1

$$\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}= 5$$

Michael Hardy
  • 1
user182480
  • 37
  • 1
  • Using https://approach0.xyz/ I found hundreds of similar or connected questions https://math.stackexchange.com/q/967397/305862 (with $5$ instead of $3$) https://math.stackexchange.com/q/369852/305862 , https://math.stackexchange.com/q/1376667/305862 , etc. – Jean Marie Aug 18 '19 at 08:56

4 Answers4

4

This is actually a very common problem in elementary algebra textbooks. When you do enough of them, you memorize the methods.

The important thing to note is that the braced part in the equation below is equal to the entire thing, or in other words, $5$.

$$\sqrt{x+\underbrace{\sqrt{x+\sqrt{x+\cdots}}}}$$

This means the equation just becomes

$$\sqrt{x+5}=5$$ Can you continue from there?

Edward Jiang
  • 3,670
  • I've never seen this problem before.... why is the braced part equal to the entire thing? – layman Oct 11 '14 at 01:35
  • The square root is supposed to be infinite right? If you write out more of it, it becomes quite obvious: $$\sqrt{x+\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}}}= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}$$ – Edward Jiang Oct 11 '14 at 01:40
  • That's weird. How would you check your solution? – layman Oct 11 '14 at 01:53
  • You could plug the value of $x$ back into the equation, if you want. – Edward Jiang Oct 11 '14 at 01:59
  • 1
    Technically, you also need to prove that $\sqrt{20 + \sqrt{20 + \sqrt{20 + \cdots}}}$ converges, and converges to $5$. It's not trivial. – Zubin Mukerjee Oct 11 '14 at 02:10
  • I agree with @ZubinMukerjee because I don't think it is obvious what the infinite nested square root would "evaluate" to, if anything. I know at the elementary algebra level they aren't at all concerned about this, but they usually are concerned about checking their solutions, so I guess this is a type of problem that doesn't require them to check the solution... – layman Oct 11 '14 at 02:26
  • 1
    @ZubinMukerjee how would you go about proving that your infinite nested radical converges? – Joao Oct 11 '14 at 03:35
4

Hint $$\sqrt{x + \color{blue}{\sqrt{x + \sqrt{x + \sqrt{x + ...}}}}}= \sqrt{x+\color{blue}{5}}$$

Graham Kemp
  • 129,094
0

In the wiki page for nested radicals you will find the formula (for $n>0$) $$ \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}} = \frac{1+\sqrt{1+4n}}{2} $$ This implies $$ \frac{1+\sqrt{1+4x}}{2} = 5, $$ showing $x = 20$.

This formula can be shown like this: Let $\phi := \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$. Then $\phi^2 = n + \phi$. This is a quadratic in $\phi$ which one can easily solve with the quadratic formula. You have to choose the positive solution since roots of positive numbers are positive.

ViktorStein
  • 4,838
0

$x+n+a = \sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\ldots}}}$

This formula was proven by Ramanujan.

Vinícius Novelli
  • 2,487