Solve 0 = 0.7x + 6.804e10 * x^4 - 1210.23 for all real x.

Step-by-step solution

1. 1

   Put it in standard form and judge the scale. Reordering:

   $6.804\times10^{10}x^{4}+0.7x-1210.23=0.$

   The leading coefficient is enormous, so the $x^{4}$ term will reach the size of the constant while $x$ is still tiny. That is the opposite situation to a quartic with a very small leading coefficient, where the roots are large.

2. 2

   Get a first estimate by dropping the linear term. Balancing only the two dominant terms:

   $6.804\times10^{10}x^{4}\approx1210.23\;\Longrightarrow\;x^{4}\approx1.7787\times10^{-8}\;\Longrightarrow\;|x|\approx0.0115485.$

   Here the neglect is justified: at that $x$ the linear term $0.7x\approx0.0081$ is five orders of magnitude below $1210.23$. Because $x^{4}$ is even, this estimate immediately suggests two roots, one near $+0.0115$ and one near $-0.0115$.

3. 3

   Confirm two sign changes. Sweeping $f(x)=6.804\times10^{10}x^{4}+0.7x-1210.23$ across $[-1,1]$ finds exactly two sign changes, near $-0.011549$ and $+0.011548$. Reporting only the positive root — the easy omission — loses half the answer.

4. 4

   Bisect each bracket to full precision. Halving until $|f|<10^{-12}$:

   $x_{+}=0.0115484893,\qquad x_{-}=-0.0115485279.$

   The two are almost, but not exactly, negatives of each other: the $0.7x$ term breaks the symmetry slightly, pushing the negative root marginally further from zero. A quoted value of $0.0115425$ is not accurate enough — there $f\approx-2.51$, not $0$.

5. 5

   Substitute back. At $x=0.0115484893$: the quartic term is $1210.222$ and the linear term is $0.008084$, together cancelling $1210.23$ to within $7\times10^{-13}$ ✓. At $x=-0.0115485279$ the quartic term is $1210.238$ and the linear term is $-0.008084$, again summing to zero ✓. The remaining two roots of the quartic are a complex conjugate pair.