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If α, β are t...

Question

If $\alpha, \beta$ are the roots of the equation $x^2-4x+3=0,$ then the value of $\sqrt{\alpha^4+\beta^4-1}$ is

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Solution

$x^2-4x+3=0$
$\Rightarrow x^2-x-3x+3=0$
$\Rightarrow (x-1)(x-3) = 0$
$\Rightarrow x =1, 3$
Now,
$\sqrt{\alpha^4+\beta^4-1}\\
=\sqrt{1^4+3^4-1}=3^2 = 9$

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