PROPERTIES AND SOLUTIONS OF QUADRATIC AND CUBIC ALGEBRAIC EQUATIONS
The simplest algebraic equations after the linear are the quadratic and cubic equations given by-
y ax 2 bx c and y ax 3 bx 2 cx d
, respectively. Here a ,b, c, and d are specified constants which may or may not be complex. Students first learn how to solve the quadratic equation in introductory high school algebra by completing the square. This produces the well known result-
b b2 4ac y 2a
This solution is about the only thing students retain into adulthood having never learned about its symmetry and the conditions on a, b, and c for the solutions to be real. The cubic equation is first introduced during freshmen calculus but the complicated formula for finding its zeros is never explicitly derived. Reference for finding the zeros is usually relegated to handbooks.
It is our purpose here to examine the properties and solutions of the general quadratic and cubic equations making full use of equation symmetry and their respective relation to a standard parabola and cubic.
We begin by rewriting the quadratic equation as-
b 2 b y c a( ) ax 2a 2a
This is recognized to be a parabola with a minimum at x=-(b/2a), where the value 2 of ymin=c-(b /4a). The line x=-(b/2a) is recognized to be a symmetry line as shown on the following graph-
The equation has two distinct roots for y=0. These are located at equal distances =sqrt(b2-4ac)/2a from the symmetry line. The roots are equal to-
b b2 4ac b b2 4ac x and x 1 2a 2 2a
For real a, b, and c, the two roots will be real only if b2>4ac. The condition where the radical vanishes corresponds to a double root of x=-(b/2a). Complex conjugate roots exist for b2<4ac.
Consider now the specific quadratic equation –
y 2x 2 3x 1
Here the radical has value sqrt(9-8)=1>0. So we expect two real roots for y=0 placed at distance =1/4 from the symmetry line at x=-b/2a=-3/4. So we have the two roots x1=-1 and x2=-1/2.
If any of the coefficients have complex form then one expects y and x to also be complex. In that case we can set-
x i and y u iv
with , , u, and v real . Such equations will still have zeroes but these will be in the complex x plane. Besides the use of the above formula, the simplest way to locate the zeros is by making a contour plot of the absolute value of y and noting where this vanishes. Take the case y=x2-2ix+3. Here we find-
y u iv ( i )2 2i( i ) 3
This yields- u 2 ( 1)2 4 and v 2 ( 1)
The roots will be found at x=2i and x=-i. A contour plot of the absolute value of y looks as follows-
The two zeros along the axis are clearly shown at =3 and = -1. This result can be considered a special case of the Gauss Fundamental Theorem of Algebra which states that any nth order polynomial y=f(x) with complex coefficients has n roots for which y=0.
Let us next go to the cubic algebraic equation. To solve it we first make the transformation x=X-(b/3a) and (y-B)=Y, where B=(2b3/27a2)-(cb/3a)+d. This produces the depressed cubic equation-
b2 Y aX 3 AX with A c ( ) 3a
This curve is a standard cubic equation which has an inflection point at X=0 and is anti-symmetric in the sense that Y(X)=-Y(-X). The first derivative reads dY/dX=3aX2+A and so vanishes at X=sqrt(-A/3a). The value of Y at that point is sqrt(-A/3){2A/3). A graph of this generic cubic equation follows-
We see the anti-symmetry about X=0 which implies anti-symmetry about x=-b/3a. The solutions corresponding to y=0 are represented by the three blue dots. We will have three real solutions provided that Y<(2A/3)sqrt(-A/3a). This fact allows us to make certain statements about any cubic without actually having obtained explicit solutions. Consider, for example, the cubic-
y x 3 4x 2 x 3
Here a=1, b=-4, c=1 and d=3. This produces A=-29/3 and B=-11/27. Since B is less than Ymax, we expect three real roots for y=0. Carrying out a numerical evaluation for the roots on our computer, we find-
x1=3.4604… , x2=1.2391… and x3=-0.6996 … in perfect agreement with the prediction.
Next take a second cubic-
y x3 x 2 1
Here we have a=1, b=1, c=0 and d=1. This produces A=-1/3 and B=29/27. Also Ymax=2/27 x1= -1.4655… , x2= 0.23278+i0.79255 , x3=0.23278-i0.79255 Note that solutions x2 and x3 are complex conjugates of each other. It is also possible to have cubics with complex coefficients where none of the roots are real. We can construct such a scenario by defining a complex function- F(z)=(z-i )(z+(sqrt(3)+i)/2)(z+(-sqrt(3)+i)2/2) This represents a complex cubic with zeros at the vertexes of an equilateral triangle of side-length L=sqrt(3). Note that none of the zeros are real. A contour plot in the z plane showing contours for the absolute value of F(z) follows- Although we have been able to glean a considerable amount of information on the properties of the cubic algebraic equation, it still remains for us to solve it explicitly. Such a solution was originally given by the Italian mathematician Cardano(1501-1576), who some say stole the idea from a contemporary Tartaglia. Although Cardano used a geometric proof, the basic idea is to introduce a couple of new variables u and v which relate to X appearing in the depressed cubic equation- Y y B aX 3 AX To find the three expected roots for which y=0, we then have- 0 X (aX 2 A) B We want to solve this non-linear algebraic equation. After doing so we will in effect have found the three roots for x=X-(b/3a) of the original cubic equation. One begins by making the substitution X=u-v. This yields - A B u3 v 3 (u v)( 3uv) ( ) a a This identity will be satisfied by letting- A 3auv and B a(u3 v3 ) One eliminating v from these definitions, one obtains the algebraic equation- B A u6 ( )u3 ( )3 0 a 3a This represents essentially a quadratic equation in u3 which can be solved exactly. It yields- 3 3 B B 2 A 3 A u ( ) ( ) ( ) 2a 2a 3a 3av Thus one now knows u and v and hence the final solution for one of the zeros of the original cubic equation is given by- b x u v 1 3a Clearly the breakthrough came with the substitution X=u-v. A few years after Cardano , his student Luddovico Ferrari(1522-1565) solved the gereral 4th order algebraic equation based on the solution of the cubic. After that mathematicians tried to go to quintic equations and higher but were unable to find a closed form solution for the general case. It was the Norwegian mathematician Niels Abel (1802-1829) who first showed in 1824 that no solutions, given in terms of radicals, are possible for quintic and higher algebraic equations. Let us now find the explicit roots of several specific cubics. Start with- 0 X 3 3X 1 where A 3, B 1, a 1, b 0, c 3 and d 1 Here u3=(sqrt(5)-1)/2=1/(golden ratio) and 1 x u 0.3221853... 1 u in perfect agreement with the computer solution for this problem which yields- x1=-0.3221853 , x2=0.16109+i1.7543 and x3=0.16109-i1.7543 The solutions x2 and x3 can be obtained by solving the quadratic- (x 3 3x 1) 0 (x x1 ) As another example consider finding the three real roots of the cubic- 0 x 3 2x 2 x 2 Here a=1, b=-2, c=-1, d=2, A==7/3. and B=20/27. This yields - (10 i 243)1/ 3 u 3 On solving, one finds- 7 2 x u 2.000000 1 9u 3 The other two roots are x=1 and x=-1 as can be readily established by dividing the original cubic by (x-2). It is really quite amazing how non-integer powers of complex numbers can lead to real values for x. This must have been especially hard for Italian mathematicians of the fifteen hundreds to understand considering that they had no idea what a complex number is. To finish, we add a graph of this last cubic using the coordinates [x,y] and [X,Y]- This more clearly shows the symmetry line passing through the X=0 (or x=2/3) vertical line. April 2015