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.

LiIIiam0216 Jun 11, 2024

#1**0 **

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**

Maxematics Jun 11, 2024