Problem
ALG-B1-M09-P022 Sum of squares of roots
#22
★★★★☆ Level 4 of 5
Let integers \(x,y\) be the roots of \(t^2-st+p=0\), where \(s,p\in\mathbb Z\), and suppose \(x^2+y^2=25\), \(xy=12\). Find \(s\).
Use \((x+y)^2=x^2+y^2+2xy\).
\((x+y)^2=25+24=49\), so \(x+y=\pm7\). By Vieta, \(s=x+y\), hence \(s=7\) or \(s=-7\). Both values are attainable: root pairs \((3,4)\) and \((-3,-4)\).
Do not forget the sign of the sum when taking the square root.