+80 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.
+128460 (2x+7)(x-5) = -43 + jx simplify
2x^2 - 3x - 35 = -43 + jx
2x^2 - 3x - jx + 8 = 0
2x^2 - (3 + j)x + 8 = 0
If we have only one real root.....the dicriminant must = 0.....so
(3 + j)^2 - 4(2)(8) = 0
(j + 3)^2 - 64 = 0 factor as a difference of squares
[ (j + 3) - 8 ] [ (j + 3) + 8 ] = 0
[ j - 5 ] [ j + 11] = 0
Setting both factors to 0 and solving for j produces
j = 5 or j = -11
So... -11, 5
