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JEE Advanced · Mathematics · 2. Quadratic Equations

Let \(\alpha\) and \(\beta\) be the roots of \(x^2-6 x-2=0\) with \(\alpha>\beta\). If \(a_n=\alpha^n-\beta^n\) for \(n \geq 1\), then the value of \(\frac{a_{10}-2 a_8}{2 a_9}\) is

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

\(\frac{a_{10}-2 a_8}{2 a_9}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^8-\beta^8\right)}{2\left(\alpha^9-\beta^9\right)}\) \(=\frac{\alpha^8\left(\alpha^2-2\right)-\beta^8\left(\beta^2-2\right)}{2\left(\alpha^9-\beta^9\right)}\) \(\left[\because \alpha\right.\) is root of \(x^2-6 x-2=0\) \(\left.\alpha^2-2=6 \alpha\right]\)
Also, \(\beta\) is root of \(x^2-6 x-2=0\)
\(\Rightarrow \beta^2-2=6 \beta \)
\( =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9\right)}=\frac{6\left(\alpha^9-\beta^9\right)}{2\left(\alpha^9-\beta^9\right)}=3\)
From JEE Advanced
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