What is #sqrt(12+sqrt(12+sqrt(12+sqrt(12+sqrt(12....)))))#?

2 Answers

Answer:

#4#

Explanation:

There is a really interesting math trick behind it.

If you see a question like this take out the number inside it (in this case is #12#)

Take consecutive numbers such as:

#n(n+1)=12#

Always remember that the answer is #n+1#

This is true because if you let the infinite nested radical function = x then realise that x is also also under the first root sign as:

#x = sqrt(12 + x)#

Then, squaring both sides: #x^2 = 12 + x#
Or: #x^2 - x = 12#
#x(x-1) = 12#

Now let #x = n + 1#
Then #n(n+1) = 12# With the answer to the infinite nested radical function (x) being equal to #n + 1#

If you solve it you get #n=3# and #n+1=4#

So,

The answer is #4#

Practice problems:

#1rArrsqrt(72+sqrt(72+sqrt(72+sqrt(72+sqrt(72....)))))#

#Solutionrarr9#

#2rArrsqrt(30+sqrt(30+sqrt(30+sqrt(30+sqrt(30....)))))#

#Solutionrarr6#

And wait!!!

If you see a question like #sqrt(72-sqrt(72-sqrt(72-sqrt(72-sqrt(72....)))))#

#n# is the solution (in this case is #8#)

Problems to solve on your own

#sqrt(1056+sqrt(1056+sqrt(1056+sqrt(1056+sqrt(1056....))))#

#sqrt(110+sqrt(110+sqrt(110+sqrt(110+sqrt(110....))))#

Better luck!

Oct 28, 2017

Answer:

There is an other method to solve this

Explanation:

First of all, consider the whole equation equals #x#

#color(brown)(sqrt(12+sqrt(12+sqrt(12....)))=x#

We can also write it as

#color(brown)(sqrt(12+x)=x#

As, the #x# is nested into it. Solve it

#rarrsqrt(12+x)=x#

Square both sides

#rarr12+x=x^2#

#rarrx^2-x-12=0#

When we simplify this, we get

#color(green)(rArr(x+3)(x-4)=0#

From this, we get, #x=4 and -3#. We need a positive value, so it is 4.