Roots of the equation \(x^3-3 x^2+3 x-9=0\) are ......
Roots of the equation \(x^3-3 x^2+3 x-9=0\) are ......
- \(3,1+2 \omega, 1+2 \omega^2\)
- \(3,-1+2 \omega,-1-2 \omega^2\)
- \(3,2-\omega, 2-\omega^2\)
- \(1,1+2 \omega, 1+2 \omega^2\)
Solution
Given equation is,
\(\begin{aligned}
& x^3-3 x^2+3 x-9 & =0 \\
\Rightarrow \quad & (x-3)\left(x^2+3\right) & =0
\end{aligned}\)
So one of root is 3 .
Now, \(x^2+3=0\)
taking \(x=1+2 w\), we have
\(\begin{aligned}
(1+2 \omega)^2 & +3 \\
& =1+4 \omega^2+4 \omega+3 \\
& =1+\left(4 \omega^2+4 \omega+4\right)-1 \\
& =1+4\left(\omega^2+\omega+1\right)-1 \\
& =0 \text {; or } \omega^2+\omega+1=0
\end{aligned}\)
Similarly, \(1+2 \omega^2\) also satisfies given equation
So roots are; \(3,1+2 \omega\) and \(1+2 \omega^2\)
Asked in: AP EAMCET 2020 (17 Sep Shift 2)
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