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

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(B) Given the quadratic equation $x^2 - 2x + 4 = 0$. The roots are $\alpha, \beta$,so $\alpha + \beta = 2$ and $\alpha \beta = 4$. Since $\alpha$ is a

Vedclass · Accepted Answer

$32$

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(B) Given the quadratic equation $x^2 - 2x + 4 = 0$. The roots are $\alpha, \beta$,so $\alpha + \beta = 2$ and $\alpha \beta = 4$. Since $\alpha$ is a root,$\alpha^2 - 2\alpha + 4 = 0$,which implies $\alpha^2 = 2\alpha - 4$. Multiplying by $\alpha^n$,we get $\alpha^{n+2} = 2\alpha^{n+1} - 4\alpha^n$. Let $S_n = \alpha^n + \beta^n$. Then $S_n = 2S_{n-1} - 4S_{n-2}$. We know $S_0 = \alpha^0 + \beta^0 = 2$ and $S_1 = \alpha + \beta = 2$. $S_2 = 2S_1 - 4S_0 = 2(2) - 4(2) = 4 - 8 = -4$. $S_3 = 2S_2 - 4S_1 = 2(-4) - 4(2) = -8 - 8 = -16$. $S_4 = 2S_3 - 4S_2 = 2(-16) - 4(-4) = -32 + 16 = -16$. $S_5 = 2S_4 - 4S_3 = 2(-16) - 4(-16) = -32 + 64 = 32$. 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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