Cubic equations always have three solutions, one of which is always a real number. The other two solutions are sometimes real and sometimes complex numbers. This problem helps you to track the solutions (roots) of cubic equations and to learn more about complex numbers.

(1) To convert the cubic equation
2t3+ 3t2 - 11t - 6=0
to an equation of the form x3+px+q = 0, make the substitution t = x - 1/2 and show that the equivalent equation in x is
x3 - 25
4
 x = 0

. Solve this equation in x and verify that the corresponding values of t are the solutions of the equation 2t3+ 3t2 - 11t - 6=0.

(2) You are not asked to do this but, in a similar way to the example in (1), by making the substitution
t = x - b
3a

, it is always possible to convert the general cubic equation at3 + bt2 + ct + d = 0 into an equation of the form x3+px+q = 0 where p and q depend on a, b, c and d. The roots of the original equation in t can be found by first finding the roots of the simplified, but equivalent, cubic equation in x and then using the relation
t = x - b
3a

.

Throughout this problem you are in control of the general cubic equation x3 + px + q = 0. You can change this equation by moving the point (p, q) in the red frame and you have to investigate what happens to the roots of this equation as you change the values of p and q.

The blue frame shows the graph of y=x3 + px + q.

Set
p=- 25
4

and q=0 and look for the three roots you found in (1).

Use either the arrow keys or the mouse to move the spot in the red frame.
.
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What happens to the graph of y=x3 + px + q if you keep p constant and change q?

(3) What happens to the graph of y=x3 + px + q if you keep q constant and change p?

(4) The green frame is called an Argand Diagram and it shows the three roots of the cubic equation. Look for three points and watch them move as you change the driving point (p,q) in the red frame and in doing so change the cubic equation and its roots.

Investigate the three roots of the equation x3+ px + q = 0 for the case p=0, q=8. Locate the three roots of the cubic equation x3 + 8 = 0 on the Argand diagram. (Note that the readout will not give you the exact values of u and v.) What are the roots of this equation? Prove that the three roots are equally spaced around a circle in the Argand diagram.

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(5) Now keep p = -3 as a constant and change q. In a table like the one below describe the features of the cubic equation and its coefficients and roots that you see in the three frames for different values of q.

p is constant p = - 3 Red frame (p,q) Blue frame y=x3+px+q Green frame Argand diagram (u,v) 
q < -2

q = -2

-2 < q < 2

q = 2

q > 2

To find out more about complex numbers see the article What Are Complex Numbers?

For more information about solving cubic equations see The Math Forum and for the history of equations see The Mac Tutor website.


NOTES AND BACKGROUND


Cardan's method for solving cubic equations depends on reducing the general cubic to an equation of the form x3 + px +q = 0. In this problem we explore the changes in the roots of cubic equations as we change the graph so as to learn more about complex numbers. The problem is not about finding actual solutions but about understanding the general case. For completeness we give Cardan's method below.

Consider the general cubic equation
at3 + bt2 + ct + d = 0.
Dividing by a this equation becomes
t3 + b
a
 t2 + c
a
 t + d
a
 = 0.
Making the substitution
t = x - b
3a

gives the equation:
(x - b
3a
 )3 + b
a
 (x - b
3a
 )2 + c
a
 (x - b
3a
 ) + d
a
 = 0.
Expanding this equation is:
x3 - 3( b
3a
 )x2 + 3( b
3a
 )2x - ( b
3a
 )3 + b
a
 (x2 - 2( b
3a
 )x + ( b
3a
 )2) + c
a
 (x - b
3a
 ) + d
a
 = 0.
Clearly the coefficient of x2 is zero and the equation reduces to x3 + px +q=0 where p and q depend on a, b, c and d.

Here, in modern notation, is Cardan's solution of x3 + px = -q .
Notice that (r - s)3 + 3rs(r - s) = r3 - s3 so if r and s satisfy 3rs = p and r3 - s3 = -q then r-s is a solution of x3 + px = -q.
But now s = p/3r so r3 - p3/27r3 = -q, i.e. r6 + qr3 - p3/27 = 0.
This is a quadratic equation in r3, so solve for r3 using the usual formula for a quadratic. Now r is found by taking cube roots and s can be found in a similar way (or using s = p/3r). Then x = r- s is the solution to the cubic.

By the Fundamental Theorem of Algebra every polynomial equation F(z) = 0, where F(z) is a polynomial of degree n has n solutions, some of which may be repeated. So cubic equations have 3 solutions, one of which must be real (why?) and the other two may be real or may be complex.

In school we only consider equations of the form F(z)=0 where the coefficients are real numbers but the methods are the same and it is really no more difficult to consider the general function with complex coefficients.

As we have learnt more about number, so we have been able to find solutions to more equations. In primary school we could find numbers to put in the boxes for equations like [¯]+ 2 = 7, 4×[¯] = 8 and [¯] ×[¯] = 9 and we were really solving the equations x + 2 = 7, 4x = 8 and x2 = 9 without using algebraic notation. We could not solve equations like x + 7 = 2 until we learned about negative numbers. We could not solve equations like 4x = 3 until we learned about fractions (rational numbers) and we could solve equations like x2 = 3 until we learned about irrational numbers. Once we understand complex numbers, which are algebraically very simple, then we are can solve all quadratic equations and we can appreciate that higher order polynomial equations will have real and complex roots.