Solve for $x $(quadratic)

Question

$$ 0=0.001 + \frac{-0.0018 x+0.009 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$

Can't seem to figure out a way how to.

Answer 1

$$ 0=0.001 + \frac{-0.0018 x+0.009 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$

$$ 0=1 + \frac{-1.8 x+9 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$

$$ 1 =\frac{1.8 x-9 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$

$$ \left(\sqrt{0.04 - x^2}\right)^3=1.8 x-9 x^2$$

$$ \left(0.04 - x^2 \right) \left(\sqrt{0.04 - x^2}\right)=1.8 x-9 x^2$$

$$ (0.04 - x^2 )^2 (0.04 - x^2)=(1.8x-x^2)^2$$

$$ (0.04 - x^2 )^3=(1.8 x-9 x^2)^2$$

$$ \{(0.2 - x)(0.2+x)\}^3 =81x^2(0.2 - x)^2$$

$$ \{(0.2 - x)(0.2+x)\}^3 -81x^2(0.2 - x)^2=0$$

Put $t=0.2-x$ and $0.2=a$

$$ t^2\cdot \{t(2a-t)^3 -81(a-t)^2\}=0$$

$$t=0 \Rightarrow x=0.2$$ But $x \neq 0.2$

So $$t(2a-t)^3 -81(a-t)^2=0$$

You will have to solve for $t$ now.

Answer 2

If you square to remove the radical, you will end up with a high degree polynomial. You are basically stuck with a numeric solution. If you plot the function as below you see there is a root near $x=0.0045$ There are many numeric root finding algorithms described in any numerical analysis text. The simplest to describe is bisection. You note the sign changes between $0.004$ and $0.005$ so there is a root in $[0.004, 0.005]$. Evaluate the function at the midpoint, $0.0045$. You find it is positive, so the root is in $[0.0045, 0.005]$ Continue until the interval is small enough.

Answer 3

Starting from where Angelo Mark finished, expanding the polynomial in $t$, using whole numbers, we end with $$125 t^4-150 t^3+10185 t^2-4058 t+405=0$$ and we are looking for the first zero of this equation (multiple squaring introduce extra solutions).

One of the most simple method is probably Newton which, starting from a guess $t_0$, will update it according to $$t_{n+1}=t_n-\frac{f(t_n)}{f'(t_n)}$$ Looking at the plot Angelo Mark gave, let us start iterating at $t_0=0.25$ (it is "far" from solution but this will show you the path to solution).

So, $$f(t)=125 t^4-150 t^3+10185 t^2-4058 t+405$$ $$f'(t)=500 t^3-450 t^2+20370 t-4058$$ So, the successive iterates will be $$t_1=0.225145590682196$$ $$t_2=0.212914221738832$$ $$t_3=0.207166952169403$$ $$t_4=0.204894106187593$$ $$t_5=0.204376815224264$$ $$t_6=0.204346921893618$$ $$t_7=0.204346821393703$$ $$t_8=0.204346821392567$$ which is the solution for fixteen significant figures.

For sure, starting closer would have required less iterations for the same accuracy.

Still looking at Angelo Mark's plot, we could have obtained an approximate solution considering that around $t=\frac 15$, the curve looks like a parabola. Expanding $f(t)$ as a Taylor series around this point, we would get $$f(t)=-\frac{1}{5}+2 \left(t-\frac{1}{5}\right)+10125 \left(t-\frac{1}{5}\right)^2+O\left(\left(t-\frac{1}{5}\right)^3\right)$$ Solving the quadratic would give two roots $$t_{\pm}=\frac{2024\pm\sqrt{2026}}{10125}$$ the value of which being $\approx 0.195456$ and $\approx 0.204347$ which are the values reported by Wolfram Alpha.

So, in a sense, the initial equation is "almost" a quadratic !