Question

If $P\left(x\right)=a{x}^2+bx+c$ and $Q\left(x\right)=-a{x}^2+dx+c$, where ac$\ne 0$, then $p\left(x\right)Q\left(x\right)=0$ has atleast

• A. Four real roots
• B. Two real roots
• C. Four imaginary roots
• D. None of these

Answer 1

The correct option is B Two real roots

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

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

Now, determinant of $P\left(x\right)=b^2-4ac$

determinant of $Q\left(x\right)=b^2+4ac$

Now, Let $ac>0;$

$b^2+4ac>0$

$\therefore Q\left(x\right)=0$ at two roots, we can't say anything about $P\left(x\right)$

$\therefore P\left(x\right)Q\left(x\right)=0$ atleast two roots

Now, let $ac<0$

$\therefore b^2-4(+a)\left(c\right)\to$ determinant of $P\left(x\right)$

$\therefore b^2-4ac>0$ (Since $b^2>0ac<0$)

$\therefore P\left(x\right)Q\left(x\right)=0$ has atleast two roots

$\therefore$ In either case $P\left(x\right)Q\left(x\right)=0$ has atleast two roots.