If $P(x)=a x^{2}+b x+c$ and $Q(x)=-a x^{2}+d x+c$, where $a c
eq 0$, then $P(x) Q(x)$ has

Question

If $P(x)=a x^{2}+b x+c$ and $Q(x)=-a x^{2}+d x+c$, where $a c 
eq 0$, then $P(x) Q(x)$ has

Professor PC Thomas Classes · Answer

True We have $P(x) Q(x)$\ $=\left(a x^{2}+b x+cight)\left(-a x^{2}+d x+cight)$ $$\begin{equation*}\Rightarrow \quad D_{1}=b^{2}-4 a c 	ag{1} \end{equation*}$$ $$\begin{equation*}D_{2}=d^{2}+4 a c 	ag{2} \end{equation*}$$ $	herefore \quad$ If $a c>0$, then from (2) $D_{2}>0$\ $	herefore \quad Q(x)=0$ has real roots.\ If $a c<0$, then $D_{1}=b^{2}-4 a c>0$\ $	herefore \quad P(x)=0$ has real roots.\ Thus $P(x) Q(x)=0$ has at least two real roots.\ $	herefore \quad$ Given statement is true.\