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

1 Answer
Stefan V.
Jul 28, 2015

#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))#

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