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Mathematics 15 Online
OpenStudy (anonymous):

Solve x2 + 8x = -3

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

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

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

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 \((x+\frac{\color{blue}{8}}{2})^2=-3+(\frac{\color{blue}{8}}{2})^2\)

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):

\[-4\pm \sqrt{{13}}\]

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

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%

OpenStudy (anonymous):

@zzr0ck3r

OpenStudy (zzr0ck3r):

thats worth 20% of what?

OpenStudy (zzr0ck3r):

close this and open a new question

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