Solve the quartic 0.0395m^4 - 1.0636m^3 + 8.1435m^2 - 11.8308m - 17.026 = 0.

Step-by-step solution

1. 1

   Estimate where the roots can live before searching. Cauchy's bound says every root satisfies $|m|\le 1+\max|a_k|/|a_4|=1+17.026/0.0395\approx 432$. That is loose, but the tiny leading coefficient is the warning sign: dividing through by $0.0395$ gives $m^4-26.927m^3+206.16m^2-299.51m-431.04$, so the roots sum to about $26.9$ and are spread over a wide range.

2. 2

   Scan for sign changes on a grid. Evaluating $f(m)$ at steps of $0.1$ from $-3$ to $60$ gives sign flips in $(-0.9,-0.8)$, $(3.3,3.4)$, $(11.1,11.2)$ and $(13.3,13.4)$. Four sign changes for a quartic means four real roots — there is no room left for a complex pair, since complex roots come in pairs and only four roots exist in total.

3. 3

   Bisect each bracket to full precision.

   $m_1=-0.86460205,\quad m_2=3.37179569,\quad m_3=11.10472698,\quad m_4=13.31466166$

   Each was refined until the residual $|f(m)|$ fell below $10^{-12}$.

4. 4

   Cross-check with Vieta's formulas. For $a_4m^4+\cdots+a_0$, the roots sum to $-a_3/a_4=1.0636/0.0395=26.92658228$ and multiply to $a_0/a_4=-17.026/0.0395=-431.03797468$. The four computed roots sum to $26.92658228$ ✓ and multiply to $-431.03797468$ ✓ — both to ten significant figures, so no root is spurious or missing.

5. 5

   Reject the values that fail substitution. A commonly circulated answer gives $m\approx-0.9365$ and $m\approx3.3065$ with a complex pair $12.2798\pm19.2988i$. Substituting: $f(-0.9365)=+2.0996$ and $f(3.3065)=-0.8396$, neither of which is zero, and the claimed complex pair would push the root sum to $26.93$ only by coincidence. The correct answer is four real roots.