What is the simple way to solve the equation $x+\sqrt{a+\sqrt{x}}=a$ for all real $x$?

Question

Solve the equation $$x+\sqrt{a+\sqrt{x}}=a$$ for all real $x$ and nonzero parameter $a$, prove that the equation has a solution if and only if $a\ge 1$.

My attempts and thoughts.

$$\sqrt {a+\sqrt x}=a-x$$

$$a+\sqrt x=(a-x)^2$$

$$\sqrt x =(a-x)^2-a$$

$$x=\big((a-x)^2-a\big)^2$$

$$x=(a^2-2ax+x^2-a)^2$$

But the last expansion seems terrible.

I haven't been able to prove why $a\ge 1$ has to be. My other concern is that not all roots of the last equation may be valid.For example, say $\sqrt {x}=-1$. We have $x=1$ but $x=1$ doesn't satisfy $\sqrt {x}=-1$.

My questions. Is there a method that doesn't square the equation and works easily in general? How can we get rid of excess roots? How can we prove that $a\ge 1$ ?

Answer 1

$$x = a - \sqrt{a +\sqrt x}$$

$$= a - \sqrt{a + \sqrt{a - \sqrt{a + \sqrt{x}}}} $$

=$ a - \sqrt{a + \sqrt{a - \sqrt{a + \sqrt{a ....}}}}$till infinity

let $y$= $\sqrt{a + \sqrt{a - \sqrt{a + \sqrt{a - ...}}}}$till infinity

$y^2 = a + \sqrt{a - \sqrt{a +.......}}$

$(y^2 - a )^2 = a - \sqrt{a + \sqrt{a...}}$

$$y^4 + a^2 -2ay^2 - a = -y$$

therefore $ y^4 - 2ay^2 + y + a^2 - a = 0$, $ y > 0$, $a - y > 0$, $a > 0$

But I'm sorry this is still an incomplete solution but I can't comment so I posted my idea. Hope it will help some other person to solve this. Sorry again:(

Answer 2

Let us know in the comments whether this answer solves your concerns and the answer can be improved later .

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Since the left-hand side of the original equation is always non-negative, we observe that $a>0$ . Then, we want to add $\sqrt x$ to both side, to check if it works for us :

$$ \begin{align}x+\sqrt x+\sqrt{a+\sqrt{x}}&=a+\sqrt x\end{align} $$

Substituting $\thinspace y:=\sqrt x$ and $z:=\sqrt {a+\sqrt x}=a-y^2$, where $y≥0\wedge z>0\thinspace $ then you have :

$$ \begin{align}&{\color{#c00}{y^2}}+{\color{#0a0}{y}}+{\color{#0a0}{z}}-{\color{#c00}{z^2}}=0\\ \implies &(y+z)(y-z+1)=0\\ \implies &y=z-1=a-y^2-1\end{align} $$

Since $\thinspace a>0$ and $y≥0$, then you get the following :

$$ \begin{align}&y^2+y-(a-1)=0\\ \implies &y=\frac {-1+\sqrt {4a-3}}{2}\\ \implies &x=\frac {\left(\sqrt {4a-3}-1\right)^2}{4}\end{align} $$

Thus, the final result becomes :

$$\bbox[5px,border:2px solid #C0A000]{x=\frac {2a-\sqrt {4a-3}-1}{2}}$$

This implies that, you have only one real root ( solution ) .

Finally, to prove that $a≥1$, observe :

$$ \begin{align}y≥0&\iff\sqrt {4a-3}≥1\\ &\iff a≥1\end{align} $$

which completes the answer .

