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).

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.

Whether real or complex (Quadratic Equation)

This post deals with the conditions to check whether the roots are real or complex. The roots are real when the graph really cuts the x-axis. When we look at the graph of a quadratic equation we see a cup like structure or cap like structure. Let us check when it is cup like and cap like. We call the cup like structure concave upward and a cap like structure convex upward. The quadratic equation is ax² + bx + c = 0. Let it be equal to y or f(x). Now the rate of change of curve is f'(x) = 2ax + b. f(x) is differentiated to obtain f'(x). We again differentiate f'(x) to obtain f''(x), f''(x) = 2a. Now we find the x-coordinate of the point where the slope is zero i.e. f'(x) = 0.

2ax₁ + b = 0
⇒ x₁ = −b/2aNow the point on the graph is concave upward or convex upward according as the point at the x-coordinate x₁ = −b/2a is maxima or minima. This thing is determined by checking the value of f''(x). If it is positive then the curve has minima and if it is negative then the curve is maxima. We have f''(x) = 2a. It is positive if a is positive and negative when a is negative. Hence we can make following conclusions

1. concave : The graph is concave upward if a is positive.
   
   
   
   The graph given above is of the function f(x) = 2x² + 4x - 3.


2. convex : The graph is convex upward if a is negative.
   
   
   The graph given above is of the function f(x) = −2x² + 4x + 3

Now, we find the y coordinate of the point where it moves upward or downward traveling from x = −∞ at x = -b/2a. It changes its direction at −b/2a to opposite direction of y-axis.
y = ax² + bx + c
at x = x₁ = −b/2a
y₁ = a(−b/2a)² + b(−b/2a) + c
= b²/4a − b²/2a + c
= − b²/4a + c
= − (b² − 4ac)/4a

Now, four conditions arise as given in the following table

a	b2−4ac	Comment
−ve	−ve	The value of y1 is −ve and the curve is concave downward. The curve will be present below y1. Hence it will cut the x-axis at no points. So the roots are imaginary.
−ve	+ve	The value of y1 is +ve and the curve is concave downward. The curve will be present below y1. Hence it will cut the x-axis at one or two points. So the roots are real and may be equal if y1 = 0.
+ve	−ve	The value of y1 is +ve and the curve is concave upward. The curve will be present above y1. Hence it will cut the x-axis at no points. So the roots are imaginary.
+ve	+ve	The value of y1 is −ve and the curve is concave upward. The curve will be present above y1. Hence it will cut the x-axis at one or two points. So the roots are real and may be equal if  y1 = 0.

Binomial theorem

We are interested in finding relation between many things. When (a + b) is raised to different powers then relation between previous expansion with the next expansion has a very interesting result. This post is about this and binomial theorem. It is called binomial as it has two variables in it. In this post I will show you how we can arrive at the coefficients of the Binomial Expansion. The foundation of binomial theorem lies in Pascals Triangle. Pascals triangle is an interesting topic and I will cover it in detail. A Pascals Triangle looks like this.

                              1
                            1   1
                          1   2   1
                        1   3   3   1
                      1   4   6   4   1
                    1   5   10  10  5   1
                      so on.  

A pascals triangle is generated from coefficients of the expansion of different powers raised to (a + b). Let us expand some of the powers.

(a + b)1 = (a + b)

(a + b)2 = (a2 + 2ab + b2)

(a + b)3 = (a3 + 3a2b + 3ab2 + b3)

(a + b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4)

(a + b)5 = (a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5)

Let us express the coefficients in other form:

1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6  1
so on

Some points to note

1. The first column is a constant.
2. The second column is an arithmetic sequence whose common difference is 1. i.e. the terms are represented by n/1.
3. The third column needs some explanation
   The sequence is 1 3 6 10 15...
   The difference between successive terms is 2 3 4 5....
   The difference of elements again is 1 1 1.
   Don't worry about the method which I am going to follow. Learn the method. If on finding successive differences you reach at a constant in 2 steps. Suppose the sequence is formed by the second power. Here, the equation is an² + bn + c. Where n is the position of term. Let us take the first three terms of the sequence. Then we get a system of simultaneous equations.
   
