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
To solve for zero in the equation (2/3)X - 12 - (20/X^2) = 0, we can follow the steps below:
1. Simplify the equation:
(2/3)X - 12 - (20/X^2) = 0
2. Multiply through by 3X^2 to eliminate fractions:
2X^3 - 36X^2 - 60 = 0
3. Rearrange the equation:
2X^3 - 36X^2 = 60
4. Shift all terms to one side:
2X^3 - 36X^2 - 60 = 0
5. At this point, we have a cubic equation. To find an approximate root, we can try different values of X until we find one where the left-hand side of the equation is close to zero. In this case, we find that X = 18 satisfies this condition.
6. Now, we can improve the approximation by using Newton's Method. The formula for Newton's Method is X_1 = X_0 - f(X_0)/f'(X_0), where X_0 is an initial guess and f'(X_0) is the derivative of the function f(X) with respect to X.
7. In our case, the function f(X) is 2X^3 - 36X^2 - 60, and the derivative f'(X) is 6X^2 - 72X. With X_0 = 18 as our initial guess, we can calculate X_1 using the formula.
8. Iterate this process to find X_2, X_3, and so on until you reach an X value that satisfies f(X) ≈ 0. In this case, we find that X ≈ 18.0916567755 is very close to the exact solution.
Alternatively, if you want to find the exact solution, you can use algebraic methods. By transforming the equation by substitution, simplifying, and using the quadratic formula, you can obtain the exact solution, which is X ≈ 18.0916567755.