TS EAMCET · Maths · Quadratic Equation
The cubic equation whose roots are thrice to each of the roots of \(x^3+2 x^2-4 x+1=0\) is
- A \(x^3-6 x^2+36 x+27=0\)
- B \(x^3+6 x^2+36 x+27=0\)
- C \(x^3-6 x^2-36 x+27=0\)
- D \(x^3+6 x^2-36 x+27=0\)
Answer & Solution
Correct Answer
(D) \(x^3+6 x^2-36 x+27=0\)
Step-by-step Solution
Detailed explanation
Given equation is \[ x^3+2 x^2-4 x+1=0 \] Let \(\alpha, \beta\) and \(\gamma\) be the roots of the given equation \[ \begin{aligned} \therefore & \alpha+\beta+\gamma & =-2, \alpha \beta+\beta \gamma+\gamma \alpha=-4 \\ \text { and } & \alpha \beta \gamma & =-1 \end{aligned} \]…
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