Problem

[Rafi_786]

Solve it,

 

[PlanetMaster]

Solve it,

View attachment 70538

Given [imath]x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}}[/imath]
We can rewrite this as [imath]x = \sqrt{6 + x}[/imath] as its iterating infinitely.

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

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...