Curious for other elegant methods for solving $x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4$

Question

Solve the equation $$x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4$$

Actually I have solved this question but I want to know other methods of solving this question. It's not that I am curious for just elementary methods but you all can give solutions using advance maths too (which I'll see and learn). I am posting my solution as an answer.

I hope many elegant solutions come :).

Answer 1

$$x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4\tag1$$ Since we have $$x+\frac12+\sqrt{x+\frac14}=x+\frac 14+\frac 14+\sqrt{x+\frac14}=\bigg(\frac 12+\sqrt{x+\frac14}\bigg)^2$$ we get $$x+\frac 12+\sqrt{x+\frac14}=4\tag2$$ From $(1)(2)$, we get $$x+\sqrt 4=4$$ i.e. $$x=2$$ which is sufficient.

Answer 2

Let $x+\frac14=a^2$ $\implies x=a^2-\frac14$ $\implies a^2-\frac14<4$ $\implies -2<a<2$

Now $$a^2-\frac14+\sqrt{a^2-\frac14+\frac12+a}=4$$ $$\implies a^2-\frac14+|a+\frac12|=4$$ $$\implies a^2+a+\frac14-4=0$$ $$\implies a=\frac32,a=\frac{-5}2$$ But $-2<a<2$ so $a=\frac32$

And hence $$\boxed{x=2}$$

Edit: Thank you @jjagmath for pointing out the flaw. I just didn't think of opening the mod both ways. But I open it both ways then we get values two more values of $a$ and they are $$a=\frac12\pm\sqrt5$$ One value is immediately discard. But for the value $\frac12-\sqrt5$ we get $x$ as $5-\sqrt5$ which should be the solution. But is it $?$ I don't know. I'm studying other answers. Thank you for pointing out the mistake :)

Answer 3

Inelegant But A Solution:

"Unraveling" it a bit, we get that

$$\left( (4-x)^2 - x- \frac 1 2 \right)^2 - x - \frac 1 4 = 0$$

or

$$x^4 - 18 x^3 + 112 x^2 - 280 x + 240 = 0$$

Though unpleasant, the rational root theorem allows us to conclude roots of $x=2$ and $x=6$. (Luckily, these are some of the first potential roots someone would check, so one reaches these relatively fast.)

What remains (when factoring out $(x-2)(x-6)$) is then the quadratic

$$x^2 - 10x + 20$$

which has roots $5 \pm \sqrt 5$ by the quadratic formula.

A trivial calculation with the derivative shows that

$$f(x) := x + \sqrt{ x + \frac 1 2 + \sqrt{x + \frac 1 4 } }$$

is increasing ($f'(x) > 0$), so only one of the four solutions ($x=2$, $x=6$, $x=5 \pm \sqrt 5$) could be satisfactory. (This fact may even be obvious since we only have compositions and sums of $x$ and $\sqrt x$, up to adding $1/2$ or $1/4$, all obviously increasing functions, so we may circumvent the derivative altogether.) Alternatively, we note that

$$4 = f(x) \ge x$$

so we can certainly eliminate $x=6$ and $x=5+\sqrt 5$ off the bat (though one still needs an argument to justify $5-\sqrt 5 \approx 2.8$ not being a solution).

One naturally will be led to try $x=2$ first since its math will likely be the least unpleasant, and acquire the desired result.

Answer 4

If $x\le 0$ then by chaining the simple inequalities $\sqrt{x+\frac14}\le \sqrt{\frac 14}\le \frac 12$ and so on, we get that $f(x)\le 1$.

So to reach the value $4$ we must have $x>0$ and we can substitute $x=\frac 14 \sinh(u)^2$

I'll use letters $s,c$ to simplify the writing and use abundantly $c^2-s^2=1$:

$f(x)=\frac {s^2}4+\sqrt{\frac {s^2}4+\frac 12+\sqrt{\frac {c^2}4}}=\frac {s^2}4+\sqrt{\frac {c^2}4+\frac c2+\frac 14}=\frac{s^2}4+(\frac c2+\frac 12)=\frac{c^2}4+\frac c2+\frac 14=\frac 14(c+1)^2$

This solves to $\frac 14(c+1)^2=4\iff c=3$ (since $c>0$) and therefore $x=\frac{s^2}4=\frac{c^2-1}{4}=2$

Note: this approach shares in fact some similarities with Mathlove's, his subtelty $x+\frac 12=(x+\frac 14)+\frac 14$ appears here as a natural change from $\frac{s^2}4+\frac 12=\frac {c^2}4+\frac 14$

Answer 5

$Hint-A:$ $$x = \frac{3}{4} + \tan^2(\theta)$$

$Hint-B:$

$$\sec^2(\theta) + \sec(\theta) + \frac{1}{4} = (\sec(\theta) + \frac{1}{2})^2$$

$Hint-C:$

$$\sec^2(\theta) = 1 + \tan^2(\theta)$$