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

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

Let's look at term by term:
The first term of is which is a perfect square. The first term of is , also a perfect square. So far we have matched the pattern with the "a" being "x".
The last term of is which is also a perfect square. The last term of 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 is 2ab, the product of 2, the "a" and the "b". In order for to match the pattern of 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.
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