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.If you think this is true write 1 otherwise write 0 ?

Answer 1

As the roots of $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$ are real, then

$4b^2{(a+c)}^2-4(b^2+c^2)(a^2+b^2)\ge 0\Rightarrow b^2(a^2+c^2+2ac)-(b^2{a}^2+b^4+a^2{c}^2+b^2{c}^2)\ge 0\Rightarrow a^2{b}^2+b^2{c}^2+2ac{b}^2-b^2{a}^2-b^4-a^2{c}^2-b^2{c}^2\ge 0\Rightarrow 2ac{b}^2-a^2{c}^2-b^4\ge 0\Rightarrow {(b^2-ac)}^2\ge 0$

$\Rightarrow b^2\ge ac\Rightarrow a,b,c$ are in $G.P.$

And if $x$ is common difference then $\frac{b}{a}=x$ and $\frac{c}{b}=x$

$\Rightarrow b=ax$ and $c=bx=a{x}^2$

$\therefore (a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=(a^2+a^2{x}^2)x^2-2ax(a+a{x}^2)x+(a^2{x}^2+a^2{x}^4)=a^2{x}^2+a^2{x}^4-2a^2{x}^2-2a^2{x}^4+a^2{x}^2+a^2{x}^4=0$