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OpenStudy (zzr0ck3r):
Do you know how to complete the square?
OpenStudy (anonymous):
no not really
OpenStudy (zzr0ck3r):
well \(x^2+bx=c\implies (x+\frac{b}{2})^2=c+(\frac{b}{2})^2\)
OpenStudy (zzr0ck3r):
so for you
\((x+4)^2=-3+4^2=13\)
OpenStudy (anonymous):
howd you get the 4
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OpenStudy (zzr0ck3r):
So you have \((x+4)^2=13\)
OpenStudy (zzr0ck3r):
well \(x^2+bx=c\) what is \(b\) in your case?
OpenStudy (anonymous):
2
OpenStudy (zzr0ck3r):
\(\huge x^2+bx=c\)
\(\huge x^2+8x=-3\)
compare these two equations, what is in the b position in the second equation
OpenStudy (anonymous):
8
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OpenStudy (zzr0ck3r):
so what is b/2?
OpenStudy (anonymous):
4
OpenStudy (zzr0ck3r):
what is c?
OpenStudy (anonymous):
13
OpenStudy (zzr0ck3r):
look back up to where I said compare the equations
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OpenStudy (zzr0ck3r):
what is c?
OpenStudy (anonymous):
-3
OpenStudy (zzr0ck3r):
right
OpenStudy (zzr0ck3r):
ok now
OpenStudy (zzr0ck3r):
ok using this
\(x^2+\color{blue}{b}x=\color{red}{c}\implies (x+\frac{\color{blue}{b}}{2})^2=\color{red}{c}+(\frac{\color{blue}{b}}{2})^2\)
and the fact that
\(\color{blue}{b=8}\) and \(\color{red}{c=-3}\)
we get \((x+\frac{\color{blue}{8}}{2})^2=-3+(\frac{\color{blue}{8}}{2})^2\)
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OpenStudy (zzr0ck3r):
you with me?
OpenStudy (anonymous):
then you you simplify
OpenStudy (zzr0ck3r):
correct, then you will get
\((x+4)^2=13\)
OpenStudy (zzr0ck3r):
now square both sides
\(x+4=\pm\sqrt{13}\iff x=-4\pm\sqrt{13}\)
OpenStudy (anonymous):
\[-4\pm \sqrt{{13}}\]
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OpenStudy (anonymous):
?
OpenStudy (zzr0ck3r):
what is your question
OpenStudy (anonymous):
is that the answer^
OpenStudy (anonymous):
and i have another question .. about a test
OpenStudy (zzr0ck3r):
yes
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OpenStudy (anonymous):
ive got a final coming up this week thats worth 275(20%) ...it has 36 multiple choice and 4 free response.. i have to get a 60% on the final or i dont get credit for the class. how many questions do you think i would have to get right to get a 60%