SOLUTION: For what real values of c is 9x^2 + 16x + c the square of a binomial? You many find more than one.

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Question 1077332: For what real values of c is 9x^2 + 16x + c the square of a binomial? You many find more than one.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The exact answer can be had by letting ax%2Bb be that binomial, so that:
9x%5E2+%2B+16x+%2B+c+=+%28ax+%2B+b%29%5E2
9x%5E2+%2B+16x+%2B+c+=+a%5E2x%5E2+%2B+2abx+%2B+b%5E2
For those to be identical polynomials, all corresponding coefficients must be equal, so:
9+=+a%5E2 .... leads to a=3 or a=-3
16+=+2ab .... leads to b+=+16%2F2a,b+=+8%2Fa; so, b+=+8%2F3 or b=-8%2F3
c+=+b%5E2 .... whether c is + or - 8%2F3, the square is 64%2F9
So, highlight%28c+=+64%2F9%29
Another approach is to use the discriminant. A quadratic is a perfect square if and only if its discriminant is zero. So:
0+=+b%5E2+-+4ac+=+16%5E2-+4%289%29%28c%29
0+=+256+-+36c
c+=+256%2F36
highlight%28c+=+64%2F9%29