Help needed

Find the value of $$x$$ while $\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\sqrt{x^{4}+...}}}}=\sqrt{2x+1}$

Note by Jason Snow
2 years, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Can you write some terms of the series in LHS ?

- 2 years, 3 months ago

- 2 years, 5 months ago

Desmos says that $$x=4$$ but I'm not sure if it's correct

- 2 years, 5 months ago

$$(n+1)=\sqrt { n^{ 2 }+2n+1 } =\sqrt { 1+\sqrt { n^{ 4 }+4n^{ 3 }+4n^{ 2 } } } =\sqrt { 1+\sqrt { 4{ n }^{ 3 }+\sqrt { n^{6}+8n^6+16n^4}}}$$

continuing the procedure we get

$$\sqrt { 1+\sqrt { 4n^{ 3 }+\sqrt { 16n^{ 4 }+\sqrt { 64{ n }^{ 12 }... } } } }$$

for n=1 we have

$$\sqrt { 1+\sqrt { 4+\sqrt { 16+\sqrt { 64... } } } }$$

squaring then subtracting one from each side we have above equation when $$x=4$$

we have the expression =3 when x=4

the other equation is also 3 when x=4

hence proved

- 2 years, 5 months ago

Thanks !! :D

- 2 years, 5 months ago

desmos is right

i'll show you why,it will take a bit of time to type it in LaTeX

- 2 years, 5 months ago

We have tried some methods that didn't seem to work, so looking forward to your solution :D

- 2 years, 5 months ago