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

How many Solutions of x + y + z = k

In this post we will consider all the positive integer solutions of the equation x + y + z = k. At the end of the post we will generalize the method.

Let us first solve the equation. There is a x, a y and a z. We can consider x 1's, y 1's and z 1's. So there are a total of (x + y + z) 1's. We can find its integer solutions by separating  1's at different positions. Consider the equation x + y + z = 9.

Some of the solutions of the equation are

11|111|1111 (2+3+4)
111|111|111 (3+3+3)
1111|111|11 (4+3+2)
1111|11|111 (4+2+3)
11|11|11111 (2+2+5)

Add all the ones in one group separated by |. The number of solutions of this type is the solution of the equation x + y + z = 9. The number of solutions of the equation is (9+2)!/(9!2!) = (11×10)/2 = 55. Hence there are 55 solutions. If we have the total as k and the number of plus sign is n. Then the total number of solutions is (k+n)!/(k!n!). This can be written in the combination symbol as C(n+k,k).

This equation is similar as taking the combination of r things in n containers of similar kind. The number of combinations is (n+r−1)!/(n−1)!r!. This can be written in the combination symbol as C(n+r−1,r).

Preliminaries in plane geometry (Part 1)

Many shapes in a plane are well represented by equations. Circles, parabolas and hyperbolas are some of the shapes which can be represented by equations.
But to deal with them we need to know the basics of geometry. The posts related to basics of coordinate geometry is given in the related posts "Preliminaries in plane geometry". The posts will be given in many parts. each part will cover two to three topics. In this post we will discuss about Distance formula and Section formula.

---

Distance formula

The Cartesian coordinates are used to represent points in a plane. Every point in a place has a one to one relationship to a coordinate point. The distance (d)  between two points A(x₁, y₁) and B(x₂, y₂) is given by the formula
d = √[(x₂ − x₁)²+(y₂ − y₁)²]

The above formula can be found with the help of Pythagoras theorem. Draw a right angled triangle as shown in the graph. When we draw a right angled triangle we draw it such that the two edges including the right angle are parallel to the two axis. This helps us to find the coordinate of the vertex. Name the right angled vertex as C. The coordinate of the point C is (x₂, y₁). a line parallel to the coordinate axis has its other coordinate same. Suppose the line is parallel to x-axis then all the points on the line has its y-coordinate equal.  The distance between the points A and B is d. The distance between the points A and C is AC =(x₂ − x₁) and distance between the points B and C is BC =(y₂ − y₁). By Pythagoras Theorem we have (AB)² = (AC)² + (BC)². Substituting the value of AC and BC we get

d² = (x₂ − x₁)²+(y₂ − y₁)²Taking square root and as distance cannot be negative, we get
d = √[(x₂ − x₁)²+(y₂ − y₁)²]

Example:
Distance between A(3,5) and B(1,2) is
AB = d
d = √[(1 − 3)²+(2 − 5)²]
= √(4+9)
= √13

---

Section Formula

A point C(x,y) is present between A(x₁,y₁) and B(x₂,y₂). C divides AB internally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx₂+nx₁)/(m+n)] , [(my₂+ny₁)/(m+n)] )

Example:
C divides AB internally in the ratio 1:2. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=1, n=2, x₁= 3, x₂ = 1, y₁ = 5, y₂ = 2.
x = (1×1 + 2×3)/(2+1)
= 7/3
  y= (1×2 + 2×5)/(2+1)
= 12/3
= 4. Hence the coordinate of C is (7/3,4).

If C divides AB externally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx₂− nx₁)/(m − n)] , [(my₂− ny₁)/(m − n)] )

Example:
C divides AB externally in the ratio 2:1. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=2, n=1, x₁= 3, x₂ = 1, y₁ = 5, y₂ = 2.
x = (2×1 − 1×3)/(2 − 1)
= −1
  y = (2×2 − 1×5)/(2 − 1)
= −1
Hence the coordinate of C is (−1,−1).

Recurrence relation (Part 2) : Linear Equations

In the post Recurrence Relation (Part 1) we studied three examples of recurrence relations. In this post we will solve them. The recurrence relation (T_n = T_n-1 + T_n-2) of Fibonacci problem is linear and has order 2 and degree 1. It is also homogeneous. The recurrence relation (T_n = 2T_n-1 + 1) of Tower of Hanoi problem is also linear. It has order 1 and degree 1 but it is non homogeneous. Recurrence relations can exist in many forms as linear or non-linear.