Answer 3

$$x+\sqrt{a+\sqrt{x}}=a$$

Move term and square it, we get

$$a+\sqrt x=(a-x)^2$$

The trick is we can treat this as a quadratic equation with respect to $a$ $$a^2-(2x+1)a+x^2-\sqrt x=0$$ Instead of solving $x$, we solve for $a$

$$a=\frac{2x+1\pm\sqrt{4x+4\sqrt x+1}}{2}$$ Note that inside square root is a complete square $(2\sqrt x+1)^2$, so we get $$a=\frac{2x+1\pm(2\sqrt x+1)}{2}$$

Let $t=\sqrt x$ and we get two equations:

$$\begin{cases} \color{blue}{a=t^2+t+1}\\ \\ \color{red}{a=t^2-t} \end{cases}$$

First let's look at the $\color{red}{\text{red-color}}$ equation. We will show it has no real roots.

$$x+\sqrt{a+\sqrt{x}}=a$$

Here, the OP implies $$\sqrt{a+\sqrt{x}}=a-x\ge0\Rightarrow a\ge x\Rightarrow \color{green}{a\ge t^2}\tag{1}$$

Plug in the $\color{red}{\text{red-color}}$ equation, we get

$$\color{red}{t^2-t=a}\ge t^2\Rightarrow -t\ge 0\Rightarrow -\sqrt x\ge0$$

This is true if and only if $x=0$, so $t=0$, the $\color{red}{\text{red-color}}$ equation gives $a=0^2+0=0$

But we require $a\neq 0$, hence we get a contradiction. Therefore, the $\color{red}{\text{red-color}}$ equation has no real roots. We can ignore it.

Now let's move to the $\color{blue}{\text{blue-color}}$ equation. Note the restriction "$\color{green}{a\ge t^2}$" in eq.(1) is automatically satisfied here.

$$\color{blue}{a=t^2+t+1}$$

Solve it and we get:

$$t=\frac{-1\pm\sqrt{4a-3}}{2}$$

Since $t=\sqrt x\ge0$, we only keep the $+$ root.

$$t=\frac{-1+\sqrt{4a-3}}{2}$$

Finally, to determine parameter $a$, we need following conditions:

$$4a-3\ge0~~\cap~~t\ge0$$

This gives the range for $a$

$$\boxed{a\ge1}$$

The solution is:

$$\boxed{x=t^2=\frac{2a-1-\sqrt{4a-3}}{2}}$$

Answer 4

I'll start with the second part first. If you're happy with graphical "proofs", you could consider the two graphs $y=a-x$ and $y=\sqrt{a+\sqrt{x}}$. You could probably make this more formal using the intermediate value theorem and other calculus arguments.

Case 1: $a>1$. Here you have $a>\sqrt{a}$, so you get exactly one intersection point where $x>0$.

Case 2: $a=1$. Here you have $a=\sqrt{a}$, so the two graphs intersect at $x=0$.

Case 3: $0<a<1$. Here you have $a<\sqrt{a}$, so since $y=\sqrt{a+\sqrt{x}}$ is increasing and $y=a-x$ is decreasing, there is no intersection point.

Case 4: $a<0$. Lastly, it's clear that when $a<0$ that there are no solutions, since we have $x\ge 0$ from the natural domain of the square root function, so the left-hand side $x+ \sqrt{a+\sqrt{x}}$ is non-negative, while the right-hand side is $a$ is negative.

The combination of the four cases shows that the equation has a solution if and only if $a\ge 1$.

For your first part, I don't have an solution which eliminates the 'squaring problem', but here's one that avoids the 'quartic problem'. One possible trick here to save yourself from solving a quartic is to solve it for $a$ rather than $x$. So if we go ahead and move the $x$ to one side and square it:

$$ a+ \sqrt{x} = (a-x)^2 \Rightarrow a^2 - (2x+1) a + (x^2 - \sqrt{x}) = 0.$$

Using the quadratic formula, you'll end up getting $$ a = x + \sqrt{x} + 1, x - \sqrt{x}.$$

Because we know that $a=1, x=0$ is a solution, we can eliminate the second option, so now we have a quadratic equation in $\sqrt{x}$, namely $(\sqrt{x})^2 + \sqrt{x} + (1-a) = 0$. Solving this, we get $$\sqrt{x} = \frac{-1\pm \sqrt{4a-3}}{2} \Rightarrow x = \frac{2a-1 \pm \sqrt{4a-3}}{2}.$$

Again, knowing that $a=1, x=0$ is a solution shows us that $x = \dfrac{2a-1 + \sqrt{4a-3}}{2}$.

