+887 For what real values of $c$ is $x^2 - 8x - 14x + c$ the square of a binomial?
+1365 First, let's combine some like terms!
\(x^2 - 22c + c\).
In order for this to be the square of a binomial, we need c to be the square of half the value of -22.
Solving this, we get c = 121.
Indeed, if we have \(x^2-22x+121=(x-11)^2\).
So we have c = 121.
Thanks! :)
+1365 First, let's combine some like terms!
\(x^2 - 22c + c\).
In order for this to be the square of a binomial, we need c to be the square of half the value of -22.
Solving this, we get c = 121.
Indeed, if we have \(x^2-22x+121=(x-11)^2\).
So we have c = 121.
Thanks! :)
