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.

ma931 Dec 25, 2018

#1**+1 **

(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

CPhill Dec 25, 2018