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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.

 Nov 19, 2020
edited by MathzSolver111  Nov 19, 2020
 #1
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-1

Looks like Math Solver can't solve this one! 

 Nov 19, 2020
 #4
avatar+116125 
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(2x + 7)  ( x  - 5)   = -43 + jx       simplify

 

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

 

2x^2  - 3x  - jx - 35 + 43  = 0

 

2x^2  - (3 + j)  + 8  =  0

 

If this has one solution the  discriminant must  =  0    ....so....

 

(3 + j)^2  - 4(2) (8)   = 0

 

j^2  + 6j  + 9  -  64  =  0

 

j^2 + 6j  - 55  =  0       factor

 

(j + 11)  ( j - 5)   =  0

 

Setting each factor to 0  and solving for  j

j + 11   = 0                   j - 5  =   0

j  = -11                           j  =  5

 

So...  j   =   -11,  5

 

BTW.....when x  = -11,  x  = -2   and when x  = 5, x  = 2

 

cool cool cool

 Nov 20, 2020

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