Answer 5

This is another form of graphical method. This answer proves $a \ge 1$ is required for solutions only, as a few other answers already target obtaining mathematical solutions to your equation. This answer may not be something you can use mathematically, but it fulfils your wish of:

I want a simpler and more descriptive answer to my concerns

Consider the LHS and RHS as separate functions, ie.

$y = a \\ y = x + \sqrt{a + \sqrt{x}}$

Looking at it this way, we need to find when the line $y = a$ has intersection(s) with the function $\sqrt{a + \sqrt{x}}$. This would be the same as when both graphs take the same value, and hence you obtain your solution for $a = x + \sqrt{a + \sqrt{x}}$.

$y = a$ is not much of an interesting function, it remains constant over $x \in [0, \infty)$. However, the function $y = x + \sqrt{a + \sqrt{x}}$ is more intriguing to analyse. Let $f(x) = x + \sqrt{a + \sqrt{x}}$.

Note that $f(x)$ only exists for $x \ge 0$.

Now see, $$\large f(x)= \underbrace{x}_{\color\green+} + \underbrace{\sqrt{a + \sqrt{x}}}_{\color\green+}$$

and hence, $\displaystyle{\color\red{f(x) > 0}}$ regardless of $a$ and strictly increasing. This suggests the $\min{(f(x))}$ occurs at $\displaystyle{\color\red{x = 0}}$

Given $f(0) = \sqrt{a}$, see that an intersection would occur if

$y = a = \sqrt{a} = f(x = 0)$

or

$y = a > \sqrt{a}$ as $f(x)$ would still exist such that $y = a = f(x)$ has a solution/intersection for some $+x$. Observe, how the info that $f(x)$ is strictly increasing plays a role here.

Overall and finally, we look for such $a$ satisfying $a \ge \sqrt{a}$; only to occur when $$\large{a \ge 1}$$

Answer 6

Reading the answers, I first thought that some answers show an incorrect proof for the claim that the equation has a solution if and only if $a\ge 1$.

However, "incorrect" is perhaps too strong a word.

I'm going to write a solution first, and then show some thoughts about a solution, and finally write another solution which tries to avoid using "squaring".

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A solution :

From $$x+\sqrt{a+\sqrt x}=a\tag1$$ we have to have $$x\ge 0\qquad\text{and}\qquad a\gt 0\tag2$$ ($a\gt 0$ because LHS $\ge 0$, and $a$ is non-zero).

Under $(2)$, we have (as some other answers did) $$\begin{align}(1)&\iff \sqrt{a+\sqrt x}=a-x \\\\&\iff a+\sqrt x=(a-x)^2\qquad\text{and}\qquad a-x\gt 0 \\\\&\iff a^2+(-2x-1)a+x^2-\sqrt x=0\qquad\text{and}\qquad a-x\gt 0 \\\\&\iff a=\frac{2x+1\pm\sqrt{4x+4\sqrt x+1}}{2}\qquad\text{and}\qquad a-x\gt 0 \\\\&\iff a=\frac{2x+1\pm\sqrt{(2\sqrt x+1)^2}}{2}\qquad\text{and}\qquad a-x\gt 0 \\\\&\iff a=\frac{2x+1\pm (2\sqrt x+1)}{2}\qquad\text{and}\qquad a-x\gt 0 \\\\&\iff a=x+\sqrt x+1,x-\sqrt x\qquad\text{and}\qquad a-x\gt 0 \\\\&\iff a=x+\sqrt x+1\tag3\end{align}$$

Now, let us prove that $(1)$ has a solution if and only if $a\ge 1$.

Since we've already seen that $(1)$ has a solution if and only if $(3)$ has a solution, it is sufficient to prove that $(3)$ has a solution if and only if $a\ge 1$.

• If $(3)$ has a solution, then $a=x+\sqrt x+1\ge 1$.

• If $a\ge 1$, then $$\begin{align}(3)&\iff \bigg(\sqrt x+\frac 12\bigg)^2=a-1+\frac 14\\\\&\iff \underbrace{\sqrt x+\frac 12}_{\color{red}{\text{positive}}}=\color{red}+\sqrt{\underbrace{a-1}_{\ge\ 0}+\frac 14}\ \ \bigg(\ge\sqrt{\frac 14}=\frac 12\bigg)\end{align}$$ so $(3)$ has a solution.

