Find all real solutions to the equation $ \frac{\sqrt{2}}{(x + 1)^2} - \frac{x}{\sqrt{x^2 + 1}}\cdot \frac{1}{x^2 + 1} = 0 $

Question

$ \displaystyle \frac{\sqrt{2}}{(x + 1)^2} - \frac{x}{\sqrt{x^2 + 1}}\cdot \frac{1}{x^2 + 1} = 0 $

If you solve it head-on, it's a $9$-degree equation, so I tried substituting

$ x = \tan a $

$\displaystyle \frac{\sqrt{2}}{(\tan a + 1)^2} - \frac{\tan a }{\sqrt{\tan ^2a + 1}}\cdot \frac{1}{\tan^2 a + 1} = 0 $

After some transformations, I got the equation

$ \displaystyle \frac{\sqrt{2} \cos^2a }{1 + 2 \sin a \cos a } - \sin a \cos ^2a = 0 $

It remains to solve the equation

$\sqrt{2} - \sin a - \sin 2a \sin a = 0 $

After which I was stacked, can anybody help to proceed, or suggest more simply approach ?

Answer 1

The starting equation:

$$ \frac{\sqrt{2}}{(x+1)^2}-\frac{x}{\sqrt{(x^2+1)^3}}=0 $$

can be transformed into this other equation:

$$ \frac{(x-1)(x^5-3x^4-3x^3-7x^2-2x-2)}{(x+1)^4(x^2+1)^3}=0 $$

which is verified for:

$$ \begin{aligned} &x_1=1\\ &x_2=4.16008\\ &x_3=-0.49156 - 1.01653\,i\\ &x_4=-0.49156 + 1.01653\,i\\ &x_5=-0.0884779 - 0.60766\,i\\ &x_6=-0.0884779 + 0.60766\,i\\ \end{aligned} $$

but only the first four verify the starting equation.

Here you can find an automated way to solve polynomial equations.

Answer 2

One has $$\frac{\sqrt2}{(x+1)^2}-\frac{x}{(x^2+1)^{3/2}}=0$$ Trying separately $f(x)=\dfrac{x}{(x^2+1)^{3/2}}$ and $g(x)=\dfrac{\sqrt2}{(x+1)^2}$ we have $$f(x)=\dfrac{x}{(x^2+1)^{3/2}}\text { and}\space f'(x)=\frac{3x^2}{(x^2+1)^{5/2}}-\frac{1}{(x^2+1)^{3/2}}$$ $$g(x)=\frac{\sqrt2}{(x+1)^2}\text { and}\space g'(x)=\frac{-2\sqrt2}{(x+1)^3}$$ this yields to:

$(1)$ $f(x)$ has a maximum at $x_0\approx0.7071$ and is strictly decreasing on $x\gt x_0$ until the limit $0$ at infinity.

$(2)$ $g(x)$ is strictly decreasing for $x$ real until the limit $0$ at infinity.

It follows that there are only two real roots wich are the almost evident $x=1$ and other whose approximate calculation is $x\approx 4.16$

Answer 3

This is just a comment, but I cannot write comments here. I have some ideas on it:

1. You can try the substitution $x=\frac{z-1/z}2$ which has the nice property that $\sqrt{x^2+1}=\frac{z+1/z}2$.
2. The answer of πρόσεχε above shows that the roots are 1 and the roots of the polynomial $x^5−3x^4−3x^3−7x^2−2x−2$.
3. The roots of this polynomial are maybe not solvable in terms of radicals. If this is true for the other real root 4.16..., then it is not true for the sum of the real roots. Do you need the sum of the real roots or the sum of all roots?
4. May it be enough for your problem to give a bound for the root?