I double dare you to expand.

Let $y=x+1$ . $x=y-1$ . Substituting and expanding, we get

$y^2+\frac{1}{y^2}-2(y+\frac{1}{y})+1=0$ .

Let $z=y+\frac{1}{y}$ .

$y^2+\frac{1}{y^2}=z^2-2$ .

Therefore, $z^2-2z-1=0$ .

Solving the quadratic equation, we get all possibilities of $z$ . By using the equation $z=y+\frac{1}{y}$ , we can find all possibilities of $y$ . The value of $x$ which satisfy the conditions in the problem can be determined by the equation $x=y-1$ .

Let $x=cos(\theta)$ and $\frac{x}{x+1}=sin(\theta)$ , hence we get our eqaution as :

$\frac{cos(\theta)}{cos(\theta)+1}=sin(\theta)$

Rewriting it we get :

$cos(\theta)-sin(\theta)=cos(\theta)sin(\theta)$

Again rewriting it we get :

$\sqrt{2}sin(\frac{\pi}{4}-\theta )=\frac{1}{2}cos(\frac{\pi}{2}-2\theta )$

Put $y=\frac{\pi}{4}-\theta$ to get the equation as :

$2\sqrt{2}sin(y)=cos(2y)$

Using double angle identities and rewriting the equation we get :

$2{sin}^{2}(y)+2\sqrt{2}sin(y)-1=0$

Solving the quadratic we get :

$sin(y)=1-\frac{1}{\sqrt{2}}$ and $cos(y)=\sqrt{\sqrt{2}-\frac{1}{2}}$

Now since $\theta=\frac{\pi}{4}-y$

hence $cos(\theta)=cos(\frac{\pi}{4}-y)=\frac{1}{\sqrt{2}}(cos(y)+sin(y))$

Put the values to get :

$\large x=cos(\theta)=\frac{1-\frac{1}{\sqrt{2}}+\sqrt{\sqrt{2}-\frac{1}{2}}}{\sqrt{2}}$

My professor taught me this technique when you have no other simple ways to do it, other than expanding.

Let $\displaystyle y = \frac{x}{x+1}$ .

The equation becomes $x^{2} + y^{2} = 1$ _(1)

and $xy = x-y$ . _(2)

Let $\large x, -y$ be the roots of this equation

$z^{2} + pz + q = 0$ _(#)

From Vieta's formula, $x+(-y) = -p$ and $x(-y) = q$ .

From (2), $-p = x-y = xy = -q$

Therefore, $p = q$ .

From (1), $x^{2} + y^{2} = (x-y)^{2} + 2xy$

$1 = (-p)^{2}+2(-q)$

$p^{2}-2p-1 = 0$

$p = 1 \pm \sqrt{2}$

Now the equation (#) becomes

$z^{2} + (1+\sqrt{2})z + (1+\sqrt{2}) = 0$ or $z^{2} + (1-\sqrt{2})z + (1-\sqrt{2}) = 0$

The first equation gives you 2 complex roots and the second gives you 2 irrational roots. So don't bother factorize the question lol.

Solve both equations we get

$\boxed{\displaystyle z = \frac{-(1+\sqrt{2})\pm i\sqrt{1+2\sqrt{2}}}{2}, \frac{2+\sqrt{2}\pm \sqrt{1+2\sqrt{2}}}{2}}$ .

And all of these roots $z$ are the roots of $x$ .

Which is approximately $x = -0.46899, 0.88320, -1.2071\pm 0.9783i$ .

The given equation can be re-written as $\left(\dfrac{1}{x}\right)^{-2}+\left(1+\dfrac{1}{x}\right)^{-2}=1$ Make the substitution $\dfrac{1}{x}=a$ $a^{-2}+(1+a)^{-2}=1$ $\Rightarrow \dfrac{1}{a^2}+\dfrac{1}{(a+1)^2}=1$ $\Rightarrow \dfrac{a^2+2a+1+a^2}{(a^2+a)^2}=1$ $\Rightarrow \dfrac{2(a^2+a)+1}{(a^2+a)^2}=1$ Make the subsitution $a^2+a=y$ $\dfrac{2y+1}{y^2}=1$ $y^2-2y-1=0$ $y^2-2y+1=2$ $(y-1)^2=2$ $y=1\pm\sqrt{2}$ substituting back to $a$ $a^2+a-(1\pm\sqrt{2})=0$ Solving the above quadratic we get real roots with only $1+\sqrt{2}$ $a=\dfrac{\sqrt{5+4\sqrt{2}}-1}{2}, \dfrac{-(\sqrt{5+4\sqrt{2}}+1)}{2}$ The second one is a negative, so we consider only the first root Substituting back to x, we get $x=\dfrac{2}{\sqrt{5+4\sqrt{2}}-1}$ Plugging the value of $x$ gives the desired result.

Using The Newton-Raphson Method $x^{4}+2x^{3}+x^{2}-2x-1=0$ Let $f(x)=x^{4}+2x^{3}+x^{2}-2x-1$ $f'(x)=4x^{3}+6x^{2}+2x^-2$ Therefore, $x_{n+1}=x_{n}-\frac{x_{n}^{4}+2x_{n}^{3}+x_{n}^{2}-2x_{n}-1}{4x_{n}^{3}+6x_{n}^{2}+2x_{n}-2}$ Substitute $x_{1}$ equal to any number which is more than 1 and it will converge to the real positive solution

$x^{2} +y^{2}=(x-y)^{2}+2xy$

so,

$x^{2}+(x/(x+1))^{2}=1$

becomes,

$(x^{2}/(x+1))^{2}+(2x^{2}/(x+1))-1=0$

let

$z=(x^{2})/(x+1)$

Then solve the quadratic equationS to get the answer!

Rewrite the equation as

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

Iterating this converges quickly to the positive root.

You can implement this quickly on a calculator by first entering a starting value by pressing (say) 0.5=, then entering

$\sqrt{1-(\frac{Ans}{Ans+1})^2}$

and pressing equals repeatedly.

Complete the square using minus instead of plus. It's counterintuitive, but the equation will become really beautiful. Call t=x^2/(x+1).