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JEE Main 2025 · Mathematics · Quadratic Equations · medium

Let α>β be the roots of x^2-6x-2=0 and an=α^n+β^n. Then (a10-2a8)/(2a9) equals

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Let α>β\alpha>\beta be the roots of x2−6x−2=0x^2-6x-2=0 and an=αn+βna_n=\alpha^n+\beta^n. Then a10−2a82a9\dfrac{a_{10}-2a_8}{2a_9} equals
  1. A. 11
  2. B. 22
  3. C. 33
  4. D. 66
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Answer: (C) 33

Since α,β\alpha,\beta satisfy x2=6x+2x^2=6x+2, multiply by xn−2x^{n-2} and add for both roots: an=6an−1+2an−2a_n=6a_{n-1}+2a_{n-2}. For n=10n=10: a10=6a9+2a8a_{10}=6a_9+2a_8, so a10−2a8=6a9a_{10}-2a_8=6a_9. Therefore a10−2a82a9=6a92a9=3\dfrac{a_{10}-2a_8}{2a_9}=\dfrac{6a_9}{2a_9}=3.
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