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

1 Answer
Aug 1, 2015

Answer:

#x_(1,2) = -1 +- sqrt(6)#

Explanation:

First, get your quadratic equation to the form

#color(blue)(x^2 + b/ax = -c/a)#

To do that, add #5# to both sides of the equation

#x^2 + 2x - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5))) = 5#

#x^2 + 2x = 5#

Next, divide the coefficient of the #x#-term by 2, square the result. then add it to both sides of the equation.

#2/2 = 1#, then #1^2 = 1#

This will get you

#x^2 + 2x + 1 = 5 + 1#

Notice that you can rewrite the left side of the equation as the square of a binomial

#x^2 + 2x + 1 = x^2 + 2 * (1) * x + 1^2 = (x+1)^2#

You will now have

#(x + 1)^2 = 6#

Take the square root of both sides

#sqrt((x+1)^2) = sqrt(6)#

#x + 1 = +- sqrt(6) => x_(1,2) = -1 +- sqrt(6) = {(x_1 = color(green)(-1 - sqrt(6))), (x_2 = color(green)(-1 + sqrt(6))) :}#