Let p, q be real numbers. If alpha is a root | vedclass

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(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$. Since $\beta$ is a root of $x^{2}+9 p

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$\alpha < \gamma < \beta$

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(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$. Since $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$, we have $\beta^{2}+9 p^{2} \beta+15 q^{2}=0$. Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$. Evaluating $f(\alpha)$: $f(\alpha)=\alpha^{2}+6 p^{2} \alpha+10 q^{2} = (\alpha^{2}+3 p^{2} \alpha+5 q^{2}) + 3 p^{2} \alpha + 5 q^{2} = 0 + 3 p^{2} \alpha + 5 q^{2}$. Since $\alpha > 0$ and $p, q$ are real numbers (assuming $p, q eq 0$ for non-trivial roots), $f(\alpha) > 0$. Evaluating $f(\beta)$: $f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2} = (\beta^{2}+9 p^{2} \beta+15 q^{2}) - (3 p^{2} \beta + 5 q^{2}) = 0 - (3 p^{2} \beta + 5 q^{2})$. Since $\beta > \alpha > 0$, $f(\beta) Since $f(x)$ is a continuous polynomial function and $f(\alpha) > 0$ while $f(\beta) < 0$, by the Intermediate Value Theorem, there must exist a root $\gamma$ such that $\alpha < \gamma < \beta$.

Let $p, q$ be real numbers. If $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$, $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$ and $0 < \alpha < \beta$, then the equation $x^{2}+6 p^{2} x+10 q^{2}=0$ has a root $\gamma$ that always satisfies:

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