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OpenStudy (anonymous):
What is the value of K of the equation x2+12x+k=0
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OpenStudy (anonymous):
Is that all that is given?
OpenStudy (gebooors):
k is a constant. Do you have any values for answer?
OpenStudy (ahsome):
\[k=-x^2-12x\]
OpenStudy (zzr0ck3r):
I am guessing them mean for it to be a perfect square? In which case \((12/2)^2=36\)
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