Fuz said:
How do you go about solving equations for x that look something like this:
x3 - x2 + x - 2 = 0
I'm stumped when I get to:
x(x2 - x + 1) = 2
All insights appreciated! :)
You should start from the "rational root" theorem: any rational number root of the polynomial equation, a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0 must be of the form \frac{m}{n} where m evenly divides the constant term a_0 and n evenly divides the leading coefficient, a_n. In this problem, the constant term is -2 which has divisors \pm 1 and \pm 2. The leading coefficient is 1 which has divisors \pm 1. That is, the only possible rational roots are \pm 1 and \pm 2. It easy to see that none of those four possibilities does, in fact, satisfy the equation so it has no rational roots.
There is, however, "cubic formula, published by Cardano in 1545.
Note that
(a- b)^3= a^2- 3a^2b+ 3ab^2- b^3
and
3ab(a- b)= 3a^2b- 3ab^3
Adding those (a- b)^3+ 3ab(a- b)= a^3- b^3
If we let x= a- b, m= 3ab, and n= a^2- b^3, we see that x= a- b satisfies the "reduced" cubic equation x^4+ mx= n ("reduced" because there is no "x^2" term).
The question is, "Suppose we know m and n. Can we recover a and b and so find x= a- b?" The answer is "Yes, we can!"
From m= 3ab, we have b= m/3a. Putting that into a^3-b^3= n, a^3- m^3/(3^3a^3)= n. Multiply through by a^3 and we have (a^3)^2- (m/3)^3= na^3 or (a^3)^2- n a^3- (m/3)^3= 0, a quadratic equation in a^3. By the quadratic equation,
a^3= \frac{n\pm\sqrt{n^2+ 4\left(\frac{m}{3}\right)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}
Since a^3- b^3= n, b^3= a^3- n= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}.
To use that formula on this equation, x^3- x^2+ x- 2= 0, first "reduce" it. Let y= x+ \alpha so that x= y- \alpha, where \alpha is yet to be determined.
x^3= y^3- \alpha y^2+ \alpha^2y- \alpha^3
-x^2= -y^2+ 2\alpha y- \alpha^2
+x = y- \alpha
so that x^3- x^2+ x- 2= y^3- (\alpha+ 1)y^2+ (\alpha^2+ 2\alpha+ 1)y- (\alpha^3- \alpha^2+ \alpha+ 2)
The coefficient of y^2, -(\alpha+ 1) will be 0 if \alpha= -1.
And in that case, the coefficient of x, \alpha^2+ 2\alpha+1= 1- 2+ 1i= 0 also, and the constant term, -(\alpha^2+ \alpha+ 2)= -(1- 1+ 2)= -2. That is, the equation is equivalent to y^3- 2= 0 or y^3= 2 so that we don't really need the cubic formula. The roots are the three complex cube roots of 2: 2^{1/3}, 2^{1/3}(-1/2+ \sqrt{3}i/2), and 2^{1/3}(-1/2- \sqrt{3}i/2).
Since x= y- \alpha and \alpha= -1, the roots to the original equation are 2^{1/3}+ 1, 2^{1/3}(1/2+ \sqrt{3}i/2), and 2^{1/3}(1/2- \sqrt{3}i/2).
There exist a similarly complicated formula for fourth roots but it can be shown that there cannot be a general formula, using roots (there can be formulas with specially defined functions) or polynomial equations of degree five or higher.
