Question

$\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } }$
​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• $\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } }$

${\sf{\green{\underline{\large{To\:find}}}}}$

• The given value of term

${\sf{\pink{\underline{\Large{Explanation}}}}}$

$\tt\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } }$

Let $\tt\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } }$ be x

$\tt\rightarrow\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } }$=x

Both side squaring

$\tt\rightarrow\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } }$=x

$\tt\rightarrow(\sqrt{5 + \sqrt{5 + \sqrt{5 + \infty } } })^2$=x²

$\tt\rightarrow{5 + \sqrt{5 + \sqrt{5 + \infty } } })$=x²

Now again after squaring the series make

Therefore,

$\tt\rightarrow{5 + \sqrt{5 + \sqrt{5 + \infty } } }$=x²

$\tt\rightarrow{5 + x)$=x²

=X²-X-5

Solving the quadratic equation

=X²-X-5

Quadratic formula

$\tt{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a}}$

a=1

b=-1

c=-5

$\rightarrow\tt{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a}}$

$\rightarrow\tt{X=\dfrac{1\pm{\sqrt{1+20}}}{2}}$

Taking x as +

$\rightarrow\tt{X=\dfrac{1+{\sqrt{1+20}}}{2}}$

$\rightarrow\tt{X=\dfrac{1+{\sqrt{21}}}{2}}$

Taking x as -

$\rightarrow\tt{X=\dfrac{1-{\sqrt{1+20}}}{2}}$

$\rightarrow\tt{X=\dfrac{1-{\sqrt{21}}}{2}}$

Hence value of x is +