SOLUTION: Find all real values of s such that x^2 + sx + 144 - 6x + 3x^2 is the square of a binomial.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find all real values of s such that x^2 + sx + 144 - 6x + 3x^2 is the square of a binomial.      Log On


   

Question 1209150: Find all real values of s such that x^2 + sx + 144 - 6x + 3x^2 is the square of a binomial.
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


x%5E2+%2B+sx+%2B+144+-+6x+%2B+3x%5E2 = 4x%5E2%2B%28s-6%29x%2B144

In the simplified polynomial, the leading term is 4x%5E2, so the leading term of the square root is sqrt%284x%5E2%29 = 2x.

The constant term of the simplified polynomial is 144, so the constant term of the square root is sqrt%28144%29 = 12.

So the possible square root polynomials are 2x%2B12 and 2x-12.

%282x%2B12%29%5E2+=+4x%5E2%2B48x%2B144 so for one solution we have

s-6=48 --> s=54

%282x-12%29%5E2+=+4x%5E2-48x%2B144 so for a second solution we have

s-6=-48 --> s=-42

ANSWERS: s=54 and s=-42