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anonymous

  • 5 years ago

How to get from this equation x^3 + 4x^2-10=0 to this equation x = x - [(x^3 + 4x^2-10)/ (3x^2+8x)]

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  1. myininaya
    • 5 years ago
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    so the original equation we divide it by -(3x^2+8x) assuming that this in not zero because you cannot divide by zero so we have (x^3+4x^2-10)/[-(3x^2+8x]=0 -(x^3+4x^2-10)/(3x^2+8x)=0 now add x no both sides x-(x^3+4x^2-10)/(3x^2+8x)=x

  2. myininaya
    • 5 years ago
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    on not no

  3. myininaya
    • 5 years ago
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    are there any questions?

  4. anonymous
    • 5 years ago
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    not really, thanks thou for help. I am still thinking about our solution

  5. myininaya
    • 5 years ago
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    you wanted to know how to write that one equation as that other equation right? you didn't want to use newton's method to find the x-intercepts? did you?

  6. anonymous
    • 5 years ago
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    f(x) = x^3 + 4x^2 -10 = 0 x^3 + 4x^2 -10 = 0 x^3 + 4x^2 = 10 x^2 = 10/(4+x) x = g4(x) = (10/(4+x))^(1/2) f(x) = x^3 + 4x^2 -10 = 0 x^3 + 4x^2 -10 = 0 4x^2 = x^3 + 10 x^2 = (-x^3 + 10)/4 x = ((-x^3 + 10)/4)^0.5 x = g3(x) = ((-x^3 + 10)/4)^0.5 f(x) = x^3 + 4x^2 -10 = 0 x^3 + 4x^2 -10 = 0 x^3 = 10 - 4x^2 x^2 = (10 - 4x^2)/x x = g2(x) = ((10 - 4x^2)/x)^(1/2) see all these equations I didn't need to add or divide any extra equations. However, your answer looks good, but you did add extra stuff there to reach the second equation. I am wondering if there will be anyway to reach the second equation without adding extra things in the original equation

  7. anonymous
    • 5 years ago
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    I am trying here to solve the original equation for x. There are five ways to solve the equation for x, I only got four ways.

  8. myininaya
    • 5 years ago
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    are you in calculus?

  9. anonymous
    • 5 years ago
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    numerical analysis

  10. myininaya
    • 5 years ago
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    so you do know newton's method?

  11. anonymous
    • 5 years ago
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    yes I do

  12. myininaya
    • 5 years ago
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    that second equation is looks like newton's method x1=x0-f(x0)/f'(x0) f'(x0) not equal to zero

  13. anonymous
    • 5 years ago
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    I know about the newton's method. I need to solve the original equation for x in five different ways, then I need to verify that each of the five equations converges with the same fixed point of original equation. I know three of them will converge, except one. I got to figure out how to find the fifth equation

  14. anonymous
    • 5 years ago
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    I know about the newton's method. I need to solve the original equation for x in five different ways, then I need to verify that each of the five equations converges with the same fixed point of original equation. I know three of them will converge, except one. I got to figure out how to find the fifth equation

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