Question

What number would you add to the equation below to complete the square? x^2 + 16x = 0
Can some one show me how to do this?

Answer 1

When completing the square, we are trying to put the equation in the form \[(x+a)^2 = (x+a)(x+a) = x*x + a*x + a*x + a*a\]\[=x^2+2ax + a^2\]Compare that with the left hand side of your equation:\[x^2+16x\]If we set \[16x=2ax\]we can see that the common factor of \(x\) can be divided out, leaving\[16=2a\]or\[a=8\]That means that if we add \(a^2= 8^2=64\) to both sides of the equation, we can rewrite the left side as \((x+a)^2 = (x+8)^2\) and the entire equation would be\[(x+8)^2 = 64\]

As a check, let's multiply it out and see if we can return it to our original form.
\[(x+8)^2 = 64\]\[(x+8)(x+8)=64\]\[x^2+8x+8x+64=64\]\[x^2+8x+64=64\]\[x^2+8x=0\checkmark\]

So, to complete the square when you have something of the form \(x^2+ax\) or \(x^2-ax\), add \((a/2)^2\) to both sides, and rewrite \(x^2+ax+(a/2)^2\) as \((x+(a/2))^2\).
Sometimes when completing the square it may be convenient not to be adding to both sides. You can accomplish this by "splitting 0" as I call it. Instead of adding \((a/2)^2\) to both sides, you both add and subtract \((a/2)^2\) from the side on which you are completing the square. The term that is added gets used in the \((x+(a/2))^2\) part, and the term that is subtracted sticks around, possibly to be incorporated into some other term. For example:

\[x^2+4x =0\]Here we take half of 4 and square it:\[(x^2+4x + (4/2)^2 )- (4/2)^2 = 0\]Now rewrite the stuff in the parentheses as our perfect square:\[(x+2)^2 - (4/2)^2 = 0\]\[(x+2)^2-4=0\]Just another way you can work the procedure if for some reason you don't wish to disturb the other side of the equality.