Question

What is the trick to solve questions like
$x = \sqrt{2 + \sqrt{2 + \sqrt{2 + ... \infty } } }$
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Answer 1

⭐Hola User_______________

⭐Here is Your Answer....!!!!

_______________________

↪Actually welcome to the concept of the SUBSTITUTION FOR INFINITE SERIES !!

↪BAsically the technique to solve this type of question is to squaring both the sides and considering the infinity series again as X

↪thus we get here after squaring..

↪Since here the whole series get totally replaced by the variable X that is already given in the question

↪the following happens to be a quadratic equation through FACTORISATION method
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〽⭐〽BEST OF LUCK 10TH CBSE MATES〽⭐〽

Answer 2

           

A simple trick.

Love this question a lot, and also the trick.

Here, the value of x is given, square root of 2 + square root of 2 + square root of 2 + ..........

Inside this we can also see another x!

x = Square root of 2 + square root of 2 + square root of 2 +......

Here, include the portion of the RHS in a bracket, except the first square root of 2, as the following:

x = Square root of 2 + (square root of 2 + square root of 2 +...... )

Which means,

$x=\sqrt{2+\sqrt{2+\sqrt{2+...\infty}}} \\ \\ \\ \Rightarrow\ x=\sqrt{2+(\sqrt{2+\sqrt{2+...\infty}})}$

Isn't the value of the sum inside the bracket, x?

So we can substitute x in the place of the portion included in the bracket, as the following:

$x=\sqrt{2+\sqrt{2+\sqrt{2+...\infty}}} \\ \\ \\ \Rightarrow\ x=\sqrt{2+(\sqrt{2+\sqrt{2+...\infty}})} \\ \\ \\ \Rightarrow\ x=\sqrt{2+x}$

Now we can find the value of x.

$x=\sqrt{2+x} \\ \\ x^2=2+x \\ \\ x^2-x-2=0 \\ \\ x^2-2x+x-2=0 \\ \\ x(x-2)+1(x-2)=0 \\ \\ (x+1)(x-2)=0 \\ \\ \\ \therefore\ x=\bold{-1}\ \ \ ; \ \ \ x=\bold{2}$

Hope this helps you. Plz mark it as the brainliest.

Plz ask me if you have any doubts.

Thank you. :-))