Quadratic EquationHard
Question
Let $a,b,c$ be real numbers, $a \neq 0$, if $\alpha$ is a root $a^{2}x^{2} + bx + c = 0,\beta$ is a root of $a^{2}x^{2} - bx - c = 0$ and $0 < \alpha < \beta$, then the equation $a^{2}x^{2} + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies
Options
A.$\gamma < \alpha$
B.$\gamma > \beta$
C.$\alpha < \gamma < \beta$
D.$\gamma = \frac{\alpha + \beta}{2}$
Solution
$${a^{2}\alpha^{2} + b\alpha + c = 0\ \Rightarrow \ \text{ }b\alpha + c = - a^{2}\alpha^{2} }{a^{2}\beta^{2} - b\beta - c = 0\ \Rightarrow \ \text{ }b\beta + c = a^{2}\beta^{2} }$$Let $f(x) = a^{2}x^{2} + 2bx + 2c$
$$\begin{matrix} & f(\alpha) = a^{2}\alpha^{2} + 2(b\alpha + c) = - a^{2}\alpha^{2} < 0 \\ & f(\beta) = a^{2}\beta^{2} + 2(b\beta + c) = 3a^{2}\beta^{2} > 0 \end{matrix}$$
$\Rightarrow \exists$ root $\gamma$, such that $\alpha < \gamma < \beta$
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