Solution of a cubic equation (Part 2)

In the post Solution of a cubic equation Part 1 we found the solution of a cubic equation for a particular condition. Then, in the post Whether real or complex (Cubic Equation) we studied when the roots of a cubic equation will be real and when it will be complex. This post deals with the solution of cubic equation. The algebraic method of solving cubic equations is supposed to be due to the Italian, del Ferro (1465-1526). But it is called Cardano's method because it became known to people after the Italian, Girolamo Cardano, published it in 1545 in his 'Ars Magna'.

Omar Khayyam gave a great deal of thought to the cubic equations. Before him, Greek mathematicians obtained solutions for third degree equations by considering geometric methods that involved the intersection of conics. Although the solution is present but I am searching for a solution of cubic equation with almost a new perspective. I will continue my search but I am giving this method so that until I find my kind of solution the traditional solution is present on my blog.

Let us consider the equation (ax³ + bx² + cx + d = 0 ; a≠0)

Step 1: Express ax³ + bx² + cx + d = 0
as x³ + px² + qx + r = 0

Step 2: Shift the middle point of the curve on the axis x=−p/3.
Read post Solution of a cubic equation. A special solution and Why -b/3a? for detail.
x³ + 3(p/3)x² + 3(p²/9)x + p³/27 − 3(p²/9)x − p³/27 +qx+ r = 0
(x + p/3)³ − [3(p²/9) − q]x − p³/27 + r = 0
(x + p/3)³ − [p²/3 − q]x − p³/27 + r = 0


Step 3:  Let x + p/3 = y then x = y − p/3.
Substitute y and (y − p/3) for (x + p/3) and x in the above equation.
y³ +[q − p²/3](y − p/3) + r − p³/27 = 0
y³ +[q − p²/3]y − [qp/3 − p³/9 − r + p³/27] = 0
y³ +[q − p²/3]y + [r − qp/3 + 2p³/27] = 0
y³ +[q − p²/3]y + [2p³/27 − qp/3 + r] = 0

Step 4: Now the equation is of the form y³ + Ay + B = 0
where A = q − p²/3 and B = 2p³/27 − qp/3 + r
Let y = (s − A/3s)
then s³ − A³/(3s)³ − As + A²/3s + As − A²/3s + B = 0
s³ − A³/(3s)³ + B = 0
Multiplying throughout by s³, we get
s⁶ + Bs³ − A³/27 = 0
Let, s³ = z
z² + Bz − A³/27 = 0
z = [−B ±√(B² + 4A³/27)]/2

s₁³ = [−B +√(B² + 4A³/27)]/2
As we know that any equation has three cube roots. Let its cube root be s₁. then the roots are s₁, ωs₁ and ω²s₁.
Similarly,
s₂³ = [−B −√(B² + 4A³/27)]/2
Let its cube root be s₂. then the roots are s₂, ωs₂ and ω²s₂.
1/s₁³ = −(3s₂/A)³
⇒ 1/s₁³ = 1/[−B +√(B² + 4A³/27)]/2
= {[−B − √(B² + 4A³/27)]/2}/{[−B −√(B² + 4A³/27)][−B +√(B² + 4A³/27)]/4}
= {[−B − √(B² + 4A³/27)]/2}/{−4A³/(27×4)}
= {[−B − √(B² + 4A³/27)]/2}/{A³/27)}
= −(3s₂/A)³
s₁s₂ = −A/3
If we consider s₁ then −A/3s₁ = s₂.
The product of s₁ and s₂ is real so the possibilities of the roots for y (= s − A/3s) are
s₁ + s₂, (ωs₁ + ω²s₂) and (ω²s₁ + ωs₂).

Step 5: Shift the graph to the original position for the roots.
The three roots are
s₁ + s₂ − p/3, (ωs₁ + ω²s₂ − p/3) and (ω²s₁ + ωs₂ − p/3).

The solution of the equation
x³ + px² + qx + r = 0 is
s₁ + s₂ − p/3, (ωs₁ + ω²s₂ − p/3) and (ω²s₁ + ωs₂ − p/3).

Where ω is a cube root of 1 i.e (−1 + i√3)/2
Where s₁ is a cube root of [−B +√(B² + 4A³/27)]/2 and
s₂ is a cube root of [−B − √(B² + 4A³/27)]/2.
and
where A = q − p²/3 and B = 2p³/27 − qp/3 + r

Whether real or complex (Cubic Equation)

This post deal with the conditions to check whether the roots are real or complex of a cubic equation. A cubic equation always has a real root. The other two roots are real or complex according as the graph cuts the x axis in one or three positions. If the graph touches the x-axis then two roots are equal. If the graph cuts the x-axis in three positions then all the roots are real. If the graph cuts the x-axis in one position then other two roots are complex. When we look at the graph of a general cubic equation we see a cup like structure and a cap like structure. The base of the cup like structure gives local minima and the top of cap like structure gives local maxima. Let us find the positions where the curve bends and changes its direction. We call the cup like structure concave upward and a cap like structure convex upward. The quadratic equation is ax³ + bx² + cx + d = 0. Let it be equal to y or f(x). y = ax³ + bx² + cx + d. Now the rate of change of curve is f '(x) = 3ax² + 2bx + c. f(x) is differentiated to obtain f '(x). We again differentiate f '(x) to obtain f ''(x),

f ''(x) = 6ax + 2b
f ''(x) = 6a(x + b/3a)

Let us check, for what values of x it may give maxima and for what values it may give minima.
There are four cases for it and is given in the following table

f ''(x)	x	a	state
> 0	x < (−b/3a)	a < 0	minima
< 0	x > (−b/3a)	a < 0	maxima
< 0	x < (−b/3a)	a > 0	maxima
> 0	x > (−b/3a)	a > 0	minima

When the local maxima and and local minima are on opposite sides of x-axis then all the roots are real.

Let us find the x-coordinate of maxima or minima. It is when f '(x)=0.

Solving 3ax² + 2bx + c = 0 we get
x = [−2b±√(4b² − 12ac)]/6a
x = [−b±√(b² − 3ac)]/3a
or
x₁ = [−b+√(b² − 3ac)]/3a
and
x₂ = [−b−√(b² − 3ac)]/3a Find the value of y for x₁ and x₂, substituting them in place of x.

For x₁ we get y₁ = [4b³ − 15abc + 27a²d + (6ac − 2b²)√(b² −3ac)]/27a²

For x₂ we get y₂ = [4b³ − 15abc + 27a²d − (6ac − 2b²)√(b² −3ac)]/27a²

Now local maxima and local minima lie on the opposite sides of the x-axis when y₁ and y₂ lie on the opposite sides of x-axis. This will happen when they are of opposite sign.

Now four cases arise:

y1	y2	Remark
y1< 0	y2< 0	One root is real and other two are complex. Graph of  y = x3 + 4x2 + 4x − 7
y1< 0	y2> 0	All the three roots are real.  Graph of y = x3 + 6x2 + 4x − 7 If y1 = 0 or y2 = 0 then there are two equal roots and it may be   [−b+√(b2 − 3ac)]/3a or [−b-√(b2 − 3ac)]/3a
y1> 0	y2< 0	All the three roots are real. Graph of y = − x3 + 4x2 + 4x − 7 If y1 = 0 or y2 = 0 then there are two equal roots. and it may be   [−b+√(b2 − 3ac)]/3a or [−b-√(b2 − 3ac)]/3a
y1> 0	y2> 0	One root is real and other two are complex. Graph of y = x3 + 4x2 + 4x + 7

When (b² − 3ac) is less than zero then y₁ and y₂ are complex and there is no maxima and minima. Then the equation has only one real root and other two are complex. See the graph for 3x² + 2x² + 3x + 4 = 0.
Where (b² − 3ac) = (4 − 27) < 0.

When y₁ or y₂ is equal to zero then there are two equal roots.

A cubic gives cubic

I ask you a question and the answer is very simple. But it has a very interesting conclusion in it. The question is "If we take a cubic equation and consider A as one number. Now If we find the difference of the roots from this point A and frame an equation which gives the values. What will we get."

The answer is very simple and is a cubic. The expression will be a cubic because we know that there are three roots and each value will be at a constant distance from the given number A. So there will be three numbers and a cubic equation only gives three values i.e. three numbers when equated to zero.

Now the question arises how can we find that equation. Suppose the equation is f(x) and the number is A. Then, the required equation is
f(x+δ) = f(x) + f'(x)δ /1!+ f''(x)(δ)² /2!+ f'''(x)(δ)³/3! +...

The above equation can be helpful for a cubic if we know one root.

Suppose the equation is f(x) = (x - 1)(x - 2)(x - 3)
= x³ - 6x² + 11x - 6

Now as 1 is a root then
f(1+δ) = f(1) + f'(1)δ/1! + f''(1)(δ)² /2!+ f'''(1)(δ)³/3!
f(1+δ) = f(1) + (3x² - 12x + 11)δ + (6x - 12)(δ)²/2 + 6(δ)³/6   x=1
f(1+δ) = 2δ - 3δ² + δ³
f(1+δ) = δ(2 - 3δ + δ²)
For the other roots f(1+δ) = 0 So,
Solving, δ(2 - 3δ + δ²) = 0
we get, δ = 0,1,2. Hence the roots are 1,2,3.

Solution of a cubic equation (Part 1)

In elementary algebra it is of prime interest to find the solution of a cubic equation.In this post and few other posts we will find the solution of a cubic equation. As from the previous post we know that the cubic is symmetric with respect to some x = a. In this post let us find the condition when the cubic expression has the middle point on line y=0 and what are the solutions of the corresponding cubic equation.

