Question

If $a,b,c,x$ are all real numbers, and $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$
then $a,b,c$ are in $G.P$, and $x$ is their common ratio.

Answer 1

Discriminant of the equation in $x$ will be found to be $-4(b^2-ac{)}^2$ which is $\le 0$

But, for real $x$, it cannot be negative and so it must have $b^2-ac=0$ showing that $a,b,c$ are in G.P.

Under this condition, the equation has equal roots say $x,x$

Therefore, the sum of the roots is $x+x=\frac{2b(a+c)}{a^2+b^2}$

$⟹x=\frac{b(a+c)}{a^2+ac}$ (since, $b^2=ac$)

$⟹x=\frac{b}{a}=\frac{c}{b}$ is the common ratio.