For what values of j does the equation $(2x+7)(x-5) = -43 + jx$ have exactly one real solution?
+128408 (2x + 7) ( x - 5) = -43 + jx
2x^2 + 7x - 10x - 35 = -43 + jx
2x^2 - 3x - 35 = -43 + jx
2x^2 - (3 + j)x + 8 = 0
If this has a single real solution , the discriminant = 0
So
(3 + j)^2 - 4(2)(8) = 0
j^2 + 6j + 9 - 64 = 0
j^2 + 6j - 55 = 0 factor
(j - 5) ( j + 11) = 0
Settng each factor to 0 and solving for j gives us
j - 5 = 0 j + 11 = 0
j = 5 j = -11
