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### Using Cardan's Formula to find Real Roots

```
Date: 04/13/2000 at 07:21:40
Subject: Cardan's formula, complex numbers, and cubic equations

How can I show that, although the cubic equation

x^3 - 6x = 4

has three real solutions, Cardan's formula can find them by
subtracting appropriate cube roots of complex numbers?

I tried to solve by simply substituting in Cardan's formula (below):

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

where p = -6, q = 4

and obtained only one answer, which was complex. Hence, it is a wrong

sqrt[3]/2 - 1/2 + i(sqrt([3]/2 - 3/2)

as the only root (and it's not even real). It doesn't even work in the
original cubic.

By using a trigonometric formula, I know that the three correct real
roots are:

x1 = (3)^0.5 + 1 =  2.732
x2 =               -2
x3 = 1 - (3)^0.5 =  0.732

```

```
Date: 04/13/2000 at 08:11:10
From: Doctor Rob
Subject: Re: Cardan's formula, complex numbers, and cubic equations

The problem is that q = -4, not +4. First you have to move all of the
terms to one side of the equation: x^3 - 6*x - 4 = 0. Then you can
read off p = -4 and q = -4.  See if that helps.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 04/14/2000 at 04:57:02
Subject: Re: Cardan's formula, complex numbers, and cubic equations

Using p = -6 and q = -4 did NOT help, for it also gives a complex
answer. Even then, what puzzles me is, how am I supposed to get 3
```

```
Date: 04/14/2000 at 09:19:18
From: Doctor Rob
Subject: Re: Cardan's formula, complex numbers, and cubic equations

Yes, you are correct. The formula above does apply with q = 4. One
problem is that the formula is not correct as written. The expression

((p/3)^3 + (q/2)^2)^(1/3)

((p/3)^3 + (q/2)^2)^(1/2)

that is, a square root, not a cube root. If you fix that and
substitute p = -6 and q = 4, you'll see that this expression becomes
2*i. Then the root given by the formula is

x = (2*i+2)^(1/3) - (2*i-2)^(1/3)

Now you have to consider which of the three cube roots each of these
complex numbers to use. One choice is

(2*i+2)^(1/3) = -1 + i

and

(2*i-2)^(1/3) = 1 + i

They lead to the root x2 = -2. Other choices will lead to the other
roots. The other cube roots are found by multiplying those above by

(-1+sqrt[-3])/2 and (-1-sqrt[3])/2

the two complex cube roots of 1.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 04/14/2000 at 05:11:58
Subject: Re: Cardan's formula, complex numbers, and cubic equations

Sorry, the question posed does have the wrong formula. However, the
corrected question is as follows:

In the form of

x^3 + px = q

Cardan's formula for the solution of the cubic equation is

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

How can I show that, although the cubic equation

x^3 - 6x = 4

(i.e. p = -6, q = 4) has three real solutions, Cardan's formula can
find them by subtracting appropriate cube roots of complex numbers?
The real question is: How can I get three answers from that one
formula?

When I tried to solve by substituting in Cardan's formula (corrected),
I still obtained only one answer, which was complex. Hence, it is a

-(sqrt[3])/2 - 1/2 + i(sqrt[3]/2 - sqrt[3]/2

as the only root (again, it's not even real and it doesn't work in the
original cubic equation either).

```

```
Date: 04/14/2000 at 05:46:40
From: Doctor Anthony
Subject: Re: Cardan's formula, complex numbers, and cubic equations

I have given a general note on Cardan and on the Trig solution (when
there are 3 real roots.) Cardan is VERY unwieldy when you have 3 real
roots because the quadratic in t has complex roots. I hope you will be
able to sort out where the difficulty is. Note too that what I have
called p and q are NOT the same as in your equation.

Look at:

http://mathforum.org/dr.math/problems/vonberwick4.30.98.html

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations
High School Polynomials

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