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        Question no. 47:√(5+√(5-√(5+..... infinity, then find the nearest value for y²-y...
5 months ago

Saurabh Koranglekar
3161 Points
							Dear studenty = sqrt( 5+y)y^2 = 5+yy^2 -y = 5Therefore answer is dRegards

5 months ago
Samyak Jain
333 Points
							y = $\dpi{80} \sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...\infty}}}}$   , Clearly y = $\dpi{80} \sqrt{5+ a positive number$  > $\dpi{80} \sqrt{5}$Now, y2 = 5 + $\dpi{80} \sqrt{5-y}$i.e. (y2 – 5)2  =  5 – y   $\dpi{100} \Rightarrow$   y4 – 10y2 + 25  =  5 – yy4 – 10y2 + y + 20  =  0. y4 – 10y2 + y + 20 gets factorised to (y2 + y – 5)(y2 – y – 4).[Method of finding above factors is lengthy. I cannot explain it in detail here.]y4 – 10y2 + y + 20 = 0   $\dpi{100} \Rightarrow$   (y2 + y – 5)(y2 – y – 4) = 0 $\dpi{100} \because$ y > $\dpi{80} \sqrt{5}$  $\dpi{100} \therefore$ y2 + y > 5 + $\dpi{80} \sqrt{5}$  or  y2 + y – 5 > $\dpi{80} \sqrt{5}$  > 0. So, y2 + y – 5 $\dpi{80} \neq$ 0  i.e.  (y2 – y – 4) = 0$\dpi{100} \therefore$  y2 – y  =  4 .

5 months ago
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