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If (x+y)^2=78  and xy=13, what is (x-y)^2?

Thanks!

 Jun 9, 2019
 #1
avatar+4330 
+2

(x+y)^2=x^2+y^2+2xy=78

 

x^2+y^2+2(13)=78

 

x^2+y^2=52

 

(x-y)^2=x^2+y^2-2xy, 52-2(13)=52-26=26,

 

-tertre

 Jun 9, 2019
edited by Guest  Jun 9, 2019
edited by Guest  Jun 9, 2019
 #3
avatar+106533 
+2

Ah, tertre....you beat me to it....sneaky  !!!!

 

 

cool cool cool

CPhill  Jun 9, 2019
 #2
avatar+106533 
+4

( x + y)^2  = 78

x^2 + 2xy + y^2  = 78

If xy = 13 .....then

x^2 + y^2 + 2(13) = 78

x^2 + y^2 + 26 = 78

x^2 + y^2  = 52

 

So

 

(x - y)^2  =  

x^2 - 2xy + y^2  =

(x^2 + y^2) - 2xy =

(52) - 2(13)  =

26

 

 

 

 

cool cool cool 

 Jun 9, 2019

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