Quadratic EquationHard
Question
The equation $x^{2} - 4ax + 1 = 0$ has real roots gien by $\alpha$ and $\beta$, where a is real. Then the complete set of values of a for which $\alpha \geq a$ and $\beta \geq 0$ is
Options
A.$\lbrack 1,\infty)$
B.$\lbrack 2,\infty)$
C.$\left( - \infty, - \frac{1}{2} \right\rbrack \cup \left\lbrack \frac{1}{2},\infty \right)$
D.$\left\lbrack \frac{1}{2},\infty \right)$
Solution
$x + \frac{1}{x} = 4a$
$$\begin{matrix} 4a & \ \geq 2 \Rightarrow a \geq \frac{1}{2} \\ \alpha,\beta & \ = 2a \pm \sqrt{4a^{2} - 1} \\ \alpha & \ \geq 2a \end{matrix}$$
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