Question

If $a,b,c$ are positive and system of equations $ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$ has non-trivial solutions. Then the roots of the equation $a{t}^2+bt+c=0$ are

• A. real and opposite in sign
• B. both positive
• C. at least one positive
• D. non real

Answer 1

The correct option is D non real

System of equations has non-trivial solutions
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0\&a^3+b^3+c^3-3abc=0$

$\&(a+b+c)(a{w}^2+bw+c)(aw+b{w}^2+c)=0$
$\because$ a,b,c are positive

$\therefore a+b+c\ne 0$
either $a{w}^2+bw+c=0$ or $aw+b{w}^2+c=0$
Hence roots of the equation
$a{t}^2+bt+c=0$ are $w$ and $w^2$ i.e. $a{t}^2+bt+c=0$ has non real roots.
Ans: D