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

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(B) Let the discriminants of $P(x) = 0$ and $Q(x) = 0$ be $D_1$ and $D_2$ respectively.$D_1 = b^2 - 4ac$ and $D_2 = d^2 - 4(-a)c = d^2 + 4ac$.Consider

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Two real roots

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(B) Let the discriminants of $P(x) = 0$ and $Q(x) = 0$ be $D_1$ and $D_2$ respectively.$D_1 = b^2 - 4ac$ and $D_2 = d^2 - 4(-a)c = d^2 + 4ac$.Consider the sum of the discriminants: $D_1 + D_2 = (b^2 - 4ac) + (d^2 + 4ac) = b^2 + d^2$.Since $b^2 \ge 0$ and $d^2 \ge 0$,we have $D_1 + D_2 \ge 0$.If both $D_1 Thus,at least one of $D_1$ or $D_2$ must be greater than or equal to $0$.If $D_1 \ge 0$,$P(x) = 0$ has at least two real roots.If $D_2 \ge 0$,$Q(x) = 0$ has at least two real roots.Therefore,the equation $P(x) \cdot Q(x) = 0$ must have 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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