Let the cubic expression be y = ax³ + bx² + cx+ d.
Differentiating y w.r.t. x we get y' = 3ax² + 2bx + c. Here the sum of the slopes is −2b/3a. Now when we take a particular slope and put it in the equation of y' then we get a quadratic equation. We know that a quadratic equation has two roots. This implies that as the middle point of the curve has both the slopes equal as that point has a unique value. So slope is half of the value −2b/3a i.e. −b/3a. Which we have seen in the previous post. Now we have to find the value of x for which the slope is −b/3a. At this point we know that the difference between the slopes is zero as it has unique value so the rate of change of slope is zero. Differentiating again w.r.t. x we get y'' = 6ax + 2b. We find that the point where rate of change of slope is zero is -b/3a. Equating 6ax + 2b to 0. x has value −b/3a at the point where the rate of change of slope is zero.

Now we will find the value of y for this point. It is
y₁ = a(−b/3a)³ + b(−b/3a)² + c(−b/3a) + d
y₁ = −b³/(27a²) + b³/9a² − bc/3a + d
y₁ = 2b³/27a² − bc/3a + d
When this value is zero then the roots can be found very easily. In this case the center of the curve lies on the axis y = 0. I will tell what are the properties of the roots at the last.
when y₁ = 0 then d = bc/3a − 2b³/27a²

Now let us find the value of x where the slopes are zero. We get
3ax² + 2bx + c = 0
x = [−2b ± √(4b² − 12ac)]/6a
x = [−b ± √(b² − 3ac)]/3a
x₁ = [−b − √(b² − 3ac)]/3a
x₂ = [−b + √(b² − 3ac)]/3a
The center root lies at the middle of the points where the slopes are zero.
So,if δ is the distance of the middle root from the points where the slopes are zero then x₁ + δ = x₂ − δ.
δ = (x₂ − x₁)/2
δ = [√(b² − 3ac)]/3a
Hence the middle root is β = −b/3a.
Let the other roots be at a distance p from the middle root. Then
(β − p)(β + p)β = −d/a
(β − p)(β + p)β =− [bc/3a − 2b³/27a²]/a
(β − p)(β + p) = − [bc/3a − 2b³/27a²]/aβ
(β² − p²) = [bc/3a − 2b³/27a²]3/b
(β² − p²) = [c/a − 2b²/9a²]
(β² − p²) = [9ac − 2b²]/9a²
p² = β² + [2b²−9ac]/9a²
p² = b²/9a² + [2b² − 9ac]/9a²
p² = [3b² − 9ac]/9a²
p = ±√[3b² − 9ac]/3a
Hence the roots are
α = { − b − √[3b² − 9ac]}/3a;
β = −b/3a;
γ = { − b + √[3b² − 9ac]}/3a

Properties of the roots

The outer roots are equidistant form the middle root.

The roots are
α = { −b − √[3b² − 9ac]}/3a;
β = −b/3a;
γ = { − b + √[3b² − 9ac]}/3a
if the equation ax³ + bx² + cx + d = 0
satisfies d = bc/3a − 2b³/27a²

Solution of a cubic equation. A special solution and Why -b/3a?

While solving a cubic equation we come across Tschirnhaus transformation. This transformation is used to express a cubic equation in depressed form. In Tschirnhaus transformation we substitute x by t−b/3a. But why it is −b/3a that is the unique value for it. In this post I will show you why such transformation is useful in case of a depressed cubic.

Look at the following figure.

By looking at the figure we may think the the curve is symmetrical about some x = a. But let us prove it. The expression of the cubic is f(x) = ax³ + bx² + cx + d. Differentiating f(x) with respect to x we get f '(x) = 3ax² + 2bx + c. This is the rate of change of curve. Differentiating again we get f ' '(x) = 6ax + 2b. This is rate of change of slope of the curve. This value is zero for x = −b/3a. f '(x) is quadratic in x so it has two roots or in other words we can express it as the product of two linear expressions. The sum of these two roots is −2b/3a. Let the two roots of the expression f '(x) be α and β. Then α + β = −2b/3a. As we can see that the sum of slopes is twice of the rate of change of slope. So the curve is symmetrical about the line x = −b/3a. On both sides of this line at equal distance the value of slope has the same modulus.

The curve given in the figure is y = x³ − 6x² + 5x − 4. Here (−b/3a) = 6/3 = 2. Shifting the origin to X = x − 2 we get
y = (x+2)³ − 6(x+2)² + 5(x+2) − 4
y = x³ + 8 + 6x² + 12x − 6x² − 24x − 24 + 5x + 10 v 4
y = x³ − 7x − 10

So we see that when we shift the origin to −b/3a we get a cubic whose graph is symmetrical about the axis x=0.