Question

if√x+√x+√x+...=√x√x√x...then what is the value of x.

Answer 1

Answer :-

Value of x is 0 or 2.

Explanation :-

$\\ \\ \mathsf{ \sqrt{x + \sqrt{x + \sqrt{ x..... } } } = \sqrt{x \sqrt{x \sqrt{x...} } } }$

$\mathsf{ let \ \ \sqrt{x + \sqrt{x + \sqrt{ x..... } } } = \sqrt{x \sqrt{x \sqrt{x...} } } = k}$

i)

$\mathsf{ \sqrt{x + \sqrt{x + \sqrt{x...} } } = k}$

Squaring on both sides

$\mathsf{ \implies \bigg(\sqrt{x + \sqrt{x + \sqrt{x...} } } \bigg)^2 = (k)^2}$

$\mathsf{ \implies x + \sqrt{x + \sqrt{x...} } = k^2}$

$\mathsf{ \implies x +k = k^2 - - - eq(1)}$

ii)

$\mathsf{ \sqrt{x\sqrt{x \sqrt{x...} } } = k}$

Squaring on both sides

$\mathsf{ \implies \bigg( \sqrt{x\sqrt{x \sqrt{x...} } } \bigg)^2 = (k) ^2}$

$\mathsf{ \implies x\sqrt{x \sqrt{x...} } = k^2}$

$\mathsf{ \implies x \bigg(\sqrt{x \sqrt{x...} } \bigg) = k ^2}$

$\mathsf{ \implies xk = k ^2}$

$\mathsf{ \implies x = \dfrac{k^{2} }{k} }$

$\mathsf{ \implies x = k}$

Substitute x = k in eq(1)

$\mathsf{ \implies k +k = k^2}$

$\mathsf{ \implies 2k= k^2}$

$\mathsf{ \implies 0= k^2 - 2k}$

$\mathsf{ \implies 0= k(k - 2)}$

$\mathsf{ \implies k(k - 2) = 0}$

$\mathsf{ \implies k = 0 \ or \ k - 2= 0}$

$\mathsf{ \implies k = 0 \ or \ k = 2}$

$\mathsf{ \implies x= 0 \ or \ x = 2} \\ \\ \\ \boxed{ \bf \because x = k}$

∴ the value of x is 0 or 2.