+73 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.
+129899 (2x + 7) ( x - 5) = -43 + jx simplify
2x^2 + 7x -10x -35 = -43 +jx rearrange as
2x^2 - 3x - jx - 35 + 43 = 0
2x^2 - (3 + j) + 8 = 0
If this has one solution the discriminant must = 0 ....so....
(3 + j)^2 - 4(2) (8) = 0
j^2 + 6j + 9 - 64 = 0
j^2 + 6j - 55 = 0 factor
(j + 11) ( j - 5) = 0
Setting each factor to 0 and solving for j
j + 11 = 0 j - 5 = 0
j = -11 j = 5
So... j = -11, 5
BTW.....when x = -11, x = -2 and when x = 5, x = 2
