For what values of j does the equation (2x + 7)(x - 5) = -43 + jx + x^2 have exactly one real solution? Express your answer as a list of numbers, separated by commas.
+592 (2x + 7)(x - 5) = -43 + jx + x2
2x2 - 3x - 35 = -43 + jx + x2
x2 - 3x - jx + 8 = 0
x2 - (3 + j)x + 8 = 0
when there is only one real solution, by the quadratic equation:
b2 - 4ac = 0
(-(3 + j))2 - 4(1)(8) = 0
(-(3 + j))2 = 32
-(3 + j) = \(±\sqrt{32}\)
3 + j = \(±4\sqrt{2}\)
j = \(±4\sqrt{2} - 3\)
\(4\sqrt{2} - 3 ≈ 2.66\)
\(-4\sqrt{2} - 3 ≈ -8.66\)
which then your answer is -8.66, 2.66
please tell me if I did anything wrong. hope this helps
