AP EAMCET · Maths · Quadratic Equation
Roots of the equation \(x^3-3 x^2+3 x-9=0\) are ......
- A \(3,1+2 \omega, 1+2 \omega^2\)
- B \(3,-1+2 \omega,-1-2 \omega^2\)
- C \(3,2-\omega, 2-\omega^2\)
- D \(1,1+2 \omega, 1+2 \omega^2\)
Answer & Solution
Correct Answer
(A) \(3,1+2 \omega, 1+2 \omega^2\)
Step-by-step Solution
Detailed explanation
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…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- \(\int_0^{\frac{\pi}{2}} \frac{\cos x d x}{\sqrt{1+\cos x \sin x}}=\)AP EAMCET 2018 Medium
- If a set \(\mathrm{A}\) has \(\mathrm{n}\) elements, then the number of functions defined from A to A that are not one-one isAP EAMCET 2023 Easy
- If the lines \(3 x+y-4=0, x-\alpha y+10=0, \beta x+2 y+4=\) 0 and \(3 x+y+k=0\) represent the sides of a square, then \(\alpha \beta(k+4)^2=\)AP EAMCET 2024 Easy
- \({ }^{34} C_{10}+3 \cdot\left({ }^{34} C_9\right)+3 \cdot\left({ }^{34} C_g\right)+{ }^{34} C_7=\)AP EAMCET 2018 Medium
- The constant \(c\) of Rolle's theorem for the function \(f(x)=(x-1)^3(x-2)^5\) in [1, 2] isAP EAMCET 2017 Medium
- \(A\) and \(B\) are playing chess game with each other. The probability that \(A\) wins the game is 0.6, the probability that he loses is 0.3 and the probability its draw is 0.1. If they played three games, then the probability that \(A\) wins atleast two games isAP EAMCET 2025 Medium
More PYQs from AP EAMCET
- Let be a function defined by . what is the value of ?AP EAMCET 2021 Easy
- The population of a city grows at the annual rate of \(3 \%\). What percentage increase is expected in \(5 \mathrm{yr}\) ?AP EAMCET 2020 Medium
- If , where . Then is increasing in the intervalAP EAMCET 2021 Hard
- If the points with position vectors \((\alpha \hat{i}+10 \hat{j}+13 \hat{k})\), \((6 \hat{i}+11 \hat{j}+11 \hat{k}),\left(\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}\right)\) are collinear then \((19 \alpha-6 \beta)^2=\)AP EAMCET 2024 Medium
- \(\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=\)AP EAMCET 2018 Easy
- If the equation of tangent at \((2,3)\) on \(y^2=a x^3+b\) is \(y=4 x-5\), then the value of \(a^2+b^2=\)AP EAMCET 2024 Easy