SOLUTION: If x^2+kx+64=(x+r)^2 and r>0, what is the value of k?

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Question 628946: If x^2+kx+64=(x+r)^2 and r>0, what is the value of k?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
We can use the %28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2 for this problem. It gives us a pattern for every square of a binomial. So for
x%5E2%2Bkx%2B64 to be equal to %28x%2Br%29%5E2, then it must fit this pattern!

Let's look at a%5E2%2B2ab%2Bb%5E2 term by term:
The first term of a%5E2%2B2ab%2Bb%5E2 is a%5E2 which is a perfect square. The first term of x%5E2%2Bkx%2B64 is x%5E2, also a perfect square. So far we have matched the pattern with the "a" being "x".
The last term of a%5E2%2B2ab%2Bb%5E2 is b%5E2 which is also a perfect square. The last term of x%5E2%2Bkx%2B64 is 64, a perfect square (the square of 8). So far we have matched the pattern with the "a" being "x" and the "b" being "8"
The middle term of a%5E2%2B2ab%2Bb%5E2 is 2ab, the product of 2, the "a" and the "b". In order for x%5E2%2Bkx%2B64 to match the pattern of a%5E2%2B2ab%2Bb%5E2 completely, its middle term, kx, must be the product of 2, the "a" (which is "x") and the "b" (which is "8") So
kx = 2*x*8
which simplifies to:
kx = 16x
So k = 16.