Question

The polynomial $({ax}^2+bx+c)({-ax}^2+dx+c)=0,ac\ne 0$ has how many roots?

Answer 1

$p\left(x\right)=a{x}^2+bx+c$

$q\left(x\right)=-a{x}^2+dx+c$

$p\left(x\right)\ast q\left(x\right)=0$

$(a{x}^2+bx+c)\ast (-a{x}^2+dx+c)=0$

Considering ${Ax}^2+Bx+C=0$

Discriminating ${=B}^2-4AC$

In above equation $A$ and $C$ are negative.

$B^2-4AC$ will be positive and have real roots.

$p\left(x\right)=a{x}^2+bx+c$

Product of co-efficient of $x^2$ and constant term is $ac$

$q\left(x\right)=-a{x}^2+dx+c$

Product of co-efficient of $x^2$ and constant term is $-ac$

If $ac$ is positive then $-ac$ will be negative and vice-versa

$(a{x}^2+bx+c)\ast (-a{x}^2+dx+c)=0$ will have atleast two real roots.