Solve x^2 (x^4 + 104x^2 + 40x) = 10000. Find every real solution to six decimal places.

Step-by-step solution

1. 1

   Expand and move everything to one side. Distributing the $x^{2}$:

   $x^{2}\cdot x^{4}+x^{2}\cdot 104x^{2}+x^{2}\cdot 40x=x^{6}+104x^{4}+40x^{3},$

   so the equation becomes

   $f(x)=x^{6}+104x^{4}+40x^{3}-10000=0.$

   Note there is no $x^{5}$, $x^{2}$ or $x$ term — the polynomial is sparse but not of a form ($x^{6}+bx^{3}+c$) that a substitution $u=x^{3}$ would reduce, because of the $104x^{4}$.

2. 2

   Rule out rational roots. The rational root theorem allows only divisors of $10000$: $\pm1,\pm2,\pm4,\pm5,\pm8,\pm10,\pm16,\pm20,\pm25,\ldots$ Testing the small ones,

   $f(2)=64+1664+320-10000=-7952,\qquad f(3)=729+8424+1080-10000=233,$

   $f(-3)=729+8424-1080-10000=-1927,\qquad f(-4)=4096+26624-2560-10000=18160.$

   None is zero, and no factorisation into rational quadratics exists either, so the roots must be found numerically.

3. 3

   Bracket the sign changes — there are two, not one. The four values above already reveal both:

   $f(-4)>0>f(-3)\quad\text{and}\quad f(2)<0<f(3),$

   so by the intermediate value theorem there is a root in $(-4,-3)$ and another in $(2,3)$. Sweeping $f$ across $[-10,10]$ in steps of $0.001$ finds exactly these two sign changes, so the sextic has two real roots (the remaining four are complex). Stopping after the positive root — the easy mistake — loses half the answer.

4. 4

   Bisect each bracket to full precision. Halving the interval repeatedly and keeping the half where $f$ changes sign:

   $x_{+}\in(2,3)\to 2.982928,\qquad x_{-}\in(-4,-3)\to-3.152475.$

   Each converges to $|f|<10^{-11}$ after about 40 halvings. A quoted value of $2.9833$ is a little coarse: $f(2.9833)\approx +5.04$, not $0$.

5. 5

   Substitute both roots back. At $x=2.982928$: $x^{2}=8.897859$ and $x^{4}+104x^{2}+40x=79.172+925.377+119.317=1123.87$, whose product is $10000.0$ ✓. At $x=-3.152475$: $x^{2}=9.938099$ and the bracket evaluates to $1006.23$, again giving $10000.0$ ✓. The two roots are not symmetric because of the odd-power $40x^{3}$ term.