If alpha and beta are the roots of the equati | vedclass

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(A) Given the quadratic equation $x^2 - 2x + 2 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt

Vedclass · Accepted Answer

$4$

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(A) Given the quadratic equation $x^2 - 2x + 2 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i$.Let $\alpha = 1 + i$ and $\beta = 1 - i$.Now,$\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{2i}{2} = i$.Alternatively,$\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = i$ and if we take $\alpha = 1 - i$ and $\beta = 1 + i$,then $\frac{\alpha}{\beta} = -i$.We need $(\frac{\alpha}{\beta})^n = 1$. If $\frac{\alpha}{\beta} = i$,then $i^n = 1$,which gives the least natural number $n = 4$.If $\frac{\alpha}{\beta} = -i$,then $(-i)^n = 1$,which also gives the least natural number $n = 4$.Thus,the least value of $n$ is $4$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 2 = 0$,then the least value of $n$ for $(\frac{\alpha}{\beta})^n = 1$ is

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