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

Question

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.

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

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

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

Answer 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.

Answer 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

Answer 4

This is a very intriguing problem. First of all, $x\geq 0$ and $a\geq\sqrt{a+x}$ must hold for real $x$.

Squaring both sides of the equation we get $x^2=a-\sqrt{a+x}$.

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

$\Rightarrow x^4-2ax^2-x+a^2-a=0$

$\Rightarrow (x^2-x-a)(x^2+x+(1-a))=0$

$\therefore x= \frac12\pm\sqrt{a+\frac14},-\frac12\pm\sqrt{a-\frac34}$

All four of these solutions may not work because some of them could be extraneous roots due to squaring. Two of the solutions give negative values for $x$, so it must be

$x= \frac12+\sqrt{a+\frac14},-\frac12+\sqrt{a-\frac34}$

But the 1st one again does not satisfy the given equation.

Hence the unique solution to the equation is $x=-\frac12+\sqrt{a-\frac34}$, and that solution only exists if $a\geq 1$.

Answer 5

If you square both sides and adjust, you get $a-x^2=\sqrt{x+a}$. The graphs of $y=a-x^2$ and $y=\sqrt{a+x}$ are congruent parabolas. One has vertex at $y=a$ with arms pointing down and the other has vertex at $x=-a$ with arms pointing to the right, well, the top half of that one. There are two points of intersection. The one in the first quadrant is the one you want. But by symmetry, it's easy to find the intersection point in the second quadrant, since it must lie on the line $y=-x$. So solve $-x = a-x^2$ to get $x= (1\pm \sqrt{1+4a})/2$. The $x$ value in that quadrant is negative so choose "minus" in the $\pm$.

So this is one solution to the quartic. You should be able to leverage that into a factorization of the quartic, and then you can find the intersection point in the first quadrant.

Answer 6

We certainly have $x>0$. Then by squaring, the original equation can be written $$a-x^2=\sqrt{x+a}.$$ The two members describe the parabolas of equations

$$\begin{cases}y=a-x^2,\\x=y^2-a\end{cases}$$ where $y>0$.

Adding these two equalities we get a pair of straight lines

$$y+x=y^2-x^2,$$ i.e. $$(y+x)(y-x-1)=0.$$

Hence, the requested solution lies in the first quadrant and is the positive root of

$$\begin{cases}y-x-1=0\\y=a-x^2\end{cases},$$ $$x^2+x+1-a=0,$$

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

Answer 7

Raise the equation to the second power and rearrange to get $$-\sqrt{x+a}=x^2-a$$ Now raise the equation to the second power again, expand and rearrange again to get $$-x^4+2ax^2+x-a^2+a=0$$ which can be factored into $$-(-x^2-x+a-1)(-x^2+x+a)=0$$ I think you can finish on your own. Don't forget to check for extraneous solutions we've obtained by squaring.