+447 For what values of $j$ does the equation $(2x + 7)(x - 4) = -31 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.
+129771 2x^2 - 1x - 28 = -31 + jx
2x^2 - (1 + j )x + 3 = 0
We will have one root when the discriminant = 0
(1 + j)^2 - 4(2)(3) = 0
j^2 + 2j + 1 - 24 = 0
j^2 + 2j - 23 = 0
j = [ -2 - sqrt [ 2^2 + 4*23 ] ] / 2
j = [ -2 - sqrt [96] ] / 2
j = [ -2 - 4sqrt 6 ] /2
j = - 1 - 2sqrt 6 and j = -1 + 2sqrt 6
