Solve 0.2 * 0.2 * x^6 + (1 + 0.2 * 0.2) x^4 + x^2 - 4 = 0.

Step-by-step solution

1. 1

   Simplify the decimal coefficients. Since $0.2\times 0.2=0.04$ and $1+0.04=1.04$, the equation is

   $0.04x^{6}+1.04x^{4}+x^{2}-4=0$

2. 2

   Clear the decimals by multiplying by 25. Because $25\times 0.04=1$ and $25\times 1.04=26$:

   $x^{6}+26x^{4}+25x^{2}-100=0$

   Working with integers removes any risk of rounding drift in the root-finding to come.

3. 3

   Substitute $y=x^{2}$ — only even powers appear. With $x^{6}=y^{3}$ and $x^{4}=y^{2}$:

   $y^{3}+26y^{2}+25y-100=0$

   The degree drops from six to three, and crucially $y=x^{2}\ge 0$, so only non-negative roots $y$ can produce real $x$.

4. 4

   Check for rational roots, then go numerical. The candidates $\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm25,\pm50,\pm100$ all fail — for example $y=1$ gives $-48$ and $y=2$ gives $+62$. So the sign change between $1$ and $2$ locates the only positive root, which must be found numerically.

5. 5

   Find the positive root accurately. Newton iteration on $f(y)=y^{3}+26y^{2}+25y-100$ starting from $y=1.5$ converges to

   $y=1.5056857750$

   A verification matters here: $f(1.5056858)=0.0000000$, whereas a nearby wrong value such as $y=1.417075$ gives $f=-9.5$, nowhere near zero. The other two roots, $y\approx-2.6747$ and $y\approx-24.8310$, are negative and give no real $x$.

6. 6

   Return to $x$. From $x^{2}=1.5056857750$,

   $x=\pm\sqrt{1.5056857750}=\pm 1.2270638838$

   Substituting back: $0.04(1.22706)^{6}+1.04(1.22706)^{4}+(1.22706)^{2}-4=0.000000$ ✓. The sextic has exactly two real roots and four complex ones.