If alpha and beta are roots of the equation x | vedclass

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(A) Given equation: $x^2 + 5|x| - 6 = 0$. Since $x^2 = |x|^2$,we can write: $|x|^2 + 5|x| - 6 = 0$. Factoring the quadratic: $(|x| + 6)(|x| - 1) = 0$.

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$\frac{\pi}{2}$

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(A) Given equation: $x^2 + 5|x| - 6 = 0$. Since $x^2 = |x|^2$,we can write: $|x|^2 + 5|x| - 6 = 0$. Factoring the quadratic: $(|x| + 6)(|x| - 1) = 0$. This gives $|x| = 1$ or $|x| = -6$. Since $|x|$ cannot be negative,we have $|x| = 1$,which implies $x = 1$ or $x = -1$. Let $\alpha = 1$ and $\beta = -1$. Then,$|	an^{-1} \alpha - 	an^{-1} \beta| = |	an^{-1}(1) - 	an^{-1}(-1)|$. $= |\frac{\pi}{4} - (-\frac{\pi}{4})| = |\frac{\pi}{4} + \frac{\pi}{4}| = |\frac{\pi}{2}| = \frac{\pi}{2}$.

If $\alpha$ and $\beta$ are roots of the equation $x^2+5|x|-6=0$,then the value of $|\tan^{-1} \alpha - \tan^{-1} \beta|$ is

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