Question

How do you solve the quadratic equation by completing the square: `x^2-2x-5=0`?

Answer 1

`x_1 = 1 + sqrt(6)`, `x_2 = 1-sqrt(6)`

Explanation:

Start by writing your quadratic in the form

`x^2 + b/ax = -c/a`

In your case, `a=1` to begin with, so you have

`x^2 - 2x = 5`

In order to solve this quadratic by completing the square, you need to write the left side of this equation as a square of a binomial by adding a term to both sides of the equation.

You can determine what this term must be by dividing the coefficient of the `x`-term by 2, then squaring the result.

In your case, you have

`(-2)/2 = (-1)`, then

`(-1)^2 = 1`

This means that the quadratic becomes

`x^2 -2x + 1 = 5 + 1`

The left side of the equation can now be written as

`x^2 - 2x + 1 = x^2 + 2 * (-1) + (-1)^2 = (x-1)^2`

This means that you have

`(x-1)^2 = 6`

Take the square root of both sides to get

`sqrt((x-1)^2) = sqrt(6)`

`x-1 = +-sqrt(6) => x_(1,2) = 1 +- sqrt(6)`

The two solutions to your quadratic will thus be

`x_1 = color(green)(1 + sqrt(6))` and `x_2 = color(green)(1 - sqrt(6))`