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OpenStudy (anonymous):
add 12 to both sides
\[x^2+4x=12\] take half of 4 which is 2 and write
\[(x+2)^2=12+2^2=12+4=16\]
\[(x+2)^2=16\]
take the square root and get
\[x+2=4\] or
\[x+2=-4\] so
\[x=2\] or
\[x=-6\]
OpenStudy (anonymous):
Gracias :)
OpenStudy (anonymous):
welcomas
OpenStudy (anonymous):
if it didnt have = 0 would it be
Ex. of a diff problem, x = 16 -4 +/- 2 sqrt3 = x + 4
OpenStudy (anonymous):
? i think you lost me
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OpenStudy (anonymous):
Nevermind lol
OpenStudy (anonymous):
k
OpenStudy (anonymous):
can yu help me with this...
how you would use the quadratic formula to solve x2 + 5x = –2. Why is the quadratic formula the best method to use?
OpenStudy (anonymous):
it is the best method to use because if you complete the square you are going to get annoying fractions
OpenStudy (anonymous):
solve
\[x^2+5x+2=0\] via
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=1,b=5,c=2\]
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OpenStudy (anonymous):
if you see
\[x^2+6x-1=0\] complete the square because half of 6 is 3
OpenStudy (anonymous):
but if you see
\[x^2+7x-2=0\] might as well use the formula because half of 7 is 7/2 and it will be kind of messy