Question 628946
We can use the {{{(a+b)^2 = a^2+2ab+b^2}}} for this problem. It gives us a pattern for every square of a binomial.  So for
{{{x^2+kx+64}}} to be equal to {{{(x+r)^2}}}, then it must fit this pattern!<br>
Let's look at {{{a^2+2ab+b^2}}} term by term:
The first term of {{{a^2+2ab+b^2}}} is {{{a^2}}} which is a perfect square. The first term of {{{x^2+kx+64}}} is {{{x^2}}}, also a perfect square. So far we have matched the pattern with the "a" being "x".
The last term of {{{a^2+2ab+b^2}}} is {{{b^2}}} which is also a perfect square. The last term of {{{x^2+kx+64}}} 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^2+2ab+b^2}}} is 2ab, the product of 2, the "a" and the "b". In order for {{{x^2+kx+64}}} to match the pattern of {{{a^2+2ab+b^2}}} 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.