Problem
ALG-B1-M03-P020 Sum, Product, and Restriction
#20
★★★★☆ Level 4 of 5
Find all real \(x,y\) if \(x+y=2\) and \(x^4+y^4=2\).
First express everything through \(p=xy\).
Let \(p=xy\). Then \(x^2+y^2=4-2p\).
\(x^4+y^4=(x^2+y^2)^2-2x^2y^2=(4-2p)^2-2p^2=16-16p+2p^2\).
Since this equals \(2\), \(16-16p+2p^2=2\), so \(p^2-8p+7=0\). Thus \(p=1\) or \(p=7\). But real \(x,y\) with \(x+y=2\) require \(p\le1\). Hence \(p=1\), and \(x,y\) are roots of \(t^2-2t+1=0\). Therefore \(x=y=1\).
It is important to reject the impossible value of \(p\).