(x+k)2=z^2+ax+a

Question

if (x+k)2=x^2+axA for some nonzero constant a, what is the value of k?

Answer 1

First, FOIL the left side.

x² + 2kx + k² = x² + ax + a

Both sides have a factor of x² so subtract it to remove it from the equation.

2kx + k² = ax + a

Now, I'm going to move everything over to the left by subtracting ax and a from both sides.

2kx - ax + k² - a = 0

Now, factor out the x from the first two terms.

(2k - a)x + (k² - a) = 0

So, look on the left. We have a constant times x, and x can assume pretty much any number and we have a constant. But on the right, we have 0. This means that no matter what number replace x with, the left has to be 0. So, I can set 2k = a and k² = a and that will guarantee that the last equation will always be zero for every value of x I can think of. Solve those equations and you'll get the numbers Robert gets above.

Answer 2

(x+k)² = x²+ax+a = (x + a/2)² - a²/4+a, after completing the square.

So, - a²/4+a = 0 => a = 4

k = a/2 = 4/2 = 2 <==Answer