If $P(x)=a x^{2}+b x+c$ and $Q(x)=-a x^{2}+d x+c$, where $a c \neq 0 \quad[a, b, c, d$ are all real], then $P(x) \cdot Q(x)=0$ has

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If $P(x)=a x^{2}+b x+c$ and $Q(x)=-a x^{2}+d x+c$, where $a c \neq 0 \quad[a, b, c, d$ are all real], then $P(x) \cdot Q(x)=0$ has

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kritika · Answer 1 · 2021-12-26T19:29:41+0000

Ans : (A)
Hint: If $P(x)=a x^{2}+b x+c, Q(x)=-a x^{2}+d x+c$
$$
\begin{aligned}
&D_{1}=b^{2}-4 a c \\
&D_{2}=d^{2}+4 a c \\
&\Rightarrow D_{1}+D_{2}>0
\end{aligned}
$$
Atleast two real roots.

If \(P(x)=a x^{2}+b x+c\) and \(Q(x)=-a x^{2}+d x+c\), where \(a c \neq 0 \quad[a, b, c, d\) are all real], then \(P(x) \cdot Q(x)=0\) has

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