We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
74
2
avatar

If (x + y)^2 = 31 and xy=6, what is the value of (x-y)^2?

 Sep 16, 2019
 #1
avatar+2361 
+2

Hint: Expand everything

 

Then use substitution

 Sep 16, 2019
 #2
avatar+104876 
+2

(x + y)^2  = 31                                        ( x - y)^2  =  x^2 - 2xy + y^2     =  

 

x^2 + 2xy + y^2  = 31                                                 (x^2 + y^2) - 2xy

 

x^2 + y^2  = 31 - 2xy                                                  (x^2 + y^2) - 2(6) 

 

x^2 + y^2  = 31 - 2(6)                                                 (x^2 + y^2)  -12      (2)

 

(x^2 + y^2) = 31 -12 =  19      (1)

 

Sub (1)  into (2)    and we have that

 

(x - y)^2   =   (x^2 +y^2) - 12  =  (19) - 12   =   7

 

 

 

cool cool cool

 Sep 16, 2019
edited by CPhill  Sep 16, 2019
edited by CPhill  Sep 16, 2019

34 Online Users

avatar
avatar
avatar
avatar