perfect squares

G

Guest

Guest
Did I do this right?

q^2 - 16q + 14
GCF = 8
q^2 = (q)^2 = perfect square
64 = (8)^2 = perfect square
16q = 2(q)(8) middle term is twice the product of q and 8
a=q , b= 8
(q-8)^2

thanks for looking at this for me! :D
Happy July 4th to you all!!!!
 

tkhunny

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Messages
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afreemanny said:
q^2 - 16q + 14
GCF = 8
q^2 = (q)^2 = perfect square
64 = (8)^2 = perfect square
16q = 2(q)(8) middle term is twice the product of q and 8
a=q , b= 8
(q-8)^2
You didn't say what it was you were supposed to do. Completing the square? You're close.

GCF? I don't know what that has to do with completing the square.

16q = 2(q)(8) middle term is twice the product of q and 8
You have what you need in this statement. 8 is the magic number for this problem.

This won't do.
(q-8)^2 = q^2 - 16q + 64 <== That is NOT where you started.

All is not lost. You are close, again.

You have from the beginning q^2 - 16q + 14, but you can't make that 14 into a 64 just because you want to. You have to account for what you do. Getting from 14 to 64 requires 50.

q^2 - 16q + 14 + 50 - 50 <== This IS still what you started with. We added 50 that we needed, but subtracted it so we wouldn't change anything.

q^2 - 16q + 14 + 50 - 50 = q^2 - 16q + 64 - 50 = (q-8)^2 - 50

or, we may need this version

(q-8)^2 - 50 = (q-8)^2 - 2*(5^2)

Well, since you didn't provide an actual problem statement, you'll have to tell me if I guessed very well.
 
G

Guest

Guest
Thanks

I appreciate what you did. I will work on it and see where I get from here. Sure appreciate your help.
 

~Midnight.Kitten~

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Jun 27, 2005
Messages
87
what are perfect circles?
 

tkhunny

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Messages
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1) Don't hijack threads.
2) "Perfect Circle" is redundant. If a circle isn't perfect, it isn't a circle.
 
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