+900 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, let's simplfy our equation. We have
2x2−3x−35=−43+jx2x2−3x−jx+8=02x2−(3+j)x+8=0
If there is only one solution, then the descriminant of the quadratic must be 0. So, we can write the equation
(3+j)2−4(2)(8)=0(j+3)2−64=0[(j+3)−8][(j+3)+8]=0[j−5][j+11]=0
Thus, we have that
j=5j=−11
So our final answer is 5 and -11.
Thanks! :)
