find the value of √6+√6+√6+......infinity

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Anshika Dixit 6 years, 6 months ago
- 1 answers
Related Questions
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 0 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Sia ? 6 years, 6 months ago
Let {tex}x = \sqrt { 6 + \sqrt { 6 + \sqrt { 6 + \dots } } }{/tex} ......(i)
{tex}\Rightarrow{/tex} x2 = {tex}( \sqrt { 6 + \sqrt { 6 + \sqrt { 6 . . } } } ) ^ { 2 }{/tex} [Squaring both sides]
{tex}\Rightarrow{/tex} x2 = {tex}6 + \sqrt { 6 + \sqrt { 6 + \sqrt { 6 + \ldots } } }{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 = 6 + x{/tex} [From (i)]
{tex}\Rightarrow{/tex} {tex}x^2 - x - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}(x - 3) (x + 2) = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x = 3, x = -2{/tex}
{tex}\therefore{/tex} x = 3 [{tex}\because{/tex} x > 0]
0Thank You