It follows that $(3)$ has a solution if and only if $a\ge 1$.

Therefore, we can say that $(1)$ has a solution if and only if $a\ge 1$.

Since we get $\sqrt x=\sqrt{a-\frac 34}-\frac 12$, we finally obtain $$x=\bigg(\sqrt{a-\frac 34}-\frac 12\bigg)^2=\frac{2a-1-\sqrt{4a-3}}{2}$$

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Next, I'm going to show some thoughts about the following solution.

"We have $x+\sqrt x+1=a$. Solving this, we get $\sqrt x=\frac{-1\pm\sqrt{4a-3}}{2}$. Since we can exclude $\frac{-1-\sqrt{4a-3}}{2}$, we get $\sqrt x=\frac{-1+\sqrt{4a-3}}{2}$. A necessary and sufficient condition for this to be real and non-negative is $4a-3\ge 0$ and $\frac{-1+\sqrt{4a-3}}{2}\ge 0$, i.e. $a\ge 1$."

I first thought that this solution is incorrect. The reason is as follows :

$\sqrt x=\frac{-1+\sqrt{4a-3}}{2}$ is valid only when $a\ge 1$ (since LHS is real and non-negative). So, when one writes $\sqrt x=\frac{-1+\sqrt{4a-3}}{2}$, one implicitly assume $a\ge 1$ (If $a\lt 1$, then $\sqrt x=\frac{-1+\sqrt{4a-3}}{2}$ does not make sense).

Then, I received the following comment :

It looks to me that you are ignoring the fact that quadratic equation always has two complex solutions. One can observe a posteriori that out of these two solutions, there is a real nonnegative solution if and only if $a≥1$.

It seems that this comment suggests the following :

"We have $x+\sqrt x+1=a$. There are two complex solutions $\sqrt x=\frac{-1\pm\sqrt{4a-3}}{2}$. Since we can exclude $\frac{-1-\sqrt{4a-3}}{2}$, we get $\sqrt x=\frac{-1+\sqrt{4a-3}}{2}$. A necessary and sufficient condition for this to be real and non-negative is $4a-3\ge 0$ and $\frac{-1+\sqrt{4a-3}}{2}\ge 0$, i.e. $a\ge 1$."

This makes sense, so "incorrect" is perhaps too strong a word.

Some comments :

• Since the "solution" in "the equation has a solution if and only if $a\ge 1$" is real, I thought that we have to solve our problem only in real numbers.

• There is a short solution only in real numbers. For example, see the above solution of mine. So, our problem can be solved only in real numbers.

• I not only prefer a solution only in real numbers especially for our problem, but also believe that we are implicitly asked to prove the claim only in real numbers, but I accept the fact that the problem does not say so.

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A solution which tries to avoid using "squaring" :

Let $t:=\sqrt{a+\sqrt x}\gt 0$, so $t^2=a+\sqrt x$.

$$(1)\iff x+t=a$$

Eliminating $a$, we get $$x+t=t^2-\sqrt x\iff t^2-t-x-\sqrt x=0$$

Solving this for $t$ gives $$t=\frac{1\pm \sqrt{(2\sqrt x+1)^2}}{2}=\frac{1\pm (2\sqrt x+1)}{2}=1+\sqrt x,-\sqrt x$$ Since $t\gt 0$, we get $$(t=)\ \ a-x=1+\sqrt x\tag3$$

Note that if $(3)$ has a solution, then $(1)$ has a solution since $$\sqrt{a+\sqrt x}=\sqrt{1+2\sqrt x+x}=1+\sqrt x=a-x$$ The rest is the same as the solution written at the top.

Answer 7

Making $x = y^2$ we have

$$ (y^2-a)^2= y+a\Rightarrow y^4-2ay^2-y+a^2-a = (y^2-a-y)(y^2-a+y)+y^2-y-a =(y^2-y-a)(y^2+y-a+1) $$

now solving $y^2-y-a = 0$ we have $y = \frac{1}{2} \left(1\pm\sqrt{4 a+1}\right)\ge 0$ then solving $1\pm\sqrt{4 a+1}=s^2$ for $a$ we have

$$ a = \frac{1}{4} \left(s^2-2\right)s^2 $$

but $a$ should be non negative, so we rule out this possibility.

