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.
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.
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.