+1501 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.
+1950 First, we have to simplify the left side of the equation.
Expanding it out, we get
2x2−x−28=−31+jx
Moving all terms to one side and combining all like terms, we achieve
2x2−x−jx−28+31=02x2−(j+1)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)j2+2j−23
Setting this to 0, we can now solve for j. We get
j2+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√6,j=−1−2√6
I'm pretty sure this is the correct answer, but you can check and verify.
Thanks! :)
