Question

Simply this...

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } }$
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Answer 1

√6+ x= x

squaring

6 + x= x^2

x^2 - x-6= 0

x^2 - 3x +2x -6= 0

x( x-3)+2( x-3)= 0

x+2)(x-3)= 0

x= -2,3

Neglect -2 as here underroot operator which gives only positive value

So x= 3

✌✌✌Dr.Dhruv✌✌✌

Answer 2

Answer:

Step-by-step explanation:

See attachement