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

Question

Nathan Axel · Accepted Answer

Just take 8 on the other side and get started!

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

Julia Adams · Answer

undefined To solve $ x^2 + 2x = 8 $ by completing the square:

1. Move the constant term to the right side: $ x^2 + 2x - 8 = 0 $.
2. Add and subtract $ (\frac{2}{2})^2 = 1 $ inside the parentheses: $ x^2 + 2x + 1 - 1 - 8 = 0 $.
3. Rearrange the expression: $ (x^2 + 2x + 1) - 9 = 0 $.
4. Rewrite the trinomial as a perfect square: $ (x + 1)^2 - 9 = 0 $.
5. Add 9 to both sides: $ (x + 1)^2 = 9 $.
6. Take the square root of both sides: $ x + 1 = \pm 3 $.
7. Solve for x: $ x = -1 \pm 3 $.
8. The solutions are $ x = -1 + 3 = 2 $ and $ x = -1 - 3 = -4 $.