Question
If \(x+ay+a^2z=1\), \(x+by+b^2z=1\), \(x+cy+c^2z=1\), where \(a \neq b \neq c\), then what is \(x+y+z\) equal to?
This question was previously asked in
NDA 2 2026 GAT Question Paper (13-Sep-2026)
The correct answer is
\(1\)
The three equations show that \(a, b, c\) are three distinct roots (in \(t\)) of \(zt^2+yt+(x-1)=0\). A quadratic can have at most two roots unless all its coefficients vanish, so \(z=0\), \(y=0\) and \(x-1=0\), giving \(x=1, y=0, z=0\). Hence \(x+y+z=1\).
Was this answer helpful?
Need Expert Advice?
Test Series
Defence
NDA 2026 Mock Test Series (Latest Pattern)
503 Tests 1 Tests Free
1066 Attempts
4.6(137)
English, Hindi
More Questions from NDA
