+1206 Real numbers \(x\) and \(y\) have a difference of 16 and a product of 23. Find \(x^2+y^2\).
+2440 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, x 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.
\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\\ \boxed{2}\hspace{5mm}+2xy\hspace{10mm}=46\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{18mm}x^2+y^2=302\)
+36916 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)
