Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$

Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$

We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of! Please help.

Solution 1:

Note that $0 \le x\le a(a-1)$,

\begin{align*} x &= \sqrt{a-\sqrt{a+x}} \\ x^2 &= a-\sqrt{a+x} \\ a+x &= (a-x^2)^2 \\ a+x &= a^2-2ax^2+x^4 \\ x^4-2ax^2-x+a^2-a &= 0 \\ (x^2-x-a)(x^2+x-a+1) &= 0 \\ \end{align*}

$$x=\frac{1 \pm \sqrt{4a+1}}{2} \quad \text{or} \quad \frac{-1 \pm \sqrt{4a-3}}{2}$$

For $a\ge 1$,

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

Alternative solution from problem $168$ in The USSR Olympiad Problem Book

Let $y=\sqrt{a+x}$, then $x=\sqrt{a-y}$

\begin{align*} a+x &= y^2 \\ a-y &= x^2 \\ x+y &= y^2-x^2 \\ x^2-y^2+x+y &= 0 \\ (x+y)(x-y+1) &= 0 \\ (x+\sqrt{a+x})(x-\sqrt{a+x}+1) &= 0 \end{align*}

Rejecting $x+\sqrt{a+x}=0$,

$$x^2+x-a+1=0$$

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

Solution 2:

The quadratic we get is $x^4-2ax^2-x+a^2-a$. We can factor it by trying to guess the coefficients (this will depend heavily on the equation and how good-written the problem is): Let's assume all coefficients will be integers. Let $p(x)=x^2-2ax^2+a^2-a$. There are two cases:

If we could factor $p(x)$ as a product of a polynomial of degree $3$ and one of degree $1$, we would be finding a root of $p(x)$. Roots of polynomials must divide the constant term, which is $a^2-a=a(a-1)$. The only "obvious" divisors of it are $a$ and $a-1$, which are not roots of $p(x)$, so this is not a case of interest.

Let's try to factor $p(x)$ as a product of two quadratics. The coefficient of $x^4$ is $1$, so the coefficients of $x^2$ in the quadratics must be either both $1$ or both $-1$. Up to sign, assume they are $1$: $$x^4-2ax^2-x+a^2-a=(x^2+b x+c)(x^2+d x+e).$$ Now, when we do the product on the right-hand side, we get a term of the form $(b+d)x^3$, so we must have $b=-d$, i.e., we are looking for a decomposition of the form $$x^4-2ax^2-x+a^2-a=(x^2+bx+c)(x^2-bx+e)$$ The constant term on the right-hand side is $ce$, which should be equal to $a^2-a=a(a-1)$. Again, since we are assuming the decomposition is "simple", we need to is decompose $a(a-1)$ as a product of two terms. There are a few options: $$a(a-1)=1\cdot(a(a-1))=(-a)(1-a)$$ Then we try $c$ to be one of the terms, $d$ to be the other one, expand the right-hand side with this option, which will give us a polynomial of degree 4, with the coefficients multiplying $x^2$ and $x$ depending on $b$. By equaling them to the respective coefficients on the left-hand side, we get a linear equation on $b$ and a quadratic equation on $b$, with coefficients depending on $a$. For the first two options, the solutions ($b$ in terms of $a$) of the linear equation is not a solution of the quadratic, so these don't work.

So let's try the third option: \begin{align*} x^4-2ax^2-x+a^2-a&=(x^2+bx-a)(x^2-bx+1-a)\\ &=x^4+(b-b)x^3+x^2-ax^2-b^2x^2-ax^2+bx-abx+abx+a^2-a\\ &=x^4+(1-b^2-2a)x^2+bx+a^2-a, \end{align*} and $b=-1$ works. All of this means that $$x^4-2ax^2-x+a^2-a=(x^2-x-a)(x^2+x+1-a)$$

The solutions of these are $$x=\frac{1\pm\sqrt{4a+1}}{2}\qquad\text{ and }x=\frac{-1\pm\sqrt{4a-3}}{2}$$

Since $x$ is a square root, it is non-negative. Since $a\geq 1$, the signs before the square roots should be $+$. Now we need to see if these are really solutions. You can try a few values of $a$, (for example, $a=6$ gives $x=3$ and the equation does not hold), just to have some ideas.

On one hand, $x=\sqrt{a-\sqrt{a+x}}<\sqrt{a}$, but $\frac{1+\sqrt{4a+1}}{2}\geq\sqrt{a}$, so this solution does not work.

(Interestingly, however, you can check that $x=\frac{1+\sqrt{4a+1}}{2}$ is a solution for $\sqrt{a+x}=x$ and for $\sqrt{a+\sqrt{a+x}}=\sqrt{a+x}=x$)

The only remaining possibility is $x=\frac{-1+\sqrt{4a-3}}{2}$. It is easy to check, in this case, that $x^2=a-(x+1)$, and since the left-hand side is non-negative the right-hand side is also non-negative, and since $x\geq 0$ it follows that $$x=\sqrt{a-(x+1)}$$ Now note, similarly, that $(x+1)^2=a+x$, and therefore, again because both $x+1$ and $a+x$ are non-negative, we have $(x+1)=\sqrt{a+x}$, from which we conclude that $x=\sqrt{a-\sqrt{a+x}}$, as we wanted, and therefore $x=\frac{-1+\sqrt{4a-3}}{2}$ is the only solution of the initial equation.

Solution 3:

Only for the existence of the solution.

We have $x \ge 0$ and $x \le a^2 -a$ for the radicals to exists.

Let $f(x)=x-\sqrt{a-\sqrt{a+x}}, x \in [0, a^2 -a]$ Then $f'(x) > 0$ so $f$ is strictly increasing. Because $f(0) \lt 0$ and $f(a^2 -a)=a^2 -a \ge 0$, it follows that there is a unique solution