Question

Prove that $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)\ge 0$ for all $xϵR$. If equality holds then find the ratio of the roots of the equation $a{x}^2+2bx+c=0$

Answer 1

Considering the equation to be: $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)\ge 0$

$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)\ge 0$

Consider

$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$

Considering the equal roots to be z and z

$\therefore z+z=\frac{-2b(a+c)}{a^2+b^2}$

$2z=\frac{-2b(a+c)}{a^2+b^2}$

$z=\frac{-b(a+c)}{a^2+b^2}$

$z^2=\frac{b^2(a+c{)}^2}{(a^2+b^2{)}^2}$...(1)

$z\times z=\frac{b^2+c^2}{a^2+b^2}$

$z^2=\frac{b^2+c^2}{a^2+b^2}$...(2)

equating (1) and (2) gives

$\frac{b^2(a+c{)}^2}{(a^2+b^2{)}^2}=\frac{b^2+c^2}{a^2+b^2}$

$b^2(a+c{)}^2=(b^2+c^2\left)\right(a^2+b^2)$

$b^2(a^2+c^2+2ac)=a^2{b}^2+b^4+a^2{c}^2+c^2{b}^2$

$a^2{b}^2+b^2{c}^2+2ac{b}^2=a^2{b}^2+b^4+a^2{c}^2+c^2{b}^2$

$b^4+a^2{c}^2-2ac{b}^2=0$

$(b^2-ac{)}^2=0$

$b^2=ac$....(3)

Given equation:

$a{x}^2+2bx+c=0$

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-2b\pm \sqrt{{\left(2b\right)}^2-4ac}}{2a}:\frac{\sqrt{b^2-ac}-b}{a}$

$x=\frac{-2b-\sqrt{{\left(2b\right)}^2-4ac}}{2a}:-\frac{b+\sqrt{b^2-ac}}{a}$

$x=\frac{\sqrt{b^2-ac}-b}{a},x=-\frac{b+\sqrt{b^2-ac}}{a}$

Ratio of roots is given by:

$=\frac{\sqrt{b^2-ac}-b}{a}\times \frac{a}{b+\sqrt{b^2-ac}}$

$=\frac{-b+\sqrt{b^2-ac}}{b+\sqrt{b^2-ac}}$

From (3) $b^2=ac$

$=\frac{-b}{b}$

$=-1$

The ratio of roots of equation is 1