Sunday
May 19, 2013

# Homework Help: math

Posted by Sam on Wednesday, January 17, 2007 at 5:54pm.

how do you solve for zero for this equation?

(2/3)X -12 -(20/X^2)

(2/3)X -12 -(20/X^2) = 0 -->

2/3 X^3 - 12 X^2 - 20 = 0 --->

X^3 - 18 X^2 - 30 = 0

Numerical solution:

You see that for X = 18 the first and second term cancel, so the left hand side is -30. For X that large this is relatively close to zero (e.g. for X = 10 the left hand side is -830). Having found an approximate root, you can improve it by using Newton's method. If X_0 is your first guess of the solution of the equation f(X) = 0, then a better solutionis:

X_1 = X_0 - f(X_0)/f'(X_0)

In our case:

f(X) = X^3 - 18 X^2 - 30 --->

f'(X) = 3 X^2 - 36X

X_1 = 18.09

Iterating by replacing X_0 by X_1 gives:

X_2 = 18.091657

Iterating again:

X_3 = 18.0916567755

f(X_3) = -1*10^(-8)

so we are already very close to the exact solution.

Exact solution:

X^3 - 18 X^2 - 30 = 0

First we need to get rid of the quadratic term. We substitute:

X = Y + 18/3 = Y + 6

Then

X^3 = Y^3 + 18 Y^2 + 108 Y + 216

X^2 = Y^2 + 12 Y + 36

And we see that in terms of Y the equation is

X^3 - 18 X^2 - 30 =

Y^3 - 108 Y - 462 = 0

An equations of the form:

Y^3 + p Y + q = 0

can be transformed into a quadratic equation by substituting:

Y = Z - p/(3Z)

Y^3 in terms of Z is:

Z^3 - 3 Z^2 p/(3Z) + 3Zp^2/(9Z^2) - p^3/(27 Z^3) =

Z^3 - p Z + p^2/(3Z) - p^3/(27 Z^3)

And we find that:

Y^3 + p Y + q =

Z^3 - p^3/(27 Z^3) + q = 0

Multiplying by Z^3 gives:

Z^6 + q Z^3 - p^3/27 = 0

This is a quadratic equation in Z^3. The solutions are:

Z^3 = -q/2 +/- sqrt[(p/3)^3 + (q/2)^2]

Note that 1/Z^3 can be simplified by multiplying the numerator and denominator by

q/2 +/- sqrt[(p/3)^3 + (q/2)^2]

the result is:

1/Z^3 = (3/p)^3 *
[q/2 +/- sqrt[(p/3)^3 + (q/2)^2]]

And it follows from this that a solution for Y is:

Y = Z - p/(3Z) =

(W - q/2)^1/3 - (W + q/2)^1/3

where

W = sqrt[(p/3)^3 + (q/2)^2]

In general the other three solutyions are found by multiplying Z by a cube root of unity exp[2 pi i n /3].

In our case p = -108 and q = -462. This gives:

W = sqrt[6705] -->

Y = [sqrt(6705) + 231]^1/3 -
[sqrt(6705) - 231]^1/3

and X = Y + 6 =

6 + [sqrt(6705) + 231]^1/3 -
[sqrt(6705) - 231]^1/3

= 18.0916567755

No one has answered this question yet.

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