Browse by Stream
QnA
Home
Search for Exam, Articles, Questions
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. Confused! kindly explain, - Complex numbers and quadratic equations - JEE Main-13
# Engineering

Let \alpha and \beta   be the roots of equation

x^{2}-6x-2=0.\; if a_{n}=\alpha ^{n}-\beta ^{n},\; for\: n\geq 1,\; then\; the \: value\; of\; \frac{a_{10}-2a_{8}}{2a_{9}}

is equal to:

  • Option 1)

    6

  • Option 2)

    -6

  • Option 3)

    3

  • Option 4)

    -3

Answers (6)
S solutionqc

x^2-6x-2=0\\ \alpha+\beta=6\\ \alpha.\beta=-2\\ Given, \frac{a_1_0-2a_8}{2a_9}, where a_n=\alpha^n-\beta^n\\ \frac{a_1_0-2a_8}{2a_9}=\frac{\alpha^1^0-\beta^1^0-2(\alpha^8-\beta^8)}{2(\alpha^9-\beta^9)}\\ Replace -2 with \alpha\beta\\ \frac{\alpha^1^0-\beta^1^0+(\alpha\beta)(\alpha^8-\beta^8)}{2(\alpha^9-\beta^9)}\\=\frac{\alpha^1^0-\beta^1^0+\alpha^9\beta-\alpha\beta^1^0}{2(\alpha^9-\beta^9)}\\ \frac{(\alpha+\beta)(\alpha^9-\beta^9)}{2(\alpha^9-\beta^9)}\\ =\frac{(\alpha+\beta)}{2}=\frac{6}{2}=3


Option 1)

6

Option 2)

-6

Option 3)

3

Option 4)

-3

A ANIKET DAS

probably option 3) 3

S surya

3

Similar Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions