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Grade: 10
        
Question no. 47:√(5+√(5-√(5+..... infinity, then find the nearest value for y²-y...
5 months ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
3161 Points
							Dear student

y = sqrt( 5+y)
y^2 = 5+y
y^2 -y = 5

Therefore answer is d

Regards
5 months ago
Samyak Jain
333 Points
							
y = \sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...\infty}}}}   , Clearly y = \sqrt{5+ a positive number  > \sqrt{5}
Now, y2 = 5 + \sqrt{5-y}
i.e. (y2 – 5)2  =  5 – y   \Rightarrow   y4 – 10y2 + 25  =  5 – y
y4 – 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   \Rightarrow   (y2 + y – 5)(y2 – y – 4) = 0
 
\because y > \sqrt{5}  \therefore y2 + y > 5 + \sqrt{5}  or  y2 + y – 5 > \sqrt{5}  > 0.
 
So, y2 + y – 5 \neq 0  i.e.  (y2 – y – 4) = 0
\therefore  y2 – y  =  4 .
5 months ago
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