AP EAMCET · Maths · Complex Number
If \(1, \omega, \omega^2\) are the cube roots of unity, then the roots of the equation \(8 z^3-12 z^2+6 z-28=0\) are
- A \(2,2 \omega, 3 \omega^2+1\)
- B \(2, \frac{3 \omega+1}{2}, \frac{3 \omega^2+1}{2}\)
- C \(2, \frac{1+3 \omega}{3}, \frac{1+3 \omega^2}{3}\)
- D \(2, \frac{1-\omega}{2}, \frac{1-\omega^2}{2}\)
Answer & Solution
Correct Answer
(B) \(2, \frac{3 \omega+1}{2}, \frac{3 \omega^2+1}{2}\)
Step-by-step Solution
Detailed explanation
\(8 z^3-12 z^2+6 z-28=0\) ...(i) Since \(z=2\) satisfies equation (i). Hence \(z=2\) is one of solution equation (i). Therefore,…
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