Problem

[Rafi_786]

Solve it,

 

[PlanetMaster]

Solve it,

View attachment 70538

Given $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}}$
We can rewrite this as $x = \sqrt{6 + x}$ as its iterating infinitely.

Squaring both sides and equating to zero
$x^{2}-x-6=0$
$x^{2}-3x+2x-6=0$
$x(x-3)+2(x-3)=0$
$(x-3)(x+2)=0$
$x=3,-2$

Therefore the answer is (b) 3 as we can't root a negative value.

 

[Cricket_Lover]

What a trash problem and it is copied. (just look at the dp)

Kindly behave yourself Hmm!

 

[piyara]

The answer is 6.
for full solution just ask me.

 

[PlanetMaster]

The answer is 6.
for full solution just ask me.

Can you please show your working?

P.S. That's not even in the options

 

[piyara]

Can you please show your working?

P.S. That's not even in the options

🫢 I done it in a wrong way, sorry.
Instead of addition sign I took it multiplication sign...