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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.

 
 Feb 3, 2025
 #1
avatar+130313 
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(2x + 7) (x - 4) + 31 - jx  = 0

 

2x^2 - ( 1 + j) x + 3 =  0

 

Thiswill have one real solutionwhen the discriminant  = 0

 

(1 + j)^2  - 4(2)(3)  =  0

 

(1 + j)^2  =  24        take  both roots

 

1 + j =  sqrt 24           and     1 + j  =-sqrt 24

 

j =  sqrt (24)  - 1                    j =  -sqrt (24) - 1

 

j = 2sqrt (6) - 1                     j = -2sqrt (6)  -1

 

 

cool cool cool

 Feb 4, 2025

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