This shouldn't be hard, but I can't ever seem to remember what to do with a problem like so:

c^2 + 12 = 7c and it asks me to solve it. Do I square root both sides? Or what?

same goes for ones like:

(y+4)^2 - 81 = 0

Thank you for the help.

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- Jan 16th 2008, 06:57 AMtmandersonNeed help solving.....squares..
This shouldn't be hard, but I can't ever seem to remember what to do with a problem like so:

c^2 + 12 = 7c and it asks me to solve it. Do I square root both sides? Or what?

same goes for ones like:

(y+4)^2 - 81 = 0

Thank you for the help. - Jan 16th 2008, 07:12 AMtmanderson
Alright, I managed to use the quadratic on those, though I am having trouble with one similar to this:

x^2 + 1 = 4x

Thanks for the help again. - Jan 16th 2008, 07:12 AMSean12345
The second is easier as there are less steps;

$\displaystyle (y+4)^2-81=0$

$\displaystyle \implies (y+4)^2=81$

$\displaystyle \implies y+4=\pm 9$

$\displaystyle \implies y=-4 \pm 9$

Now for the first;

$\displaystyle c^2+12=7c$

$\displaystyle \implies c^2-7c+12=0$

We want this in the same form as above i.e. $\displaystyle A(c+B)^2+C$

$\displaystyle \implies A(c^2+2Bc+B^2)+C$

$\displaystyle \implies Ac^2+2ABc+AB^2+C$

Now compare co-effecients;

$\displaystyle c^2=Ac^2 \implies A=1$

$\displaystyle 7c=2ABc \implies B=\frac {-7}{2}$

$\displaystyle 12=AB^2+C \implies C=12-(\frac{-7}{2})^2 \implies C=12-\frac {49}{4}$

$\displaystyle \implies C=-\frac{1}{4}$

Hence $\displaystyle c^2-7c+12=(c-\frac{7}{2})^2-\frac{1}{4}=0$

Now solve like number 2. - Jan 16th 2008, 07:15 AMSean12345