Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$

Question

I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.

Answer 1

$$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$ substitute $x=\sqrt{y}$ $$\sqrt y\left(1+\sqrt{1-y}\right)=\sqrt{1-y}$$ divide both sides by $\sqrt{y}$ $$1+\sqrt{1-y}=\sqrt{y^{-1}-1}$$ square both sides $$2-y+2\sqrt{1-y}=y^{-1}-1$$ isolate the square root $$2\sqrt{1-y}=y+y^{-1}-3$$ square both sides $$4-4y=y^2+y^{-2}-6y-6y^{-1}+11$$ multiply both sides by $y^2$ $$4y^2-4y^3=y^4-6y^3+11y^2-6y+1$$ subtract the left hand side from the right $$y^4-2y^3+7y^2-6y+1=0$$ Factoring $y^4-2y^3+7y^2-6y$ we get $$y^4-2y^3+7y^2-6y=y(y^3-2y^2+7y-6)=y(y-1)(y^2-y+6)=(y^2-y)(y^2-y+6)$$ Substitute $y^2-y=z$ $$y^4-2y^3+7y^2-6y=z^2+6z$$ Plugging it back in we get $$z^2+6z+1=0$$ $$(z+3)^2-8=0$$ $$z=-3±\sqrt{8}$$ Undoing the substitution we get $$y^2-y+3±\sqrt{8}=0$$ $$y=\frac{-(-1)±\sqrt{(-1)^2-4(1)(3±\sqrt{8})}}{2(1)}=\frac{1±\sqrt{-11±4\sqrt{8}}}{2}$$ Knowing $x=\sqrt{y}$ we get the following $$x=\sqrt\frac{1±\sqrt{-11±4\sqrt{8}}}{2}$$ Three of these solutions are extraneous

Answer 2

Observe $\sqrt {1-x^2}≥0$ implies that, $x≥0.$

Thus, under the restriction $0≤x≤\frac {\sqrt 2}{2}$ we have:

$$\begin{align}x\sqrt{1-x^2}&=\sqrt{1-x^2}-x\\ \left(x\sqrt {1-x^2}\right)^2&=1-2x\sqrt {1-x^2}\end{align}$$

Letting $x\sqrt {1-x^2}=u$, we have:

$$\begin{align}&u^2=1-2u\\ \implies &u^2+2u-1=0\end{align}$$

This leads to:

$$\begin{align}&u=\sqrt{1-x^2}-x\\ \implies &ux=u-x^2,\,x≠0\\ \implies &x^2+ux-u=0.\end{align}$$

Finally, note that all possible roots of the original equation must satisfy the inequality $0≤x≤\frac {\sqrt 2}{2}.$

Answer 3

Another variation.

We observe that, $\sqrt {1-x^2}≥0$ and $1+\sqrt {1-x^2}>0$ imply $x≥0$.

Letting $\sqrt {1-x^2}=u,~u≥0$ we have:

$$\begin{cases} ux=u-x\\u^2+x^2=1\end{cases}$$

Then setting $u=a+b,~ x=a-b$ under the restriction $a≥b≥0$, we have:

$$\begin{align}&\begin{cases}a^2-b^2=2b\\a^2+b^2=\frac 12\end{cases}\\ \implies &4b^2+4b-1=0\\ \implies &x=\frac{\sqrt {4b+1}}{2}-b.\end{align}$$

This completes the answer.

Answer 4

$$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$

$$x^2\left(1+2\sqrt {1-x^2}+1-x^2\right)=1-x^2$$

$$x^2\left(2+2\sqrt {1-x^2}-x^2\right)=1-x^2$$ Substitute $x^2=y$,

$$y\left(2+2\sqrt {1-y}-y\right)=1-y$$

$$2y+2y\sqrt {1-y}-y^2=1-y$$

$$2y\sqrt {1-y}=1-y+y^2-2y$$

$$4y^2(1-y)=(y^2-3y+1)^2$$

$$y^4-2y^3+7y^2-6y+1=0$$

$$(y^2-y+3)^2=8$$

$$y^2-y+3=\pm 2\sqrt 2$$

$$y = \frac {1}{2}\bigg(1 \pm \sqrt{8 \sqrt 2 - 11}\bigg)$$

$$x=\pm\sqrt {\frac {1}{2}\bigg(1 \pm \sqrt{8 \sqrt 2 - 11}\bigg)}$$

Hope this helps.

