21Let x2-y2=x+y
or, (x+y)(x-y-1)=0
→ x+y=0 or x=y+1
So, ketan your assumption was wrong.
Find the value of \sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-....\propto }}}} ?
my method...
let \sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-....\propto }}}} =x
and {\sqrt{7-\sqrt{7+\sqrt{7-....\propto }}}} =y
now, x2=7+y and
y2=7-x
subtracting the equations we get
x2-y2=x+y
therefore x-y=0 and thus x=y
putting it in equation 1 i m getting x=1+√292 which is 3.1925
but without taking y and solving it as given in the book...
x2=7+√(7-y)
x4-14x2+y+42=0
(x-3)(x3+3y2-5y-14)=0
therefore x=3
why is my answer incorrect.....is it bcoz i assumed x+y≠0
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4 Answers
21
1708\hspace{-16}$Let $\mathbf{x=\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7.............\infty}}}}}$\\\\ $\mathbf{x=\sqrt{7+\sqrt{7-x}}\Leftrightarrow (x^2-7)=\sqrt{7-x}}$\\\\ $\mathbf{(x^2-7)^2=7-x}$\\\\\ $\mathbf{x^4+7^2-2.7.x^2=7-x}$\\\\ $\mathbf{x^4-2.7.x^2+x+(7^2-7)=0}$\\\\ Now Let $\mathbf{7=a}$. Then \\\\ $\mathbf{a^2-a.(2x^2+1)+(x^4+x)=0}$\\\\ So $\mathbf{a=\frac{(2x^2+1)\pm \sqrt{(2x^2+1)^2-4(x^4+x)}}{2}}$\\\\ $\mathbf{a=\frac{2x^2+1\pm (2x-1)}{2}}$\\\\ Now taking positive Sign::\\\\ $\mathbf{a=\frac{2x^2+1+2x-1}{2}=x^2+x}$\\\\ So $\mathbf{x^2+x=a=7}$\\\\ $\mathbf{x^2+x-7=0}$\\\\ $\mathbf{x=\frac{-1\pm \sqrt{29}}{2}}$\\\\ So $\mathbf{\boxed{\bold{x=\frac{-1 +\sqrt{29}}{2}\;,x=\frac{-1- \sqrt{29}}{2}}}}$\\\\\\
\hspace{-16}$Similarly Now taking Negative Sign::\\\\ $\mathbf{a=\frac{2x^2+1-2x+1}{2}=x^2-x+1}$\\\\ So $\mathbf{x^2-x+1=a=7}$\\\\ $\mathbf{x^2-x-6=0}$\\\\ $\mathbf{x=\frac{1\pm 5}{2}=-2\;,3}$\\\\ So $\mathbf{\boxed{\bold{x=-2\;,3}}}$\\\\ So $\mathbf{\boxed{\bold {x=3}}}$
62yes arnab.. quite a good solution...
something like
http://www.targetiit.com/iit-jee-forum/posts/you-have-all-the-time-in-the-world-try-this-17266.html