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Hindi
4-2.Quadratic Equations and Inequations
hard

$x^2 - 6x - 2 = 0$ ના બીજ $\alpha$ અને $\beta$ લો. જ્યાં $\alpha$ > $\beta$ જો બધા $n \geq 1$ માટે $a_n = \alpha^n - \beta^n$ હોય, તો $\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$ નું મૂલ્ય કેટલું થાય ?

A

$1$

B

$2$

C

$3$

D

$4$

Solution

${x^2}\, – \,\,6x\,\, – \,\,2\,\, = \,\,0$ ના બીજ $\alpha {\rm{,}}\,\,\beta \,$ છે.

$ \Rightarrow \,\,{\alpha ^{\rm{2}}}\, – \,\,6\alpha \,\, – \,\,2\,\, = \,\,0\,\,\& \,\,{\beta ^2}\, – \,\,6\beta \,\, – \,\,2\,\, = \,\,0$

$\frac{{{a_0}\, – \,\,2{a_8}}}{{2{a_9}}}\,\, = \,\,\frac{{{\alpha ^{10}} – \,\,{\beta ^{10}}\, – \,\,2({\alpha ^8}\, – \,\,{\beta ^8})}}{{2({\alpha ^9}\, – \,\,{\beta ^9})}}\,\,\,\,$

$ = \,\,\,\,\frac{{{\alpha ^8}({\alpha ^2}\, – \,\,2)\,\, – \,\,{\beta ^8}({\beta ^2}\, – \,\,2)}}{{2({\alpha ^9}\, – \,\,{\beta ^9})}}$

$ = \,\,\frac{{{\alpha ^8}\,.\,\,6\alpha \,\, – \,\,{\beta ^8}\,.\,\,6\beta }}{{2({\alpha ^9}\, – \,\,{\beta ^9})}}\,\, = \,\,3$

Standard 11
Mathematics

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