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avatar+840 

Real numbers \(x\) and \(y\) have a difference of 16 and a product of 23. Find \(x^2+y^2\).

 Sep 30, 2018
 #1
avatar+2341 
+2

This problem is easier than it looks at first glance. If you utilize clever algebraic manipulation, this problem becomes simpler. 

 

\(x-y=16\) \(xy=23\)
\((x-y)^2=16^2\)  
\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\) \(\boxed{2}\hspace{1mm}2xy=46\)
   

 

Notice what I have done. I have manipulated the information I know about these real numbers, and y , and I am manipulating it in a way that is much more convenient for this particular problem. The only thing left to do is add the equations together.

 

\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\\ \boxed{2}\hspace{5mm}+2xy\hspace{10mm}=46\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{18mm}x^2+y^2=302\)


 

 Sep 30, 2018
 #2
avatar+18495 
+3

Another method (though just a bit more difficult than x-factor's answer)

x-y=16

 

x = 16+y

 

xy = 23

(16+y)y=23

y^2 +16y -23=0    use quadratic formula to find y = 1.3273      or    -17.3273

                                                           then        x = 17.3273     or      -1.3273

 

then   x^2 + y^2  = 302     (pick either x,y pair above....same result)

 Sep 30, 2018

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