For what values of $j$ does the equation $(2x + 7)(x - 4) = -31 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.

learnmgcat Aug 16, 2024

#1**+1 **

First, we have to simplify the left side of the equation.

Expanding it out, we get

\(2x^2-x-28=-31+jx\)

Moving all terms to one side and combining all like terms, we achieve

\(2x^2-x-jx-28+31=0\\ 2x^2-\left(j+1\right)x+3=0\)

If there's only one solution, then that must mean that the descriminant is 0.

The descriminant of this quadratic is

\((j+1)^2 - 4(2)(3)\\ j^2+2j-23\)

Setting this to 0, we can now solve for j. We get

\(j^2+2j-23=0\)

Unfortunately, j cannot be factored conveniently, so we must use the quadratic equation. (I'm going to leave that for you).

From that, we find that

\(j=-1+2\sqrt{6},\:j=-1-2\sqrt{6}\)

I'm pretty sure this is the correct answer, but you can check and verify.

Thanks! :)

NotThatSmart Aug 16, 2024