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Find the product of all real values of \(r\) for which \(\dfrac{1}{2x}=\dfrac{r-x}{7}\) has exactly one real solution. 

 

THANKS

 Jan 14, 2019
 #1
avatar+111437 
+3

Cross-multiply and we get that

 

2xr - 2x^2 = 7     rearrange as

 

2x^2 - 2rx + 7 =  0

 

Single real solutions will occur when the dicriminant = 0

 

So

 

(2r)^2 - ( 4) (7) (2)  = 0

 

4r^2  -  56 = 0     

 

4r^2 =  56    divide both sides by 4

 

r^2 = 14      take both roots

 

r =  ±√14

 

And the product of these r's  =   (√14) ( - √14)  =    -14

 

 

cool cool cool

 Jan 14, 2019

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