TS EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+3 x^2-x-3=0\), then \(\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right)=\)
- A 16
- B 24
- C 36
- D 40
Answer & Solution
Correct Answer
(D) 40
Step-by-step Solution
Detailed explanation
Given equation, \(x^3+3 x^2-x-3=0\) \(\Rightarrow \quad x^2(x+3)-1(x+3)=0\) \(\Rightarrow \quad(x-1)(x+1)(x+3)=0\) So, roots \(\alpha=-3, \beta=-1\) and \(\gamma=1\) Therefore, \(\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right)\) \(=(1+9)(1+1)(1+1)=40\)
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