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

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(B) Let the roots of $P(x) = 0$ and $Q(x) = 0$ be considered.For $P(x) = ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$.For $Q(x) = -ax^2 +

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Two real roots

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(B) Let the roots of $P(x) = 0$ and $Q(x) = 0$ be considered.For $P(x) = ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$.For $Q(x) = -ax^2 + dx + c = 0$,the discriminant is $D_2 = d^2 - 4(-a)(c) = d^2 + 4ac$.If all four roots were imaginary,then both $D_1 Adding these inequalities: $(b^2 - 4ac) + (d^2 + 4ac) Since $b^2$ and $d^2$ are non-negative,$b^2 + d^2 If $b=0$ and $d=0$,then $P(x) = ax^2 + c = 0$ and $Q(x) = -ax^2 + c = 0$.This gives $x^2 = -c/a$ and $x^2 = c/a$.Since $ac 
eq 0$,one of $c/a$ or $-c/a$ must be positive,ensuring at least two real roots.Thus,in all cases,$P(x) \cdot Q(x) = 0$ has at least two real roots.

If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$ where $ac \neq 0$,then $P(x) \cdot Q(x) = 0$ has at least

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