For what real values of c is 4x^2 + 14x - c - 3x^2 - 12x the square of a binomial?

Guest Dec 31, 2022

#1**+1 **

We start by simplifying

\(4x^2+14x-c-3x^2-12x\)

\(x^2+2x-c\)

To find all values of c, we take 2x the middle term, divide by 2, then x, which gives us 1,

So, a square root of some number should equal 1, which and that number is one, so -c = 1, **c = - 1 **

This may sound complicated, dividing by so many numbers, but it's really not, the reason we do that

is because when (ax+b)^{2} is expanded, you get ax^{2}+2abx+b^{2} so to get b when provided the first term and the middle term, we divide by 2ax, and square the result.

hairyberry Dec 31, 2022