A linear recurrence relation looks like this

T_n = f₁(n)T_n-1 + f₂(n)T_n-2 + f₃(n)T_n-3 + ...+ f_r(n)T_n-r+...+f_n-1(n)T₁ + f(n)
The order of the recurrence relation is r if f_i(n) = 0 for i > r.
It is called homogeneous if f(n) = 0 and non-homogeneous if f(n) ≠ 0.
We will consider linear homogeneous and non-homogeneous recurrence relation with constant coefficients. A linear recurrence relation with constant coefficients is given by

T_n = C₁T_n-1 + C₂T_n-2 + C₃T_n-3 + ...+ C_rT_n-r+...+C_n-1T₁ + f(n) where C_i is a constant, 0 < i < n.
The order of the recurrence relation is r if C_i = 0 for i > r.
It is called homogeneous if f(n) = 0 and non-homogeneous if f(n) ≠ 0.

---

We will solve the linear homogeneous recurrence relation with constant coefficients first. The example of this type of equation which we are going to solve is Fibonacci Problem.
T_n = T_n−1 + T_n−2
Substitute the Terms with the r^{(the subscript)}
We get,
rⁿ = rⁿ⁻¹ + rⁿ⁻²
Divide throughout by the lowest power term which is r^n-2.
We get,
r² = r + 1
r² − r − 1 = 0
The above equation is called the characteristic polynomial of the recurrence relation and its roots are called characteristic roots.
The solution of the polynomial is r₁ = (1 + √5)/2 and r₂ = (1 − √5)/2

The solution of the recurrence relation can be written like this
T_n = A(r₁)ⁿ + B(r₂)ⁿ = A[(1 + √5)/2]ⁿ + B[(1 − √5)/2]ⁿ

T₁ = 1 = A[(1 + √5)/2] + B[(1 − √5)/2] = (A + B)/2 + (√5)/2(A − B)
T₂ = 1 = A[(1 + √5)/2]² + B[(1 − √5)/2]²
= A [1 + 5 + 2√5]/4 + B [1 + 5 − 2√5]/4
= A [6 + 2√5]/4 + B[6 − 2√5]/4

We get two relations
(A + B) + (√5)(A − B) = 2 ----(i)
and
3(A + B) + (√5)(A − B) = 2 ----(ii)

Solving the equations simultaneously, we get
3(i) - (ii)
2√5(A − B) = 4
A − B = 2/√5 and substituting in (i)  we get
A + B = 0

Again solving simultaneously,
2A = 2/√5
A = 1/√5
B = -1/√5
Hence the solution of the recurrence is
T_n = [1/√5][(1 + √5)/2]ⁿ - [1/√5][(1 − √5)/2]ⁿ

---

Next we will solve the recurrence relation T_n = 2T_n−1 + 1 which belongs to the tower of Hanoi problem.

Linear non homogeneous recurrences require us to solve for their homogeneous part then substitute a relation which satisfies the non homogeneous recurrence. There are many methods which are different from it and can be used to solve many recurrences. These methods are

1. Method of iteration
2. Method of substitution
3. Method of telescoping sums
4. Method of inspection

Here, we will use method of iteration to solve the recurrence.
In this method we substitute the earlier term in the recurrence and look at the pattern and solve the series.
T_n = 2T_n−1 + 1
T_n = 2(2T_n−2 + 1) + 1
T_n = 2(2(2T_n−3 + 1) + 1) + 1
As T₁ = 1, so we get
T_n = 2^n-1 + 1 + 2 + 2² +....+ 2^n-2
T_n = 2^n-1 + (2^n-1 − 1) = 2ⁿ − 1

---

Now we will solve the recurrence T_n = T_n-1 + k. This problem is multiplication problem. This recurrence will be solved by inspection. In inspection we write down enough terms such that we can guess the solution.
T₁ = k. Hence,
T_n  = T_n−2 + k + k
T_n  = T_n−3 + k + k + k
T_n  = T_n−4 + k + k + k + k
We can guess the solution to be T_n = nk as the number of k's increases as we use the lower terms of the recurrence and it increases by a definite number..

