KCET · Maths · Quadratic Equation
If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+x+1=0\), then \(\alpha^{16}+\beta^{16}\) is equal to
- A 1
- B \(-1\)
- C 2
- D 0
Answer & Solution
Correct Answer
(B) \(-1\)
Step-by-step Solution
Detailed explanation
Given, equation is \(x^{2}+x+1=0\)
\[
\begin{aligned}
&\Rightarrow \quad \mathrm{x}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2} \\
&\Rightarrow \quad \quad \mathrm{x}=\omega, \omega^{2} \\
&\text { Since, } \alpha \text { and } \beta \text { are the roots of } \mathrm{x}^{2}+\mathrm{x}+1=0 \\
&\begin{aligned}
\therefore \quad \alpha=& \text { and } \beta=\omega^{2} \\
\text { Now, } \alpha^{16}+\beta^{16} &=(\omega)^{16}+\left(\omega^{2}\right)^{16} \\
&=\omega^{16}+\omega^{32} \\
&=\omega+\omega^{2} \\
&=-1 \quad\left(\because \omega^{3}=1\right)
\end{aligned}
\end{aligned}
\]
\[
\begin{aligned}
&\Rightarrow \quad \mathrm{x}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2} \\
&\Rightarrow \quad \quad \mathrm{x}=\omega, \omega^{2} \\
&\text { Since, } \alpha \text { and } \beta \text { are the roots of } \mathrm{x}^{2}+\mathrm{x}+1=0 \\
&\begin{aligned}
\therefore \quad \alpha=& \text { and } \beta=\omega^{2} \\
\text { Now, } \alpha^{16}+\beta^{16} &=(\omega)^{16}+\left(\omega^{2}\right)^{16} \\
&=\omega^{16}+\omega^{32} \\
&=\omega+\omega^{2} \\
&=-1 \quad\left(\because \omega^{3}=1\right)
\end{aligned}
\end{aligned}
\]
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