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n^12+6n^10+10n^6+20n+4=0. How do I solve for n?

Update:

N is the same as n

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  • 9 months ago
    Favorite 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.

    • Doctor9 months agoReport

      thanks a lot, I don't need to be this precise though :D

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  • nbsale
    Lv 6
    9 months ago

    Once you get up to 5th power polynomials, there is no general method to get an exact answer. Of course, some of those may have an answer expressible in an exact form using radicals, like x^5 - 10 = 0 has exact solution 10^(1/5). But IN GENERAL you cannot find an exact solution using radicals.

    Therefore the best you can do is to use a numerical solution, like llaffer did. You can pursue such methods to whatever degree of accuracy you require.

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  • Amy
    Lv 7
    9 months ago

    copypaste it into wolframalpha.

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  • Anonymous
    9 months ago

    You have 1 equation and two variables, N and n, there is an infinite set of solutions.

    • Doctor9 months agoReport

      itş all n, yahoo just puts capital letter at the begining of a sentence.

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  • 9 months ago

    graphical is the only way I know of.

    assuming n and N are the same.

    n = –0.2, –0.991

    Attachment image
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