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.

Quadratic equation

A quadratic expression when equated to zero is called a quadratic equation. A quadratic equation looks like this ax² + bx + c = 0. In this post I will show you a method which is very good to find the solution of a quadratic equation. I found this method.Let us look at the method.

ax² + bx + c = 0

We can write it as,
(x + b/a)x + c/a = 0
or (x + b/a)x = -c/a

Let A = x + b/a and B = x
Then,A-B = b/a, AB = -c/a and A+B = 2x + b/a
Applying the identity (A+B)² = (A-B)² + 4AB
(2x + b/a)² = (b/a)² - 4c/a = (b² - 4ac)/a²
2x + b/a = ±(1/a)√(b² - 4ac)
x = (-b ± √(b² - 4ac))/2a

As we can see there are two values which satisfy the equation hence the number of solutions is two and there are two roots. As the solutions of a quadratic equation are called roots.

Let us analyze the roots i.e. when they are real. The value under the square root is positive if b² - 4ac is positive. When such condition arises then the roots are real. The value b² - 4ac is called the discriminant. If the discriminant is equal to zero then both the roots are equal. If the discriminant is negative then both the roots are imaginary and they occur in conjugate pairs. If the roots are real and distinct then the graph cuts the x-axis at two different points. If the roots are real and equal then the graph cuts the x-axis at one point. If the roots are imaginary then the graph does not cut the x-axis.

The graph below shows two real roots.
x² + 5x - 2 = 0

The graph below shows two real roots
x² + 5x + 6.25 = 0

The graph below represents when roots are imaginary.

x² + 5x + 8 = 0

Linear equations in two variables

Let us first look at the form of linear equations in two variables.

ax + by = c

Such equations arise when we have two things changing at the same time. The simplest is to watch it on a graph. The x-axis forms one variable and the y-axis forms the other variable. A line has infinite number of solutions but we have to get a unique solution. And this is possible only when the number of lines is two and they intersect or in other words we have two linear equations in two variable. Hence to solve a system of two variable linear equation the required equations are

ax+ by = c
dx + ey = fA system of two variable linear equation is solved either by Substitution method or by Elimination method.

Substitution method
In substitution method we find the value of one variable in terms of the other and substitute it in the other equation.

ax + by = c ----(i)
dx + ey = f ----(ii)From (i) x = (c - by)/a

Substituting in second we get
d(c - by)/a + ey = f
dc/a - dby/a + ey= f
dc/a - (db - ea)y/a = f
dc - (db - ea)y = fa
(db - ea)y = (dc - fa)
y = (dc - fa)/(db - ea)

Substituting in (i) we get
ax + b(dc - fa)/(db - ea) = c
ax = c - b(dc - fa)/(db - ea)
ax = [c(db - ea) - b(dc - fa)]/(db - ea)
x = [cdb - cea - bdc + bfa]/a(db - ea)
x = (bf - ce)a/a(db - ea)
x = (bf - ce)/(db - ea)

x = (bf - ce)/(db - ea) = (ce - bf)/(ea - db)
y = (dc - fa)/(db - ea) = (fa - dc)/(ea - db)

Elimination method
In elimination method we eliminate one variable by equaling the other variable in both the equations. Then we substitute the value of first variable to get the value of other variable.

ax + by = c ----(i)
dx + ey = f ----(ii)
Multiplying (i) by e and (ii) by b, we get
aex + bey = ce ----(iii)
dbx + bey = fb ----(iv)

Subtracting (iv) from (iii) we get
(ae - db)x = (ce - fb)
x = (ce - fb)/(ae - db)
As substituted above in substitution method we get,
y = (fa - dc)/(ea - db)
x = (ce - bf)/(ea - db)
y = (fa - dc)/(ea - db)

Linear Equations in one variable

Linear Equations are the starting point of two branches of mathematics. One deals with roots and other deals with solutions. Linear equation can be single variable or multi variable. The complexity of the solutions increases as we move to higher number of variables. As we will see every degree equation can be represented as multi variable linear equations. The point here is that the equations are related and a quadratic and cubic has 3 and 4 unknowns in it. Let us talk about the single variable linear equation. A single variable linear equation is very simple and is taught from the very basic grades or classes. A single variable linear equation looks like this:

ax + b = 0.

