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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.

 Dec 25, 2018
 #1
avatar+98005 
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(2x+7)(x-5) = -43 + jx     simplify

 

2x^2 - 3x - 35  = -43 +   jx

 

2x^2 -   3x - jx + 8 =  0

 

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

 

If we have only one real root.....the dicriminant must = 0.....so

 

(3 + j)^2 - 4(2)(8)  = 0

 

(j + 3)^2 - 64  =  0       factor as a difference of squares

 

[ (j + 3) - 8 ]   [ (j + 3) + 8 ]  = 0

 

[ j - 5 ]  [ j + 11]  =  0

 

Setting both factors to 0  and solving for j  produces 

 

j = 5    or     j  = -11

 

So...    -11, 5

 

 

cool cool cool

 Dec 25, 2018

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