Solve 0.2 * 0.2 * x^6 + x^4 + x^2 - 4 = 0 for the real values of x.

Step-by-step solution

1. 1

   Simplify the coefficient and clear the decimal. Since $0.2\times 0.2=0.04$, the equation is $0.04x^{6}+x^{4}+x^{2}-4=0$. Multiplying through by $25$ gives integer coefficients:

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

2. 2

   Substitute to drop the degree. Only even powers of $x$ appear, so set $y=x^{2}$ (hence $x^{6}=y^{3}$, $x^{4}=y^{2}$):

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

   Crucially $y=x^{2}\ge 0$, so only non-negative roots $y$ can produce real $x$.

3. 3

   Rule out rational roots. By the rational root theorem the candidates are the divisors of $100$: $\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm25,\pm50,\pm100$. Substituting each gives a nonzero value, so the cubic does not factor over the rationals and the root must be found numerically.

4. 4

   Bracket the positive root. Let $g(y)=y^{3}+25y^{2}+25y-100$. Then

   $g(1)=1+25+25-100=-49<0,\qquad g(2)=8+100+50-100=58>0$

   so by the intermediate value theorem a root lies in $(1,2)$. Since $g'(y)=3y^{2}+50y+25>0$ for $y>0$, $g$ is strictly increasing there and this root is the only non-negative one.

5. 5

   Refine with Newton's method. Iterating $y\leftarrow y-\dfrac{g(y)}{g'(y)}$ from $y_{0}=1.5$ converges to

   $y\approx 1.526734392725$

   (The other two roots of the cubic, $y\approx -2.75539$ and $y\approx -23.77135$, are negative and give no real $x$.)

6. 6

   Undo the substitution and check. From $x^{2}=y$,

   $x=\pm\sqrt{1.526734392725}\approx\pm 1.235610939060$

   Substituting $x=1.235610939060$ back into $0.04x^{6}+x^{4}+x^{2}-4$ gives $0$ to machine precision. The equation is even, so the two roots are symmetric about the origin.