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(D) Given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$,we have $a^2\alpha^2 + b\alpha + c = 0$,which implies $b\alpha + c = -a^2\alpha^2$.Given th

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$\alpha < \gamma < \beta$

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(D) Given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$,we have $a^2\alpha^2 + b\alpha + c = 0$,which implies $b\alpha + c = -a^2\alpha^2$.Given that $\beta$ is a root of $a^2x^2 - bx - c = 0$,we have $a^2\beta^2 - b\beta - c = 0$,which implies $b\beta + c = a^2\beta^2$.Let $f(x) = a^2x^2 + 2bx + 2c$.Evaluating $f(x)$ at $\alpha$: $f(\alpha) = a^2\alpha^2 + 2(b\alpha + c) = a^2\alpha^2 + 2(-a^2\alpha^2) = -a^2\alpha^2$. Since $a e 0$ and $\alpha > 0$,$f(\alpha) Evaluating $f(x)$ at $\beta$: $f(\beta) = a^2\beta^2 + 2(b\beta + c) = a^2\beta^2 + 2(a^2\beta^2) = 3a^2\beta^2$. Since $a e 0$ and $\beta > 0$,$f(\beta) > 0$.Since $f(\alpha) 0$,by the Intermediate Value Theorem,there exists a root $\gamma$ of $f(x) = 0$ such that $\alpha < \gamma < \beta$.

Let $a, b, c$ be real numbers with $a \ne 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$,$\beta$ is a root of $a^2x^2 - bx - c = 0$,and $0 < \alpha < \beta$,then the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies:

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