+2729 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.
+135 For a polynomial with degree two to have one solution, you must be able to write it as (x + y)^2. We can rewite the equations like this:
(2x+7)(x−4)=jx−31
2x2+7x−8x−28=jx−31
2x2−x−jx+3=0
2x2−(j+1)x+3=0
Now we must make an equation in the form of x^2 + 2xy + y^2 or x2 - 2xy + y^2
√22x2−(j+1)x+√32=0
Therefore, x = sqrt(2) and y = sqrt(3). Now we see that j+1 = 2(sqrt(2))(sqrt(3)) or -2(sqrt(2))(sqrt(3)). We can now sove for both values of j.
j+1=−2√2√3
j=−1−2√6
j+1=2√2√3
j=2√6−1
So the only two values of j that make the equation have one solution are 2sqrt(6)-1 and -2sqrt(6)-1
