AP EAMCET · Maths · Quadratic Equation
If the number of real roots of \(x^9-x^5+x^4-1=0\) is \(n\), the number of complex roots having argument on imaginary axis is \(m\) and the number of complex roots having argument in \(2^{\text {nd }}\) quadrant is k , then \(m \cdot n \cdot k \cdot=\)
- A 6
- B 9
- C 12
- D 24
Answer & Solution
Correct Answer
(A) 6
Step-by-step Solution
Detailed explanation
\begin{aligned} & x^9-x^5+x^4-1=0 \Rightarrow\left(x^5+1\right)\left(x^4-1\right)=0 \\ & \Rightarrow x^4=1 \Rightarrow x=1,-1, i,-i \\ & \Rightarrow x^5=-1 \Rightarrow x=-1, e^{-\frac{i \pi}{5}}, e^{\frac{i \pi}{5}}, e^{-\frac{3 \pi i}{5}}, e^{\frac{i 3 \pi}{5}} \end{aligned}…
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