How do you solve the quadratic equation by completing the square: x^2-2x-5=0x2−2x−5=0?

1 Answer
Stefan V.
Jul 28, 2015

x_1 = 1 + sqrt(6)x1=1+√6, x_2 = 1-sqrt(6)x2=1−√6

Explanation:

Start by writing your quadratic in the form

x^2 + b/ax = -c/ax2+bax=−ca

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

x^2 - 2x = 5x2−2x=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 xx-term by 2, then squaring the result.

In your case, you have

(-2)/2 = (-1)−22=(−1), then

(-1)^2 = 1(−1)2=1

This means that the quadratic becomes

x^2 -2x + 1 = 5 + 1x2−2x+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)^2x2−2x+1=x2+2⋅(−1)+(−1)2=(x−1)2

This means that you have

(x-1)^2 = 6(x−1)2=6

Take the square root of both sides to get

sqrt((x-1)^2) = sqrt(6)√(x−1)2=√6

x-1 = +-sqrt(6) => x_(1,2) = 1 +- sqrt(6)x−1=±√6⇒x1,2=1±√6

The two solutions to your quadratic will thus be

x_1 = color(green)(1 + sqrt(6))x1=1+√6 and x_2 = color(green)(1 - sqrt(6))x2=1−√6

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