TS EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma\) are the real roots of the equation \(18 x^3-15 x^2-\) \(4 x+4=0\) such that \(\alpha=\beta\) and \(\alpha>\gamma\), then \(\alpha+\beta^2+\gamma^3=\)
- A \(\frac{71}{72}\)
- B \(\frac{53}{54}\)
- C \(\frac{89}{90}\)
- D \(\frac{59}{60}\)
Answer & Solution
Correct Answer
(A) \(\frac{71}{72}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text 18 x^3-15 x^2-4 x+4=0 \\ & \alpha=\beta, \alpha>\gamma \\ & \alpha+\alpha+\gamma=\frac{15}{18}=\frac{5}{6} \\ & \Rightarrow \quad 2 \alpha+\gamma=\frac{5}{6}\end{aligned}\)…
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