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For what values of  j does the equation ( 2 x + 7 )( x - 5) = - 43 + jx  have exactly one real solution?

 Jan 9, 2021
 #1
avatar+53 
+2

 ( 2 x + 7 )( x - 5) = - 43 + jx 

 2x2-3x-35-jx=-43

 2x2-3x-35-jx+43=-43+43

2x2-(j+3)x+8=0

 Jan 9, 2021
 #2
avatar+114361 
+3

(2x + 7)  ( x - 5)   =  -43 +  jx         expand

 

2x^2  -3x  - 35  =  -43 + jx         rearrange as

 

2x^2 - ( 3 + j)x  + 8  = 0

 

This will have one real  root  if the discriminant   =  0   ....so....

 

(3 + j)^2    - 4(2) ( 8)   =  0

 

j^2  + 6j + 9   - 64   = 0

 

j^2  + 6j  - 55    =   0       factor  as

 

(j  + 11)  ( j - 5)   =   0

 

Set  both factors to 0 and solve for j

 

We will  have one real root whenever  j   =  -11   or when   j   =  5

 

 

cool cool cool

 Jan 9, 2021

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