Completing the square or not

jinx24

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Jan 23, 2006
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When is "completing the square" used? All of my math books say it is used to make a quadratic equation factorable and a perfect square.

This is my problem:

"Write the function, f(x) = -2x^2 - 4x + 15
in the standard form, f(x) = a ( x - h)^2 + k

I seem to remember that you use completing the square for this, but I can't find it anywhere. If I am wrong, then what else should I use? I am going to double check my work before I post where I think I am going wrong.

As always, thank you.
 
You sure can complete the square.

\(\displaystyle {-2}x^{2}-4x+15\)

\(\displaystyle {-2}x^{2}-4x=-15\)

\(\displaystyle {-2}(x^{2}+2x)=-15\)

\(\displaystyle {-2}(x^{2}+2x+1)=-15-2\)

\(\displaystyle {-2}(x+1)^{2}+17\)
 
I thought so! Here is where I am going wrong:

-2(x^2 +2x + 1) = -15 - 2

Where did you get the -2 on the right side. I thought to complete the square you take half of the b term (2x), square it, then add it to both sides:

-2(x^2 + 2x +1) = -15 +1

What am I doing wrong?
 
You do indeed use half the coefficient of b squared, but you forgot about factoring

out the -2. So, you must add -2 to both sides. See?.

Inside the parentheses you used the 1, so on the other side you added

-2. Anytime you factor out in a 'completing the square' problem you must

compensate. See what I mean?.
 
I think I do:

-2(x^2 + 2x +1) = -15 +1 <---Incorrect; not balanced
= -2x^2 - 4x - 2 = -15 +1 (I did not add the full value to both sides)

I need to remember that when I factor something out when completing the square, I have to multiply it to whatever I am adding to the other side.

Thanks for clearing that up for me.

Jenny
 
Jenny, can you complete the square on this expression:

\(\displaystyle \mbox{ }\)x^2 + 3x + 2
 
oh, wow...you are making me nervous now! I honestly don't know if you can or not. Doesn't " x^2 + 3x +2 " have to be equal to something in order to do anything with it?
 
jinx24 said:
oh, wow...you are making me nervous now! I honestly don't know if you can or not. Doesn't " x^2 + 3x +2 " have to be equal to something in order to do anything with it?

No, it doesn't have to be "equal to something", as long as whatever we ADD to complete the square, we also subtract.

I won't spoil Unco's fun by doing the problem he proposed, but here's another similar problem:

x^2 - 5x + 4
(x^2 - 5x + ?) + 4 - ?

Put the same number in place of each of the question marks:
(x^2 - 5x + 25/4) + 4 - (25/4)
(x - 5/4)<SUP>2</SUP> - (9/4)

There.....we completed the square.
 
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