3.For what values of j does the equation (2x+7)(x-5)=-43+jx have exactly one real solution?
2x^2 - 3x - 35 = -43 + jx rearrange
2x^2 - (3 + j)x + 8 = 0
This will have one solution when the disriminant = 0 .......so....
(3 + j)^2 - 4(2)(8) = 0
(3 + j)^2 - 64 = 0
(3 + j)^2 = 64 take both roots
3 + j = ±8
3 + j = 8 or 3 + j = -8
j = 5 j = -11
