If P(x)=ax^{2}+bx+c and Q(x)=-ax^{2}+dx+c, wh | vedclass

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(A) Given $P(x) = ax^{2} + bx + c$ and $Q(x) = -ax^{2} + dx + c$. For $P(x) = 0$, the discriminant is $D_{1} = b^{2} - 4ac$. For $Q(x) = 0$, the discr

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at least two real roots

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(A) Given $P(x) = ax^{2} + bx + c$ and $Q(x) = -ax^{2} + dx + c$. For $P(x) = 0$, the discriminant is $D_{1} = b^{2} - 4ac$. For $Q(x) = 0$, the discriminant is $D_{2} = d^{2} - 4(-a)(c) = d^{2} + 4ac$. Adding the two discriminants, we get $D_{1} + D_{2} = b^{2} + d^{2} \geq 0$. Since the sum of the discriminants is non-negative, at least one of the discriminants must be non-negative ($D_{1} \geq 0$ or $D_{2} \geq 0$). If $D_{1} \geq 0$, $P(x)$ has real roots. If $D_{2} \geq 0$, $Q(x)$ has real roots. Therefore, the equation $P(x) \cdot Q(x) = 0$ must have at least two real roots.

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

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

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