There can be many variant of it. ax + b = c or ax + c = bx + d or ax + c = bx, etc.

All we need to know is basic arithmetic. All the laws of basic arithmetic also works in algebra. To solve a linear equation in one variable we perform operations like this:

Let us solve different types of linear equations:

1. ax + b = 0
   Transpose b to right hand side and divide by a.
   x = -b/a
2. ax + b = c
   Transpose b to RHS and divide by a.
   x = (c - b)/a
3. ax + b = cx
   Transpose cx to LHS and b to RHS
   ax - cx = -b
   (a-c)x = -b
   x = -b/(a-c)

Solution

The simplest way to solve a linear equation is to transpose all the constants to RHS and terms containing variable to LHS and divide RHS by the coefficient of resulting LHS.

Solution is a solution

Let us start with two equations

1. ax + b = 0
2. ax² + bx + c = 0

Among the above two equations one is linear and the other is quadratic. The solution of the linear equation is x = -b/a and the solution of the quadratic is (1/2a)(-b±√(b² - 4ac)). The point which I want to state is that the solution of a quadratic equation contains the solution of a linear equation. When we substitute c for zero in the quadratic equation then we get ax² + bx = 0. On factorizing we get (ax + b) = 0 and x = 0. Whose solutions are x = -b/a and 0. -b/a is the solution of ax + b = 0. And x = 0 comes because x is multiplied throughout to get the equation ax² + bx = 0.

Now according to me if solution of cubic exist then it must have the solution of quadratic contained in it. Similar is the case for the others. Let us compute the solution of a cubic equation when one root is known

The equation is ax³ + bx² + cx + d =0. If the roots are α, β and γ and γ is known then
α + β + γ = -b/a
αβ + βγ + γα = c/a
αβγ = -d/a
Expressing sum of two roots in terms of third
α + β = -b/a - γ
Expressing product of two roots in terms of third
αβ = c/a - (βγ + γα)
αβ = c/a - γ(β + α)
αβ = c/a - γ(-γ - b/a)
αβ = c/a + γ² + bγ/a
(α - β)² = (α + β)² - 4αβ
(α - β)² = (-b/a - γ)² - 4(c/a + γ² + bγ/a)
(α - β)² = (-b/a)² + (γ)² + 2bγ/a - 4c/a - 4γ² - 4bγ/a
(α - β)² = b²/a² - 4c/a - 3(γ)² - 2bγ/a
α - β = ±√[(b² - 4ac)/a² - 3(γ)² - 2bγ/a]
adding α - β and α + β
2α or 2β = -b/a - γ ± √[(b² - 4ac)/a² - 3(γ)² - 2bγ/a]
α or β = -b/2a - γ/2 ± (1/2a)√[(b² - 4ac) - 3a²(γ)² - 2baγ]

As we can see when one root γ = 0 and the equation is independent of d then the solution reduces to the solution of a quadratic equation.

Now it can be the solution of 5^th degree equation exist and it contains the solution of 4^th degree in it.

Bisection Method

There are many methods to find the roots by approximation and one of it is bisection method. This method uses the intermediate value theorem which states that if a continuous curve changes sign then it must have taken zero at some point.

We take the given function f(x) and two values a and b such that f(a)·f(b) < 0. Then we bisect the values a and b. Let it be c. c = (b-a)/2. If f(c)=0 then c is the root else if f(a)·f(c)<0 then b=c else a=c. We continue the same procedure again. This method converges very slowly.

The property of this method is that it does not use the value of f(x) as the formula to compute the root.

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

Linear equations

Linear equations are those equations which are linear in its variables i.e. the power of the variables are 1. There can be many kinds of linear equations depending upon the number of variables used. a single variable linear equation has one variable, a double variable linear equation has two variables, a triple variable linear equation has three variables and so on. The linear equations with two or more variables is called a multi-variable linear equation. A single variable linear equation looks like this: ax = b where x is the variable. A double variable linear equation looks like this: ax + by = c and a triple variable linear equation looks like this: ax + by + cz = d. The number of variables can extend to any number of variables.

