+0

# Endless root

+5
687
6

Find x

Oct 15, 2015

#4
+109812
+20

$$\sqrt{X+\sqrt{X+\sqrt{X+...}}}=5\\ Let\\ Y= \sqrt{X+\sqrt{X+\sqrt{X+...}}}\qquad where\;\;\;Y=5\\ Now\\ \sqrt{X+Y}=5\\ X+Y=25\\ but\; Y=5\\ X+5=25\\ X=20$$

.
Oct 15, 2015

#1
+94
+5

X is equal to 20

Oct 15, 2015
#2
+10

And why?

Oct 15, 2015
#3
+5

x=x^(1/2^y)=5, where y=the number of nested squares.E.g., if you have 5 nested squares, then y would be 2^5=32. So x^1/32=5.

Oct 15, 2015
#4
+109812
+20

$$\sqrt{X+\sqrt{X+\sqrt{X+...}}}=5\\ Let\\ Y= \sqrt{X+\sqrt{X+\sqrt{X+...}}}\qquad where\;\;\;Y=5\\ Now\\ \sqrt{X+Y}=5\\ X+Y=25\\ but\; Y=5\\ X+5=25\\ X=20$$

Melody Oct 15, 2015
#5
0

Thanks YangShizzle,Melody and Guest for anweser my question!

Melody ,that is a very cool way to do it!

Oct 15, 2015
#6
+2498
0

i think this way is easear

sqrt{X+\sqrt{X+\sqrt{X+...}}}=5

sqrt{5x4+\sqrt{5x4+\sqrt{5x4+...}}}=5 if it is + you will bring 5 if it is - you will bring 4

let s say:

sqrt{12+\sqrt{12+\sqrt{12+...}}}=x

we can write it like

sqrt{4x3+\sqrt{4x3+\sqrt{4x3+...}}}=x  so x will be 4 because + we choosing bigger one

Jan 3, 2016
edited by Solveit  Jan 3, 2016