Question

How do you solve by completing the square: `x^2- 4x-11=0`?

Answer 1

• First, we Transpose the Constant to one side of the equation.
  Transposing `-11` to the other side we get:
  `x^2-4x = 11`

• Application of `(a-b)^2 = a^2 - 2ab + b^2`
  We look at the Co-efficient of `x`. It's `-4`
  We take half of this number (including the sign), giving us `–2`
  We square this value to get `(-2)^2 = 4`. We add this number to BOTH sides of the Equation.
  `x^2-4x+4 = 11+4`
  `x^2-4x+4 = 15`
  The Left Hand side `x^2-4x+4` is in the form `a^2 - 2ab + b^2`
  where `a` is `x`, and `b` is `2`

• The equation can be written as
  `(x-2)^2 = 15`

So `(x-2)` can take either `sqrt15` or `-sqrt15` as a value. That's because squaring either will give us 15.

`x-2 = sqrt15` (or) `x-2 = -sqrt15`
`x = 2+sqrt15` (or) `x = 2-sqrt15`

• Solution : `x = 2+sqrt15,2-sqrt15`

• Verify your answer by substituting these values in the Original Equation `x^2- 4x - 11 = 0`
  You will see that the Solution is correct.