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For what values of j does the equation (2x+7)(x4)=31+jx have exactly one real solution? Express your answer as a list of numbers, separated by commas.

 Aug 16, 2024
 #1
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First, we have to simplify the left side of the equation. 

Expanding it out, we get

2x2x28=31+jx

 

Moving all terms to one side and combining all like terms, we achieve

2x2xjx28+31=02x2(j+1)x+3=0

 

If there's only one solution, then that must mean that the descriminant is 0. 

The descriminant of this quadratic is 

(j+1)24(2)(3)j2+2j23

 

Setting this to 0, we can now solve for j. We get

j2+2j23=0

 

Unfortunately, j cannot be factored conveniently, so we must use the quadratic equation. (I'm going to leave that for you). 

From that, we find that

j=1+26,j=126

 

I'm pretty sure this is the correct answer, but you can check and verify. 

 

Thanks! :)

 Aug 16, 2024
edited by NotThatSmart  Aug 16, 2024

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