For what values of j does the equation \((2x+7)(x-5) = -43 + jx\) have exactly one real solution? Express your answer as a list of numbers, separated by commas.
+128474 (2x + 7) ( x - 5) = -43 + jx
2x^2 -3x - 35 + 43 - jx = 0
2x^2 - (3 + j)x + 8 = 0
This will have one real solution when the discriminant = 0
So
(3 +j)^2 - 4* 2 * 8 = 0
(j + 3)^2 - 64 = 0
(j + 3)^2 = 64 take both roots
j + 3 = -sqrt (64) j + 3 = sqrt (64)
j + 3 = -8 j + 3 = 8
j = -11 j = 5
