If alpha, beta are the roots of x^2 - 2x + 4 | vedclass

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Question

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.For this equation,the sum of roots $\alpha + \beta = 2$ and the product of roots $\alpha \beta = 4

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.For this equation,the sum of roots $\alpha + \beta = 2$ and the product of roots $\alpha \beta = 4$.We know that $\alpha$ and $\beta$ satisfy the equation,so $\alpha^2 = 2\alpha - 4$ and $\beta^2 = 2\beta - 4$.Using the recurrence relation $S_n = \alpha^n + \beta^n$,we have $S_n - 2S_{n-1} + 4S_{n-2} = 0$.For $n=1$,$S_1 = \alpha + \beta = 2$.For $n=2$,$S_2 = (\alpha + \beta)^2 - 2\alpha \beta = (2)^2 - 2(4) = 4 - 8 = -4$.For $n=3$,$S_3 = 2S_2 - 4S_1 = 2(-4) - 4(2) = -8 - 8 = -16$.For $n=4$,$S_4 = 2S_3 - 4S_2 = 2(-16) - 4(-4) = -32 + 16 = -16$.For $n=5$,$S_5 = 2S_4 - 4S_3 = 2(-16) - 4(-16) = -32 + 64 = 32$.Thus,$\alpha^5 + \beta^5 = 32$.

If $\alpha, \beta$ are the roots of $x^2 - 2x + 4 = 0$,then $\alpha^5 + \beta^5$ is equal to

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