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If $\mathrm{P}(\mathrm{x})=a \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ and $\mathrm{Q}(\mathrm{x})=-\mathrm{ax}^{2}+\mathrm{dx}+\mathrm{c}$, where $\mathrm{ac} \neq 0 \quad[\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are all real], then $\mathrm{P}(\mathrm{x}) \cdot \mathrm{Q}(\mathrm{x})=0$ has
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Verified Answer
The correct answer is:
at least two real roots
Hint:
If $P(x)=a x^{2}+b x+c, Q(x)=-a x^{2}+d x+c$
$\mathrm{D}_{1}=\mathrm{b}^{2}-4 \mathrm{ac}$
$\mathrm{D}_{2}=\mathrm{d}^{2}+4 \mathrm{ac}$
$\Rightarrow \mathrm{D}_{1}+\mathrm{D}_{2}>0$
Atleast two real roots.
If $P(x)=a x^{2}+b x+c, Q(x)=-a x^{2}+d x+c$
$\mathrm{D}_{1}=\mathrm{b}^{2}-4 \mathrm{ac}$
$\mathrm{D}_{2}=\mathrm{d}^{2}+4 \mathrm{ac}$
$\Rightarrow \mathrm{D}_{1}+\mathrm{D}_{2}>0$
Atleast two real roots.
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