The solution of a linear requires as much equations as there are variables in it. A single variable linear equation requires one equation. a double variable linear equation requires two variables and a three variable linear equation requires three equations. The number of variables is equal to the number of equations. When we find the solution of a set of linear equations then the solutions can be unique or dependent or no solution. The solution of a linear equation can be found with the help of substitution, elimination or with the help of determinants. The solution using determinants is very cumbersome so to simplify the steps we use Gauss algorithms which use matrix to solve such equations. Let us look at the solutions of some linear equations.

Single variable linear equation
A single variable linear equation ax = b has solution x = b/a. When we plot it on a graph then we get a straight line parallel to y-axis. As the value of y is not present in the equation therefore the solution is independent of y.

Double variable linear equation

A double variable linear equation can be solved either by substitution method or elimination method. The solution on a graph is the point of intersection of the lines represented by the two equations.

The equations are 2x − y = 3 and 4x + y = −6

Let us solve the above two equations by substitution method and elimination method.

Substitution method

In substitution method we express one variable in terms of the other and substitute it to get value of one variable. Then substitute it in any equation to get value of other variable.

Step 1:Express any one equation in terms of one variable.
2x−y = 3 ⇒ y = 2x − 3
Step 2:Substitute it in next equation.
4x + y = −6
⇒ 4x + 2x − 3 = −6
⇒ 6x = −3
⇒ x = −(1/2) = −.5
Step 3:Substitute value in the first equation.
y = 2x − 3 = 2(−.5) − 3 = −4
So solution is x = −.5 and y = −4

Elimination method

The elimination method is based on the fact that if one variable is removed from the equation then the reduced equation contains only one variable. After we can find the solution. Substitute one solution in one equation to get the other solution.

Step 1: Eliminate one variable by making coefficient of that variable equal in two equations.

2x − y = 3
4x + y = −6
Add the two equations
6x = −3
x = −(1/2) = −.5
Substitute in anyone
2(−.5) − y = 3
y = − 4

the linear equations of multi variables is found by Gauss elimination method. We will discuss about such methods in some other posts.

Quadratic Equation

A quadratic equation is an expression with highest power in it equal to 2 and equated to zero. The term with the power 2 must not have coefficient equal to zero. If the coefficient is equal to zero then the equation is not quadratic anymore. Usually a quadratic equation has three coefficients but when all the coefficients are divided by the coefficients of x² then the coefficients left are two. The three coefficients of a quadratic equation are a,b and c. When they are put in the equation the equation looks like ax² + bx + c = 0.
A quadratic graph looks like this

The equation of the above graph is y = x² + x − 6

When all the terms of the quadratic equation are made negative then the equation looks like below

The equation of the above graph is y = − x² − x + 6

The solution of a quadratic equation by algebra is visible to all. Let us find the solution of the quadratic equation by calculus and graph. As we can observe from the graph that the graph of the curve is symmetrical about the line x = -b/2a. we will prove this when we arrive at the solution of the quadratic equation. When we use calculus to arrive at the solution of a quadratic equation we make use of derivatives. Let y = ax² + bx + c. When the expression is differentiated for the first time then we get ^dy/_dx = 2ax + b. We have got the rate of change of slope as a linear function. Differentiate it again. we get ^d2x/_dx² = 2a. We have found the rate of change of slope of the slope as constant. This implies that the change in slope of the quadratic expression is linear. As the second derivative is positive so the point where ^dy/_dx = 0 is point of minimum value of y. The point is x = ^−b/_2a. Let us shift the y axis to this point. i.e. X = x + b/2a

The equation becomes y = a(X − ^b/_2a)² + b(X − ^b/_2a) + c
y = aX² − bX + ^b²/_4a + bX − ^b²/_2a + c
y = aX² − ^(b²−4ac)/_4a

The roots for the above equation is when y = 0 i.e.
aX² − ^(b²−4ac)/_4a = 0
aX² = ^(b²−4ac)/_4a
X² = ^(b²−4ac)/_4a²
X = ±√(b²−4ac)/2a

As X = x + b/2a
x + b/2a = ±√(b²−4ac)/2a
x = [− b ±√(b²−4ac)]/2a
The two solutions are x₁ and x₂ where x₁ = [− b + √(b²−4ac)]/2a
x₂ = [− b − √(b²−4ac)]/2a
The above method also tells us that the two roots are equidistant from the axis x = −b/2a. This also gives us an idea that the graph is symmetrical about x = −b/2a because the rate of change of slope is constant. And change is slope is linear.