   a + b + c = 1
   4a + 2b + c = 3
   9a + 3b + c = 6
   solving we get, a = 1/2, b = 1/2 and c = 0
   The equation is (1/2)(n²+n). In the table the sequence start from row 2 so we will replace n by (n-1).
   Then the equation becomes (1/2)((n-1)² + (n-1)).
   Reducing into factors we get n(n-1)/2 = n(n-1)/(1x2)
   The above thing in combinations is represented as ⁿC₂.
4. Following the above process and assuming the sequence depends on third power and solving and replacing n by (n-2) we get ⁿC₃.
5. Following the above process and assuming the sequence depends on fourth power and solving and replacing n by (n-3) we get ⁿC₄.
6. Following the above process we get ⁿC₅.
7. Following the above process we get ⁿC₆.

Hence the expansion of (a+b) raised to n^th power is

The binomial theorem
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b + ⁿC₂a^n-2b² + ... + ⁿC_n-1ab^n-1 + ⁿC_nbⁿ

Finite and Infinite Sequences

If a set of numbers are such that they follow a specific pattern is called a sequence. If all the terms are added in order then it is called a series. The main types of series and sequences which are taught till intermediate are Arithmetic series and sequences, Geometric Series and Sequences and Harmonic Sequence.

Look at the following sequences:
1. 2, 4, 6, 8, 10, 12,...
2. 3, 9, 27, 81, 243,...
3. 1/2, 1/4, 1/6, 1/8, 1/10, 1/12,...

• Arithmetic sequence: The first sequence (2, 4, 6, 8, 10, 12,...) among the three sequences is an arithmetic sequence. 2+4+6+8+10+12+.... is an arithmetic series. A general arithmetic sequence looks like this.
  a, a+d, a+2d, a+3d,... and its corresponding series is a + a+d + a+2d + a+3d +....
  a is called the first term and d is the common difference. The first term of the example sequence is 2 and common difference is 2.
  
  The n^th term of an A.S. is
  t_n = a + (n-1)d
  and sum of first n terms is
  S_n = (n/2)(2a + (n-1)d)
  n is the number of terms.
  
  How to check whether a sequence is arithmetic sequence.
  
  Difference is found by subtracting any term with its successor term. When all the differences are the same then the sequence is arithmetic sequence and the difference is called common difference.
  


• Geometric sequence: The second series (3, 9, 27, 81, 243,...) among the three sequences is a geometric sequence. 3+9+27+81+243+.... is a geometric series. A general geometric sequence looks like this.
  a, ar, ar², ar³,... and its corresponding series is a + ar + ar² + ar³ + ....
  a is called the first term and r is the common ratio. The first term in the example is 3 and common ratio is 3.
  
  The n^th term of an G.S. is
  t_n = ar^(n-1)
  and sum of first n terms is
  S_n = a(1-rⁿ)/(1-r) if r<1
  else
  S_n = a(rⁿ-1)/(r-1) if r>1
  n is the number of terms.
  
  How to check whether a sequence is geometric sequence.
  
  Ratio is found by dividing any term with its previous term. When all the ratios are the same then the sequence is geometric sequence and ratio is called common ratio.
  


• Harmonic sequence: The third series (1/2, 1/4, 1/6, 1/8, 1/10, 1/12,...) among the three sequences is a harmonic sequence. 1/2+1/4+1/6+1/8+1/10+1/12+.... is a series. A general Harmonic sequence looks like this.
  1/a, 1/(a+d), 1/(a+2d), 1/(a+3d),... and its corresponding series is 1/a + 1/(a+d) + 1/(a+2d) + 1/(a+3d) +....
  
  (1/a) is called the first term. There is no formula to find the sum of n terms of the series.
  
  How to check whether a sequence is harmonic sequence.
  
