+456 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.
+130493 (2x + 7) (x - 4) + 31 - jx = 0
2x^2 - ( 1 + j) x + 3 = 0
Thiswill have one real solutionwhen the discriminant = 0
(1 + j)^2 - 4(2)(3) = 0
(1 + j)^2 = 24 take both roots
1 + j = sqrt 24 and 1 + j =-sqrt 24
j = sqrt (24) - 1 j = -sqrt (24) - 1
j = 2sqrt (6) - 1 j = -2sqrt (6) -1
