Question

How do you solve this problem x^2+4x-12=0 by completing the square?

Answer 1

@whpalmer4

Answer 2

When we complete the square, we are trying to rearrange the terms involving \(x\) into the form \((x+a)^2\) where \(a\) is a constant. We shovel everything not involving \(x\) off to the other side of the = sign, and take the square root of both sides in order to solve for \(x\).

To do this, we need to find out what the correct constant term is going to be.\[(x+a)^2= (x+a)(x+a) = x^2 + 2ax + a^2\]This shows that If we take the coefficient of \(x\) (not \(x^2\)!), divide it by 2, and then square the result, that's the number that we need to set aside to be the constant term in our squared binomial. Now, if that number can be carved out of something we already have in the equation, it's a piece of cake. However, that often isn't the case.

Here's how I like to do it:

Instead of adding the same value to both sides of the equation, I prefer to simultaneously add AND subtract the same value on the same side of the equation. Let me do an example.

\[x^2+6x+3 = 0\]The coefficient of \(x\) here is \(6, 6/2 = 3, 3^2 = 9\), so I'm going to insert \(+9-9\) on the left side.
\[(x^2+6x+9 -9) + 3 = 0\]Now I move the \(-9\) outside of the parentheses:\[(x^2+6x+9) - 9 + 3 = 0\]I rewrite the expression inside the parentheses as a perfect square. Because we determined that \(a\) is one-half the coefficient of \(x\), we can rewrite it as \((x+a)^2 = (x+3)^2\):
\[(x+3)^2 -6 = 0\]Now I move the remaining detritus to the other side of the = sign and take square roots:\[(x+3)^2 = 6\]\[\sqrt{(x+3)^2} = \sqrt{6}\]\[x+3 = \pm\sqrt{6}\]\[x=-3\pm\sqrt{6}\]

Many teachers will have you instead add the value of \(a^2\) to both sides of the equation instead of doing what I did. This is fine, and completely equivalent. Where my approach pays off for me is when I'm completing the square in something like \[x^2+ 4x+y^2 + 3x + 7 =0\]when I have to complete the square on more than one variable in succession. Do it whichever way works most comfortably for you.

Answer 3

If you move everything that doesn't have an \(x\) or \(x^2\) in it to the opposite side of the = sign as a first step, you may find it a bit easier at first while you are still getting comfortable with the process.

Answer 4

I don't understand what you're trying to say?

Answer 5

your equation. x^2+4x-12=0

what do you need to do to make a left side a perfect square?
to get a perfecet square you would want something like (x+2)^2 because with 2 times 2 the middle term is 4.
what is the last term going to be?
what do you need to do to obtain this last term? (add/subtract WHAT to both sides?)

Answer 6

you add 12 to both sides. Then you take half of 4 then square it and add it to both sides.