SOLUTION: What are the zeros of this function: f(x)=(x+6)^2+2

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Question 672872: What are the zeros of this function:
f(x)=(x+6)^2+2
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%28x%2B6%29%5E2%2B2...set it equal to zero
0=%28x%2B6%29%5E2%2B2

0=x%5E2%2B12x%2B36%2B2

x%5E2%2B12x%2B38=0..use quadratic formula to find zeros

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-12+%2B-+sqrt%28+12%5E2-4%2A1%2A38+%29%29%2F%282%2A1%29+

x+=+%28-12+%2B-+sqrt%28+144-152%29%29%2F2+

x+=+%28-12+%2B-+sqrt%28+-8%29%29%2F2+

x+=+%28-12+%2B-+sqrt%28%28-1%294%2A2%29%29%2F2+

x+=+%28-12+%2B-+2i%2Asqrt%282%29%29%2F2+

complex solutions:

x+=+%28-12+%2B2i%2Asqrt%282%29%29%2F2+

x+=+-12%2F2+%2B2i%2Asqrt%282%29%2F2+

highlight%28x+=+-6+%2Bi%2Asqrt%282%29%29+

or
x+=+%28-12+-2i%2Asqrt%282%29%29%2F2+

x+=+-12%2F2+-2i%2Asqrt%282%29%2F2+

highlight%28x+=+-6+-i%2Asqrt%282%29%29+