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.