Q54 / 160 · TS EAMCET 2013 · Mathematics

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2 x+4=0$, then $\alpha^9+\beta^9$ is equal to

  • A
    $-2^8$
  • B
    $2^9$
  • Correct answer
    C
    $-2^{10}$
  • D
    $2^{10}$

Solution

Given quadratic equation is $$ x^2-2 x+4=0 $$ whose roots are $\alpha$ and $\beta$. $$ \therefore \quad \alpha+\beta=2 \text { and } \alpha \beta=4 $$ Now, $$ \begin{aligned} & \alpha^9+\beta^9=\left(\alpha^3\right)^3+\left(\beta^3\right)^3 \\ & =\left(\alpha^3+\beta^3\right)\left(\alpha^6+\beta^6-\alpha^3 \beta^3\right) \\ = & (\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)\left[\left(\alpha^2\right)^3+\left(\beta^2\right)^3-\alpha^3 \beta^3\right] \\ = & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right] \\ & \quad\left[\left(\alpha^2+\beta^2\right)\left(\alpha^4+\beta^4-\alpha^2 \beta^2\right)-\alpha^3 \beta^3\right] \\ = & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]\left[\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}\right. \\ & \left.\left\{\left(\alpha^2+\beta^2\right)^2-3 \alpha^2 \beta^2\right\}-\alpha^3 \beta^3\right] \\ = & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]\left[\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}\right. \\ & \left.\quad\left[\left\{(\alpha+\beta)^2-(2 \alpha \beta)\right]^2-3 \alpha^2 \beta^2\right\}-\alpha^3 \beta^3\right] \\ = & 2[4-12]\left[\{4-8\}\left\{(4-8)^2-48\right\}-64\right] \end{aligned} $$ [from Eq. (i)] $$ \begin{aligned} & =2(-8)\{(-4)(-32)(-64)\} \\ & =2(-8)(128-64) \\ & =2(-8)(64)=-2^{10} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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