SOLUTION: I am not for sure how to start this one at all. x^2+6x+6=0 and f(x)=16-x^2

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Question 258271: I am not for sure how to start this one at all.
x^2+6x+6=0
and
f(x)=16-x^2
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Are you solving for the zeros?
If that's not what you're asking please re-post.
x%5E2%2B6x%2B6=0
Use the quadratic formula,
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A6+%29%29%2F%282%2A1%29+
x+=+%28-6+%2B-+sqrt%28+36-24+%29%29%2F%282%29+
x+=+%28-6+%2B-+sqrt%28+12+%29%29%2F%282%29+
x+=+%28-6+%2B-+2%2Asqrt%28+3+%29%29%2F%282%29+
x+=+-3+%2B-+sqrt%28+3+%29+ or approximately,
x=-1.26, -4.73
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+graph%28+300%2C+300%2C+-6%2C+2%2C+-2%2C+6%2C+x%5E2%2B6x%2B6%29+
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f(x)=16-x^2
This one you can solve directly.
16-x%5E2=0
x%5E2=16
x=0+%2B-+4
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+graph%28+300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C+x%5E2-16%29+