+750 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.
+28 We can first expand the left side to get 2x^2-x-28=-31+jx
Then we shift everything to one side to get 2x^2-x-jx+3=0 and then to get 2x^2-(1+j)+3=0
We know that if the it only has one real solution then we can use the discriminant condition from teh quadratic equation to get
sqrt(b^2-4ac)=0
we blu everything in to get sqrt((1+j)^2-24)=0
(1+j)^2-24=0 by squaring both sides
(1+j)^2=24 by adding 24 to both sides
1+j=sqrt24 by square rooting both sides
j=-1±sqrt24
j=-1±2sqrt6
