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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.

 Mar 7, 2024
 #1
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2x^2 - 1x - 28  = -31 + jx

 

2x^2 - (1 + j )x + 3  = 0

 

We will have one root when  the discriminant =  0 

 

(1 + j)^2  - 4(2)(3)   =  0

 

j^2 + 2j + 1  - 24  =  0

 

j^2 + 2j - 23  = 0

 

j = [ -2 - sqrt [ 2^2 + 4*23 ] ] / 2

 

j =  [ -2 - sqrt [96] ] / 2

 

j = [ -2 - 4sqrt 6 ]  /2

 

j = - 1 - 2sqrt 6          and     j  =  -1 + 2sqrt 6

 

 

cool cool cool

 Mar 7, 2024

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