Taking now $y^2+y-a+1=0$ we have $y = \frac{1}{2} \left(-1\pm\sqrt{4 a-3}\right)$ then analogously we need $\left(-1\pm\sqrt{4 a-3}\right)=s^2$ that solved for $a$ gets

$$ a = \frac{1}{4} \left(4+(s^2 + 1)s^2\right) $$

or $a\ge 1$

Answer 8

As already proposed in the answer for the linked question, we have that by $x=(y^2-a)^2$ with $y^2-a\ge 0$ and $y\ge 0$

$$x+\sqrt{a+\sqrt{x}}=a \iff ( y^2 + y-a) ( y^2 - y + 1-a) = 0$$

which lead respectevely to

• $\color{red}{y=\frac{-1- \sqrt{4a+1}}{2}<0\;\text{(no solution)}}$
• $\color{red}{y=\frac{-1+ \sqrt{4a+1}}{2}\ge 0 \;\text{with}\;a\ge 0\implies y^2-a=\frac{1- \sqrt{4a+1}}{2}<0\;\text{(no solution)}}$
• $\color{red}{y=\frac{1- \sqrt{4a-3}}{2}\;\text{with}\;\frac34 \le a\le1\implies y^2-a=\frac{-1- \sqrt{4a-3}}{2}<0\;\text{(no solution)}}$

and finally

• $\color{green}{y=\frac{1+ \sqrt{4a-3}}{2}>0\;\text{with}\;a\ge \frac34 \implies y^2-a=\frac{-1+ \sqrt{4a-3}}{2}\ge 0\;\text{with}\;a\ge 1}$

which leads to the unique solution

$$\boxed{x=(y^2-a)^2=a-\frac{1+\sqrt{4a-3}}{2}}$$

for $a\ge 1$.

Answer 9

Let $\quad u=-\sqrt{a+\sqrt x}\le0,\quad v=\sqrt x\ge0,\quad$ then $$u^2-v=v^2-u=a,$$ $$\left(u+\frac12\right)^2 = \left(v+\frac12\right)^2=a+u+v+\frac14,\quad u\le 0,\quad v\ge 0,$$ $$u+\dfrac12=-\left(v+\frac12\right),\quad u+v=-1,$$ $$v^2+v+1-a=0,\quad v\ge0\Rightarrow\quad v=\sqrt{a-\frac34}-\frac12,\quad a\ge1,$$ $$\color{brown}{\mathbf{x=v^2=a-\frac12-\sqrt{a-\frac34},\quad a\ge1}.}$$

Answer 10

Here is a solution that comes (arguably) under the conditions of the question, namely "do not square", and while searching for solutions $x\ge 0$ of the given equation $$ f(x) = 0\ ,\qquad\ f:\Bbb R_{\ge 0}\to \Bbb R\ ,\ f(x) = x+\sqrt{a+\sqrt x} - a\ , $$ how to "eliminate excess roots", i.e. roots that would pop up after getting an equivalent polynomial equation in $x$ of higher degree after taking squares. Also, we want a formula for the solution. The idea is to use the Galois conjugated expressions, so we get some structure during the computations.

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• $f$ is continuous, strictly increasing, and $f(0)=\sqrt a-a$. The value $a=0$ was excluded. If $a\in(0,1)$, then $f(0)>0$, so there is no solution. Also, for $a=1$ we have only the solution $x=0$, $f$ being monotone.
• So we do assume from now on $a> 1$. Then $f(0)<0<f(a)$, so there exists a unique solution in $(0,a)$.
• Let $x$ be this solution for the equation $f=0$. We "do not square", but use the conjugate expressions $E_\pm\pm$, and $f=0$ is equivalent to $E_{-+}=0$, getting: $$ \begin{aligned} E_{++}(a,x) &= (a-x) + \sqrt{a+\sqrt x}\ ,\\ E_{-+}(a,x) &= (a-x) - \sqrt{a+\sqrt x}\ ,\\ E_{+-}(a,x) &= (a-x) + \sqrt{a-\sqrt x}\ ,\\ E_{--}(a,x) &= (a-x) - \sqrt{a-\sqrt x}\ ,\qquad\text{ and compute products:}\\[3mm] E_{++} E_{-+}\ (a,x) &= (a-x)^2 - (a+\sqrt x)\ ,\\ E_{+-} E_{--}\ (a,x) &= (a-x)^2 - (a-\sqrt x)\ ,\qquad\text{ and finally:}\\[3mm] \Pi E(a,x):= E_{++} E_{-+} E_{+-} E_{--}\ (a,x) &= \Big(\ (a-x)^2 - a\ \Big)^2 - x\ . \end{aligned} $$ Yes, equation to zero, this is also the true polynomial equation obtained when twice building squares, but there is a further plus in this "Galois-theoretic" view.

