Quadratic Equations - Solutions by Completing the Square Method |
|
Consider the following identities |
x2+2ax+a2=(x+a)2
x2-2ax+a2=(x-a)2
These trinomials are perfect squares. The highest power x2 has unity (i.e., 1) as its coefficient. In this case, the term without an x (the constant term) must equal the square of half the coefficient of x. Therefore, if the terms x2 and x are given, the square may be completed by adding the square of half the coefficient of x.
If ax2+bx+c=0
Transposing ax2+bx=-c
Dividing by a to make the coefficient of x2 = 1 we get |
| |
 |
| |
| To complete the square, we add the square of half the coefficient of x, or |
| |
 |
= |
 |
, to both sides of the equation |
|
| |
We obtain |
|
|
| |
| Taking the square roots on both sides |
| |
 |
| |
| which is, as you know, the formula. |
| |
Example 1 |
Solve using the completing the square method
x2+14x=32
a=1, b=14, c= -32 |
| |
 |
| |
(x + 7)2 = 32 + 49
(x + 7)2 = 81
 x + 7 = ± 9
i.e., x = 9 - 7 or x = -9 - 7
 x = 2 or x = -16 |
| |
Example 2 |
| Solve the equation x2-2x=8 by completing the square, |
| |
comparing x2 - 2x = 8 with x2+ b / a x = c / -a
a = 1, b = -2, c = -8
|
| |
 |
| |
x2 - 2x + (1)2 = 8 + (1)2
(x - 1)2 = 8 + 1
(x - 1)2 = 9
Taking square roots on both the sides
x2 - 1 = ± 3
 x-1=3 or x-1=-3
x=1+3 or x=-3+1
x=4 or x=-2 |
| |
Example 3 |
Solve by completing the square
3x2-5x= 2 |
| |
 |
| |
| Taking the square roots on both sides |
| |
 |
| |
Try these questions: |
Solve by completing the square |
1. |
(2x-3) (3x+4) = 0 |
2. |
3x2+7x = 6 |
3. |
4x2-2x = 3 |
4. |
x2 = 0.6x - 0.05 |
5. |
5(x+2) = 4(2x-1)(x+1)-16 |
6. |
2x2-7x = 4 |
7. |
12x2-cx-20c2 = 0 |
|
| |
Answers to Practice Problems |
| Solutions: To solve by completing the square. |
| |
1. |
(2x-3) (3x+4) = 0
On expanding we get
2x(3x+4) -3 (3x+4) = 0
6x2 +8x -9x -12 = 0
6x2-x-12 = 0
⇒ 6x2-x = 12
Dividing throughout by 6 we get
x2 - x / 6= 12/6
x2 - x / 6 = 2 (1)
Comparing with x2 b / -a x = c / -a we get
a=6 b= -1, c= -2 |
| |
|
| |
 |
| |
|
| |
 |
| |
|
2. |
To solve 3x2 + 7x = 6 by completing the square
Dividing throughout by 3 we get |
| |
|
| |
 |
| |
|
| |
 |
| |
|
|
∴ x = 2/3, -3 are the roots of the equation. |
| |
|
3. |
To solve 4x2-2x = 3 by completing the square
Dividing throughout by a = 4 we get
x2 2 / -4 x = 3/4
Comparing with x2+ b/a x= c/-a
a=4 b=-2, c=-3
|
| |
|
| |
 |
| |
|
| |
Taking square roots on both the sides we get |
| |
|
| |
 |
| |
|
4. |
To solve by completing the square
x2 = 0.6x - 0.05
Transposing
x2-0.6x = -0.05
Comparing with x2+b / a x = c / -a
a = 1, b = - 0.6, c = - 0.05
Adding (b / 2a)2 = (-0.6/2)2 = (-0.3)2 = .09 to both sides we get
x2-0.6x+0.09 = -0.05+0.09
x2-0.6x+(0.3)2 = 0.04
(x-0.3)2 = (0.2)2
Taking square roots on both the sides
x - 0.3 = ± 0.2
⇒ x-0.3 = 0.2 or x - 0.3 = - 0.2
⇒ x = 0.2 + 0.3 or x = - 0.2 + 0.3
⇒ x = 0.5 or x = 0.1
Roots of the equations are x = 0.1, 0.5 |
| |
|
5. |
To solve 5(x+2) = 4(2x-1)(x+1) -16 by completing the square
We first obtain the form ax2 + bx + c = 0
5x+10 = 4[(2x -1)x+(2x - 1) * 1] -16
5x+10 = 4[2x2-x+2x-1]-16
5x+10 = 4[2x2 + x - 1] -16
5x+10 = 8x2 + 4x - 4 - 16
⇒ 8x2 + 4x - 20 - 5x - 10 = 0
⇒ 8x2 - x - 30 = 0
⇒ 8x2 - x = 30
Dividing throughout by 8 |
| |
|
| |
 |
| |
|
| |
 |
| |
|
6. |
To solve 2x2-7x = 4 by completing the square
Dividing throughout by a = 2 |
| |
|
| |
 |
| |
|
| |
 |
| |
|
7. |
To solve 12x2-cx-20c2 = 0 by completing the square
Transposing and dividing by a = 12 we get
x2c / -12 x = 20 / 12 c2
Comparing with x + (b/-a) x = -c/a
a = 12, b = -c, c = - 20c2
|
| |
|
| |
 |
to both sides
|
|
| |
|
| |
Taking square roots on both sides |
| |
|
| |
 |
| |
|
| |
Roots of the equations are |
| |
|
| |
 |
|
| |
| |
|
|