TS EAMCET · Maths · Quadratic Equation
\(\alpha, \beta\) are the real roots of the equation \(x^2+a x+b=0\). If \(\alpha+\beta=\frac{1}{2}\) and \(\alpha^3+\beta^3=\frac{37}{8}\), then \(a-\frac{1}{b}=\)
- A \(\frac{-1}{6}\)
- B \(\frac{3}{2}\)
- C \(\frac{-3}{2}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{-1}{6}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \alpha+\beta=-a=\frac{1}{2} \Rightarrow a=\frac{-1}{2} \\ & \alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)=\frac{1}{2}\left((\alpha+\beta)^2-3 \alpha \beta\right) \\ & \frac{1}{2}\left(\frac{1}{4}-3 b\right)=\frac{37}{8}…
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