The equation $x^{2} - 4ax + 1 = 0$ has real roots gien by $\alpha$ and $\beta$, where a is real. The

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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

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$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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