Observe that if we introduce correspondingly instead of $a$ values $A_\pm$ given by $$\bbox[yellow]{A_\pm=x\pm\sqrt x} $$ we obtain: $$ \begin{aligned} E_{\pm-}(A_+,x) &= (A_+-x) \pm \sqrt{A_+-\sqrt x}\\ &=((x+\sqrt x)-x)\pm \sqrt{(x+\sqrt x)-\sqrt x}=\sqrt x\pm\sqrt x\ ,\\ E_{\pm+}(A_-,x) &= (A_--x) \pm \sqrt{A_-+\sqrt x}\\ &=((x-\sqrt x)-x)\pm \sqrt{(x-\sqrt x)+\sqrt x}=-\sqrt x\pm\sqrt x\ ,\\ \end{aligned} $$ so two factors above are vanishing. I hope this observation is a fair one, this is the key point in the solution. Without this point, the solution is missing the point. The final product $\prod E_{\pm\pm}$ is a polynomial, so we expect in this product the polynomial factor $$ (a-A_+)(a-A_-)=a^2-(A_++A_-)a+A_+A_-=a^2-2xa+(x^2-x)=((a-x)^2-x)\ . $$ And of course, if we know this factor, we can force it: $$ \begin{aligned} \Pi E(a,x) &= \Big(\ (a-x)^2 - a\ \Big)^2 - x \\ &=(a-x)^4\color{gray}{-x(a-x)^2+x(a-x)^2}-2a(a-x)^2+a^2 - x \\ &=((a-x)^2-x)\Big((a-x)^2-(2a-x)\Big) \color{gray}{-x(2a-x)}+a^2 - x \\ &=((a-x)^2-x)\color{blue}{\Big((a-x)^2-(2a-x)+1\Big) } \ . \end{aligned} $$ So the given equation $f=E_{-+}=0$ is equivalent to the vanishing of the blue polynomial, $0= x^2-(2a-1)x+(a-1)^2$. Among the two solutions $$ \bbox[lightblue]{\frac 12\Big((2a-1)-\sqrt{4a-3}\Big)}\ ,\qquad \frac 12\Big((2a-1)+\sqrt{4a-3}\Big)>\frac 12\Big((2a-1)+\sqrt{4-3}\Big)=a\ , $$ we have to refuse the last one, since it lives outside $(0,a)$. So the one in the light blue color remains.

$\square$

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Note: The exposition would have been shorter, if allowed to observe that plugging in instead of $a$ the values $$\bbox[yellow]{B_\pm=x\pm\sqrt x+1} $$ we obtain annihilation in two cases, and then consider only these cases: $$ \begin{aligned} E_{\pm+}(B_+,x) &= (B_+-x) \pm \sqrt{B_+ + \sqrt x}\\ &=((x+\sqrt x+1)-x)\pm \sqrt{(x+\sqrt x+1)+\sqrt x}=\sqrt x+1\pm\sqrt {(\sqrt x +1)^2}\ ,\\ E_{\pm-}(B_-,x) &= (B_--x) \pm \sqrt{B_- - \sqrt x}\\ &=((x-\sqrt x+1)-x)\pm \sqrt{(x-\sqrt x+1)-\sqrt x}=-\sqrt x+1\pm\sqrt {(\sqrt x -1)^2}\ . \end{aligned} $$