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For what values of j does the equation \((2x+7)(x-5) = -43 + jx\) have exactly one real solution? Express your answer as a list of numbers, separated by commas.

 Jul 1, 2021
 #1
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(2x + 7) ( x - 5)   =  -43  + jx

 

2x^2  -3x  - 35  +  43  - jx  =  0

 

2x^2  - (3 + j)x  + 8  =   0

 

This will have one  real solution when the discriminant  =   0 

 

So

 

(3 +j)^2  - 4* 2 * 8  =  0

 

(j + 3)^2   - 64 =  0

 

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

 

j + 3  =  -sqrt (64)                    j + 3 = sqrt (64)

 

j +  3  =  -8                               j +  3  = 8

 

j = -11                                        j = 5

 

 

cool cool cool

 Jul 1, 2021

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