Best Answer:
n¹² + 6n¹⁰ + 10n⁶ + 20n + 4 = 0

If we set this up as a function, we can use Newton's method to get an estimation for the roots.

We know that there are two real roots at or near -0.2 and -0.99 from a previous answer. We can use those as starting points.

Newton's method has us start with an initial guess and every loop into the process gets us closer. The equation we use is:

x₁ = x₀ - f(x₀) / f'(x₀)

We have an expression for f(x), so now we need the first derivative:

f(n) = n¹² + 6n¹⁰ + 10n⁶ + 20n + 4

f'(n) = 12n¹¹ + 60n⁹ + 60n⁵ + 20

If we substitute this into my equation above, we get:

x₁ = x₀ - f(x₀) / f'(x₀)

x₁ = x₀ - (x₀¹² + 6x₀¹⁰ + 10x₀⁶ + 20x₀ + 4) / (12x₀¹¹ + 60x₀⁹ + 60x₀⁵ + 20)

Using x₀ = -0.99, solve for x₁:

x₁ = -0.99 - ((-0.99)¹² + 6(-0.99)¹⁰ + 10(-0.99)⁶ + 20(-0.99) + 4) / (12(-0.99)¹¹ + 60(-0.99)⁹ + 60(-0.99)⁵ + 20)

I'll round numbers to 7DP during the processes, so may be off by a little bit due to rounding errors:

x₁ = -0.99 - (0.8863849 + 6(0.9043821) + 10(0.9414802) - 19.8 + 4) / (12(-0.8953383) + 60(-0.9135172) + 60(-0.9509900) + 20)

x₁ = -0.99 - (0.8863849 + 5.4262926 + 9.414802 - 19.8 + 4) / (-10.7440596 - 54.811032 - 57.0594 + 20)

x₁ = -0.99 - (-0.0725205) / (-102.6144916)

x₁ = -0.99 - 0.0007067

x₁ = -0.9907067

If we put this as the input into a second loop, we should get a closer value. You'll see this since the number that is added or subtracted from this value will be closer to zero than the last time (and it was pretty close the first time):

x₂ = x₁ - (x₁¹² + 6x₁¹⁰ + 10x₁⁶ + 20x₁ + 4) / (12x₁¹¹ + 60x₁⁹ + 60x₁⁵ + 20)

x₂ = -0.9907067 - ((-0.9907067)¹² + 6(-0.9907067)¹⁰ + 10(-0.9907067)⁶ + 20(-0.9907067) + 4) / (12(-0.9907067)¹¹ + 60(-0.9907067)⁹ + 60(-0.9907067)⁵ + 20)

x₂ = -0.9907067 - (0.8940076 + 6(0.9108587) + 10(0.9455197) - 19.814134 + 4) / (12(-0.9023938) + 60(-0.9194030) + 60(-0.9543892) + 20)

x₂ = -0.9907067 - (0.8940076 + 5.4651522 + 9.455197 - 19.814134 + 4) / (-10.8287256 - 55.16418 - 57.263352 + 20)

x₂ = -0.9907067 - 0.0002258 / (-103.2562576)

x₂ = -0.9907067 + 0.0000022

x₂ = -0.9907045

This is slightly closer. And if we want to get more precise (which would require more than 7 DP, we can loop through again.

We can then use the same equation with our -0.2 to see what we get:

x₁ = x₀ - (x₀¹² + 6x₀¹⁰ + 10x₀⁶ + 20x₀ + 4) / (12x₀¹¹ + 60x₀⁹ + 60x₀⁵ + 20)

x₁ = -0.2 - ((-0.2)¹² + 6(-0.2)¹⁰ + 10(-0.2)⁶ + 20(-0.2) + 4) / (12(-0.2)¹¹ + 60(-0.2)⁹ + 60(-0.2)⁵ + 20)

x₁ = -0.2 - (0.000000004096 + 6(0.0000001024) + 10(0.000064) + 20(-0.2) + 4) / (12(-0.00000002048) + 60(-0.000000512) + 60(-0.00032) + 20)

x₁ = -0.2 - (0.000000004096 + 0.0000006144 + 0.00064 - 4 + 4) / (-0.00000024576 - 0.00003072 - 0.0192 + 20)

x₁ = -0.2 - (0.000640618496) / 19.98076903424

x₁ = -0.2 - 0.000032062

x₁ = -0.200032062

We can drill down again if we want, but we can see the answer isn't exactly -0.2, but very close to it.

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