How do you solve #x^2 + 2x = 8# by completing the square?

1 Answer
Jul 15, 2016


Just take 8 on the other side and get started!


Well, as i said take 8 to other side;
#x^2 + 2x-8 =0#

for completing the square method,here are the steps you apply to every quadratic equation mostly:

(1) First multiply the coefficient of #x^2# with the term containing no x.
In this case its : (-8*1)=-8

(2) Now break this term in such a way that the sum or difference of those breaked terms give you the coefficient of the term which has x and the multiplication of those terms give you the original term.

In this case its: 4 ans -2 because the multiplication of these two terms gives you 8 (4*-2=-8) and the sum of these two terms gives you the coefficient of the term containing x i.e. 2 (4-2=2).

(3) Now write the equation with those terms, in the form of x.

In this case: #x^2+4x-2x-8=0#

(4). Now take common such that the terms inside the bracket are same :
In this case: #x(x+4)-2(x+4)=0#
which reveals the same equation.

(5). now write the bracket terms in one bracket and combine the terms outside in one bracket.

In this case: #(x-2)(x+4)#.

Hence the answer is this and thus the roots are 2,-4.