JEE Main · 2016 · 03 Apr 2016 · Offline · Quadratic Equations and Inequalities

A power $a^b=1$ in three ways: the base equals 1; the base equals $-1$ with an even exponent; or the exponent is 0 with a non-zero base. Setting the base to 1, $x^2-5x+5=1$ gives $x^2-5x+4=0$, so $x=1$ or $4$. Setting the exponent to 0, $x^2+4x-60=0$ gives $x=-10$ or $6$. Setting the base to $-1$, $x^2-5x+6=0$ gives $x=2$ or $3$, but the exponent must then be even: at $x=2$ it is $-48$, even and allowed, while at $x=3$ it is $-39$, odd, so $x=3$ is rejected. The valid roots are $1,4,-10,6,2$, whose sum is $3$.