Solve the equation $\sqrt{x+3}+\sqrt{x^2+2x+7}-\sqrt{x^2+3}=(x+1)^2$ [closed]

$\sqrt{x+3}+\sqrt{x^2+2x+7}-\sqrt{x^2+3}=(x+1)^2$

The equation I'm trying to solve is $x_1=-2$. But the equation has 2 roots. Can any one find $x_2$?

Thanks!

Solution 1:

Almost everything was said in comments.

If you are allowed to plot the function $$f(x)=\sqrt{x+3}+\sqrt{x^2+2x+7}-\sqrt{x^2+3}-(x+1)^2$$ you notice that is looks like a parabola.

So, as a try, make a series expansion around $x=-2$ to obtain $$f(x)=\left(\frac{5}{2}+\frac{1}{\sqrt{7}}\right) (x+2)+\left(\frac{3}{14 \sqrt{7}}-\frac{9}{8}\right) (x+2)^2+O\left((x+2)^3\right)$$

$$\frac{f(x)}{x+2}=\left(\frac{5}{2}+\frac{1}{\sqrt{7}}\right) +\left(\frac{3}{14 \sqrt{7}}-\frac{9}{8}\right) (x+2)+O\left((x+2)^2\right)$$

Using the above, an approximation $$x=\frac{2 \left(49+40 \sqrt{7}\right)}{3 \left(147-4 \sqrt{7}\right)}=\frac{2 \left(1189+868 \sqrt{7}\right)}{9213}=0.756651$$ Repeat the expansion around $x=\frac 34$ $$f(x)=\frac{1}{16} \left(-49+8 \sqrt{15}-4 \sqrt{57}+4 \sqrt{145}\right)+$$ $$\left(-\frac{7}{2}-\sqrt{\frac{3}{19}}+\frac{1}{\sqrt{15}}+ \frac{7}{\sqrt{145}}\right) \left(x-\frac{3}{4}\right)+O\left(\left(x-\frac{3}{4}\right)^2\right)$$ from which $$x =\frac 34+\frac{8265 \left(-49+8 \sqrt{15}-4 \sqrt{57}+4 \sqrt{145}\right)}{8 \left(57855-1102 \sqrt{15}+870 \sqrt{57}-798 \sqrt{145}\right)}=0.748997$$ while the solution obtained by Newton method is $x=0.748996$.

Notice that this simplistic process is very close to the one Newton used to invent Newton method.