Class 10 Math All topics Quadratic Equations

Find the smallest solution in positive integers of $x_{2}=41y_{2}−1$.

Solution:

We can write $41 =6+41 −6$

$=6+41 +65 $

$541 +6 =2+541 −4 $

$=2+41 +45 $

$541 +4 =2+541 −6 =2+41 +61 $

Therefore, $41 =6+2+1 2+1 12+1 ....$

Here penultimate is $532 $, and since the no. of quotients in the period is odd,$x=32,y=5$ is a solution.

Penultimate convergent

$=6+2+21 1 $

$=6+52 =532 $

Similar topics

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

polynomials

introduction to trigonometry

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quadratic equations

polynomials

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