+0  
 
-1
37
2
avatar+284 

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

Logic  Sep 30, 2018
 #1
avatar+2248 
+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.

 

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


 

TheXSquaredFactor  Sep 30, 2018
 #2
avatar+13014 
+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)

ElectricPavlov  Sep 30, 2018

38 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.