Question

If the equation ${ax}^2+2bx+c=0$ has real roots, $a,b,c$ being real numbers and if $m$ and $n$ are real number such that $m^2>n>0$ then show that the equation ${ax}^2+2mbx+nc=0$ has real roots.

Answer 1

Given roots of the equation
${ax}^2+2bx+c=0$ are real.

$\therefore {\left(2b\right)}^2-4ac\ge 0$

$\therefore b^2-ac\ge 0$ .............(1)

Discriminant of ${ax}^2+2mbx+nc=0$ is

$D={\left(2mb\right)}^2-4anc$

$D=4m^2{b}^2-4anc$ .............(2)

From (1) $b^2\ge ac$ ........(3)
and given $m^2>n$ .......(4)
So, from (3) and (4),
${\therefore b}^2{m}^2\ge anc$
$\Rightarrow 4m^2{b}^2-4anc\ge 0$
$\Rightarrow D\ge 0$ {from (2)}
Hence roots of equation ${ax}^2+2mbx+nc=0$ are real.