+2653 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:
\((2x+7)(x-4)=jx-31\)
\(2x^2+7x-8x-28=jx-31\)
\(2x^2-x-jx+3=0\)
\(2x^2-(j+1)x+3=0\)
Now we must make an equation in the form of x^2 + 2xy + y^2 or x2 - 2xy + y^2
\(\sqrt{2}^2x^2 - (j+1)x+\sqrt{3}^2=0\)
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.
\(j+1 = -2\sqrt{2}\sqrt{3}\)
\(\mathbf{j = -1 - 2\sqrt{6}}\)
\(j+1 = 2\sqrt{2}\sqrt{3}\)
\(\mathbf{j = 2\sqrt{6}-1}\)
So the only two values of j that make the equation have one solution are 2sqrt(6)-1 and -2sqrt(6)-1
