Question

$\rm{If \: x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + } } } }...then,}$
$\textsf{Find the positive value of x is}$​

Answer 1

$\rm{ x = \dfrac{3.23}{2} \: or \: x = \dfrac{1 + \sqrt{5} }{2} }$

Step-by-step explanation:

★Given -

• $\rm{ \small\: x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}}} \: \: \: (1)$

★ To find -

• $\rm{Positive \: value \: of \: x.}$

★Solution -

$\rm{x \: = \sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}}$

$\rm{Putting \: in \: (1),}$

$\rm{x = \sqrt{1 + x} }$

$\rm{ \small{Now, on \: squaring \: both \: sides \: we \: get,}}$

$\longmapsto\rm{ {x}^{2} = x + 1}$

$\longmapsto\rm{ {x}^{2} - x - 1= 0}$

$\rm{Where, a = 1 , b = - 1 , c = - 1}$

$\rm{D = b² - 4ac\: (D = discriminant)}$

$\longmapsto \rm{D = { (- 1)}^{2} - 4(1)( - 1)}$

$\longmapsto \rm{D = 1 + 4}$

$\longmapsto \rm{D = 5}$

$\rm{ {\large{x}} = \dfrac{ - b ± \sqrt{D} }{2a} }$

$\longmapsto \rm{{ \large{x} }= \dfrac{ - ( - 1) ± \sqrt{5} }{2(1)} }$

$\longmapsto \rm{{ \large{x} }= \dfrac{ 1 ± \sqrt{5} }{2} }$

$\longmapsto \rm{{ \large{x} }= \dfrac{ 1 + \sqrt{5} }{2} } \: \: ,\rm{{ \large{x} }= \dfrac{ 1 - \sqrt{5} }{2} }$

$\mapsto \rm{{ \large{x} }= \dfrac{ 1 + 2.23 }{2} } \: \: ,\rm{{ \large{x} }= \dfrac{ 1 - 2.23 }{2} }$

$\mapsto \rm{{ \large{x} }= \dfrac{ 3.23 }{2} } \: \: ,\rm{{ \large{x} }= \dfrac{ - 1.23 }{2} }$

From (1) we can see that the value of 'x' is inside a root. Hence, the value of 'x' cannot be negative.

$\rm{ {\large{\therefore}} \: x = \dfrac{3.23}{2} \: or \: x = \dfrac{1 + \sqrt{5} }{2} }$

★More to know -

• $\large( + )( + ) = ( + )$
• $\large( - )( + ) = ( - )$
• $\large( + )( - ) = ( - )$
• $\large( - )( - ) = ( + )$