Answer 5

Let us make the substitution $y := \sqrt{1-x^2}$; then we get the system of polynomial equations \begin{align*} x^2 + y^2 & = 1, \\ x(1+y) & = y. \end{align*} From here, if we solve the second equation for $y$, $y = \frac{x}{1-x}$, and substitute into the first equation, it is straightforward to get the polynomial equation $$x^4 - 2x^3 + x^2 + 2x - 1 = 0.$$ However, let us go back to the original system of two equations in $(x, y)$. Here, the first equation is the unit circle, and the second equation is equivalent to $(x-1)(y+1) = -1$, so it is a hyperbola with center $(1, -1)$. We see that both curves are symmetric about the line $y = -x$; therefore, if $(x, y)$ is a solution to the system of equations, then so is $(-y, -x)$. This implies that if $\alpha$ is a root of the quartic polynomial, then so is $-\frac{\alpha}{1 - \alpha} = \frac{\alpha}{\alpha - 1}$.

From here, we can hypothesize that if $K$ is the splitting field of the quartic over $\mathbb{Q}$, then there might be an element $\sigma \in \operatorname{Gal}(K / \mathbb{Q})$ which permutes the roots according to $\alpha \mapsto \frac{\alpha}{\alpha - 1}$. If this is the case, then it is easy to see that $\sigma^2 = \operatorname{id}$ since it fixes each root of the quartic; however $\sigma \ne \operatorname{id}$ since the only fixed point of $\alpha \mapsto \frac{\alpha}{\alpha - 1}$ is 2, and 2 is not a root of the quartic. Then setting $L$ to be the fixed field of $\sigma$, we will have $\operatorname{Tr}^K_L(\alpha) = \alpha + \frac{\alpha}{\alpha - 1} = \frac{\alpha^2}{\alpha - 1}$. This suggests searching for a polynomial which will have $\frac{\alpha^2}{\alpha - 1}$ as a root, and hopefully finding the roots of this polynomial will be simpler because $L$ has smaller degree over $\mathbb{Q}$.

However, from this point, we fairly quickly see that $$\left( \frac{\alpha^2}{\alpha-1} \right)^2 - 2 \left( \frac{\alpha^2}{\alpha-1} \right) - 1 = \frac{\alpha^4 - 2\alpha^2(\alpha-1) - (\alpha-1)^2}{(\alpha-1)^2} = \\ \frac{\alpha^4 - 2\alpha^3 + \alpha^2 + 2\alpha - 1}{(\alpha-1)^2} = 0.$$

But then, that means that $\frac{\alpha^2}{\alpha-1} = 1 \pm \sqrt{2}$. Another application of the quadratic formula to the equation $\alpha^2 - (1 \pm \sqrt{2}) \alpha + (1\pm\sqrt{2}) = 0$ gives the roots of the quartic. (Then from here, two of the roots of the quartic will be complex, and one of the real roots will not satisfy the original equation because it does not satisfy $\sqrt{1-x^2} = \frac{x}{1-x} \ge 0$.)

Answer 6

I wanted to provide the trigonometric approach as a $3$rd answer, thinking it might be useful to some readers.

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Substitute $x=\sin\theta$, where $0\leqslant \theta\leqslant \frac {\pi}{2}$ based on the fact $x\geqslant 0$, then we have:

$$\begin{align}&\sin\theta+\sin \theta\cos \theta=\cos \theta\\ \implies &\cos \theta-\sin \theta=\sin \theta\cos \theta\\ \implies &1-\sin (2\theta)=\frac{\sin ^2 (2\theta)}{4}\end{align}$$

Since $\sin (2\theta)\geqslant 0$ we obtain:

$$\begin{align}&\sin^2(2\theta)+4\sin (2\theta)-4=0\\ \implies &\sin (2\theta)=2\sqrt 2-2\\ \implies &\theta=\frac 12\arcsin \left(2\sqrt 2-2\right)\end{align}$$

Finally, the answer is:

$$\bbox[5px,border:2px solid #C0A000]{x=\sin\left(\frac 12\arcsin \left(2\sqrt 2-2\right)\right)}$$

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Important note:

The final answer assume that, for $0\leqslant \theta\leqslant \frac {\pi}{2}$ the inequality $\cos\theta\geqslant \sin\theta$ holds to avoid extraneous roots. No other root satisfies the inequality $\cos\theta\geqslant \sin\theta$, under the restriction $0\leqslant \theta\leqslant \frac {\pi}{2}.$