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For what values of $j$ does the equation $(2x+7)(x-5) = -43 + jx have exactly one real solution? If you find more than one, then list the values separated by commas.

 Jan 26, 2019
 #1
avatar+111326 
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(2x + 7) ( x - 5)  =   -43 + jx      simplify

 

2x^2 + 7x - 10x - 35 = -43 + jx

 

2x^2  - 3x - 35  =  -43 + jx

 

2x^2 - (3 + j)x + 8 = 0

 

This will have one real root when the discriminant = 0

 

So

 

(3 + j)^2 -   (4)(2)(8)  =  0

 

(3 + j)^2  =  64       take both roots

 

3 + j  =  8               or           3 + j = - 8

 

j = 5                      or           j = -11

 

 

cool cool cool

 Jan 26, 2019

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