Completing the Square


To solve a quadratic with completing the square method, you will add and deduct: ¼ (coefficient of x) 2

Example 1:

Solve the quadratic: x2 + 8x + 4 = 0

Solution:

You cannot find two numbers whose product is 4 and sum is 8. So, we will use completing the square method to solve the given quadratic. Express the given quadratic in the form of the identity:

a2 + 2ab + b2, x2 + 8x + 4 = 0, (x2 + 2 × x × 4 + 42) – 42 + 4 = 0 so, add 42tox2 + 8xto complete the square.

Important Tip:

To solve a quadratic with completing the square method, you will add and deduct: ¼ (coefficient of x) 2

Subtract 42 to keep the given quadratic unchanged. Now, the quadratic:

x2 + 8x + 4 = 0 is (x2 + 2 × x × 4 + 42) – 42 + 4 = 0,
(x + 4)2 – 12 = 0, i.e.
(x + 4)2 – [√ (12)] 2 = 0, i.e.
(x + 4)2 = [√ (12)] 2, so,
x + 4 = ± √ (12), so,
x + 4 = √12, or x + 4 = – √12, so
x = 4 + √12 or x = 4 – √12


Example 2:

Solve the quadratic: 2×2 + x – 4= 0

Solution:

You cannot find two numbers whose product is: (2) × (- 4) i.e. – 8 and whose sum is +1. Use the completing square method.

Use the tips:

  1. Divide the quadratic with the leading coefficient (2, here)
  2. Add and subtract ¼ (x coefficient) 2

½ (2×2 + x – 4) = 0, i.e. x2 + x/2 – 2 = 0
Now add and subtract ¼ (x coefficient) 2 i.e. ¼ (1/2)2 = ¼(1/4) = 1/16
The quadratic becomes: x2 + x/2 + 1/16 – 1/16 – 2 = 0

Write x/2 as 2 × x × ¼ to express in complete the square form:

x2 + x/2 + 1/16 – 1/16 – 2 = 0,
x2 + 2 × x × ¼ + 1/16 = 1/16 + 2,
x2 + 2 × x × ¼ + (¼)2 = 33/16
(x + ¼)2 = 33/16,
(x + ¼) = ±√ (33/4)
x + ¼ = +√ (33/4) or x + ½ = – √ (33/4)
x = ¼ + √ (33/4) or x = ¼ – √ (33/4)
x = (1 + √ 33)/4 or x = (1- √ 33)/4

Insert math as
Block
Inline
Additional settings
Formula color
Text color
#333333
Type math using LaTeX
Preview
\({}\)
Nothing to preview
Insert