots on both sides, we get
(x - 7) = ± √3
x = 7 ± √3
Problem 5 :
x2 = 4x + 3
Solution:
x2 = 4x + 3
x2 - 4x = 3
Half of the coefficient of x(4/2) is 2.
x2 - 4x + 22 = 3 + 22
(x - 2)2 = 7
Taking square roots on both sides, we get
(x - 2) = ± √7
x = 2 ± √7
Problem 6 :
x2 + 6x + 7 = 0
Solution:
x2 + 6x + 7 = 0
x2 + 6x = -7
Half of the coefficient of x(6/2) is 3.
x2 + 6x + 32 = -7 + 32
(x + 3)2 = 2
Taking square roots on both sides, we get
(x + 3) = ± √2
x = -3 ± √2
Problem 7 :
x2 = 2x + 6
Solution:
x2 = 2x + 6
x2 - 2x = 6
Half of the coefficient of x(2/2) is 1.
x2 - 2x + 12 = 6 + 12
(x - 1)2 = 7
Taking square roots on both sides, we get
(x - 1) = ± √7
x = 1 ± √7
Problem 8 :
x2 + 6x = 2
Solution:
x2 + 6x = 2
Half of the coefficient of x(6/2) is 3.
x2 + 6x + 32 = 2 + 32
(x + 3)2 = 11
Taking square roots on both sides, we get
(x + 3) = ± √11
x = -3 ± √11
Problem 9 :
x2 + 10 = 8x
Solution:
x2 - 8x = -10
Half of the coefficient of x(8/2) is 4.
x2 - 8x + 42 = -10 + 42
(x - 4)2 = 6
Taking square roots on both sides, we get
(x - 4) = ± √6
x = 4 ± √6
Problem 10 :
x2 + 6x = -11
Solution:
x2 + 6x = -11
Half of the coefficient of x(6/2) is 3.
x2 + 6x + 32 = -11 + 32
(x + 3)2 = -2
Taking square roots on both sides, we get
(x + 3) = ± √-2
x = -3 ± √-2
So, there is no real solutions.
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