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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.

 Jun 22, 2024
 #1
avatar+1075 
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First, let's simplfy our equation. We have

\(2x^2 - 3x - 35 = -43 + jx \\ 2x^2 - 3x - jx + 8 = 0 \\ 2x^2 - (3 + j)x + 8 = 0\)

 

If there is only one solution, then the descriminant of the quadratic must be 0. So, we can write the equation

\((3 + j)^2 - 4(2)(8) = 0 \\ (j + 3)^2 - 64 = 0 \\ [ (j + 3) - 8 ] [ (j + 3) + 8 ] = 0 \\ [ j - 5 ] [ j + 11] = 0\)

 

Thus, we have that 

\(j = 5 \\ j = -11\)

 

So our final answer is 5 and -11. 

Thanks! :)

 Jun 22, 2024

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