Question

If $a>b>c$ and the system of equations $ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$ has a non-trivial solution then prove that both the roots of the quadratic equation $a{t}^2+bt+c=0$ are real.

Answer 1

(a) For non-trivial solution.

$triangle=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

$\Rightarrow -(a^3+b^3+c^3-3abc)=0$

$\Rightarrow a+b+c=0orb=-(a+c)$

$a{t}^2+bt+c=0$

$△=b^2-4ac=(a+c{)}^2-4ac$

or $△=(a-c{)}^{2}i.e+ive\therefore$ Real.