SOLVING QUADRATIC EQUATIONS USING COMPLETING THE SQUARE METHOD
In this section, you will learn how to solve a quadratic equation by completing the square method..
The following steps will be useful to solve a quadratic equation by completing the square.
Step 1 :
In the given quadratic equation ax2 + bx + c = 0, divide the complete equation by a (coefficient of x2).
If the coefficient of x2 is 1 (a = 1), the above process is not required.
Step 2 :
Move the number term (constant) to the right side of the equation.
Step 3 :
In the result of step 2, write the "x" term as a multiple of 2.
Examples :
6x should be written as 2(3)(x).
5x should be written as 2(x)(5/2).
Step 4 :
The result of step 3 will be in the form of
x2 + 2(x)y = k
Step 4 :
Now add y2 to each side to complete the square on the left side of the equation.
Then,
x2 + 2(x)y + y2 = k + y2
Step 5 :
In the result of step 4, if we use the algebraic identity
(a + b)2 = a2 + 2ab + b2
on the left side of the equation, we get
(x + y)2 = k + y2
Step 6 :
Solve (x + y)2 = k + y2 for x by taking square root on both sides.
Example 1 :
Solve by completing the square method
x2 + 6 x – 7 = 0
Solution :
x2 + 6x – 7 = 0
x2 + 6x = 7
x2 + 2 (x) (3) = 7
x2 + 2 (x) (3) + 32 = 7 + 32
(x + 3)2 = 7 + 9
(x + 3)2 = 16
x + 3 = ±√16
x + 3 = ±4
|
x + 3 = 4 x = 4 - 3 x = 1 |
x + 3 = -4 x = -4 - 3 x = -7 |
Example 2 :
Solve by completing the square method
x2 + 3x + 1 = 0
Solution :
x2 + 3x + 1 = 0
x2 + 3x = -1
x2 + 2⋅x⋅(3/2) = -1 + (3/2)2
x2 + 2⋅x⋅(3/2) + (3/2)2 = -1 + (3/2)2
[x + (3/2)]2 = -1 + (9/4)
[x + (3/2)]2 = 5/4
[x + (3/2)] = √(5/4)
[x + (3/2)] = ±√5/2
|
[x + (3/2)] = √5/2 x = (√5/2) - (3/2) x = (√5 - 3)/2 |
[x + (3/2)] = -√5/2 x = (-√5/2) - (3/2) x = (-√5 - 3)/2 |
Example 3 :
Solve by completing the square method
2x2 + 5x - 3 = 0
Solution :
2x2 + 5x - 3 = 0
2x2 + 5x = 3/2
Divide the equation by 2.
x² + (5/2)x = (3/2)
Half of coefficient of x is 5/4.
By adding (5/4)2 on both sides, we get
x² + (5/2)x + (5/4)2 = (3/2) + (5/4)2
x² + (5/2)x + (5/4)2 = (3/2) + (5/4)2
[ x + (5/4) ]2 = (3/2) + (25/16)
[ x + (5/4) ]2 = 49/16
[ x + (5/4) ]2 = √(49/16)
x + (5/4) = ± 7/4
|
x + (5/4) = 7/4 x = 7/4 - (5/4) x = (7 - 5)/4 x = 2/4 x = 1/2 |
x + (5/4) = -7/4 x = -7/4 - (5/4) x = (-7 - 5)/4 x = -12/4 x = -3 |
Example 4 :
Solve by completing the square method
4x2 + 4bx – (a2 - b2) = 0
Solution :
4x2 + 4bx – (a2 - b2) = 0
4x2 + 4bx = (a2 - b2)
x2 + bx = (a2 - b2)/4
Half of coefficient of x is b/2. By adding (b/2)2 on both sides, we get
x2 + bx + (b/2)2 = (a2 - b2)/4 + (b/2)2
[x + (b/2) ]2 = (a2 - b2)/4 + (b2/4)
[x + (b/2) ]2 = a2/4
[x + (b/2) ] = √(a2/4)
[x + (b/2)] = ±(a/2)
|
x + b/2 = a/2 x = a/2 - b/2 x = (a - b)/2 |
x + b/2 = -a/2 x = -a/2 - b/2 x = -(a + b)/2 |
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