  Find the reciprocal of all the terms. If the sequence formed is an arithmetic sequence then the sequence is harmonic sequence.

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.

Factors, LCM and HCF or GCD

Factors
2a²bc has 2,a,b and c as factors. Factors are used to find the L.C.M. and H.C.F. of expressions. L.C.M. and H.C.F. also play an important role in mathematics. Suppose you are given some sticks of different sizes and you are asked to find a length whose length will be divisible by all the sticks of different sizes. You need the knowledge of L.C.M. Again consider you are given many sticks and you are asked to find a length which will measure all of them in integer terms. You will need the knowledge of H.C.F.

L.C.M.

Rules to find L.C.M. (Lowest common multiple)

1. Find the L.C.M. of the constant terms.
2. Find the power of all the expressions for each variable or symbolic constant.
3. Find the maximum power for each variable or symbolic constant among each expression.
4. List the variables with the maximum power of each.
5. Multiply it with the L.C.M. of the constant term.

Evaluate L.C.M. of 3a²b⁶c and 8a³b²cd

1. L.C.M. of the constant terms is 24.
2. Power of all the expressions for each variable or symbolic constant.
   a:2 and 3
   b:6 and 2
   c:1 and 1
   d:0 and 1
3. Maximum power for each variable or symbolic constant among each expression.
   a:2 and 3 max = 3
   b:6 and 2 max = 6
   c:1 and 1 max = 1
   d:0 and 1 max = 1
4. List the variables with the maximum power of each.
   a³b⁶cd
5. Multiply it with the L.C.M. of the constant term.
   24a³b⁶cd

L.C.M. is 24a³b⁶cd

H.C.F. or G.C.D.

Rules to find G.C.D. or H.C.F. (Greatest common Divisor or Highest Common Factor)

1. Find the G.C.D. of the constant terms.
2. Find the power of all the expressions for each variable or symbolic constant.
3. Find the minimum power for each variable or symbolic constant among each expression.
4. List the variables with the minimum power of each.
5. Multiply it with the G.C.D. of the constant term.

Evaluate G.C.D. of 3a²b⁶c and 8a³b²cd

1. G.C.D. of the constant terms is 1.
2. Power of all the expressions for each variable or symbolic constant.
   a:2 and 3
   b:6 and 2
   c:1 and 1
   d:0 and 1
3. Minimum power for each variable or symbolic constant among each expression.
   a:2 and 3 max = 2
   b:6 and 2 max = 2
   c:1 and 1 max = 1
   d:0 and 1 max = 0
4. List the variables with the maximum power of each.
   a²b²cd⁰
5. Multiply it with the G.C.D. of the constant term.
   a²b²c

G.C.D. is a²b²c

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.

Operations in algebra (Multiplication and Division)

Multiplication and division in algebra rests on exponents and ordinary division. The ordinary division is applied to coefficients and the exponents are applied to symbolic constants and variables. Exponents are result of successive multiplication. Some rules of exponents that are useful in multiplication and division of algebra are:

1. a^maⁿ = a^m+n
2. a^m/aⁿ = a^m−n
3. 1/a^m = a^−m

Here the first rule is useful in multiplication and the second rule is useful in division. One more important point is to be noted here. The law of multiplication for exponents is applied on similar symbolic constants and variables. Similar is the case for division. All the symbolic constants and variables are grouped and the law of exponents are applied. Then the different symbolic constants are multiplied or divided as the case may be.

Procedure for division and multiplication in algebra

1. Group the symbolic constants and variables.
2. Perform the division or multiplication on coefficients.
3. Apply the rule of exponents for similar symbolic constants and variables.
4. Solve.

Example 1:
Multiply 2a²b³ and 5a³b²
2a²b³×5a³b²
= 2×5a²a³b³b²
= 2×5a⁽²⁺³⁾b⁽³⁺²⁾
= 10a⁵b⁵

Divide 2a²b³ by 5a³b²
2a²b³/(5a³b²)
= (2/5)a²b³×a⁻³b⁻²
= (2/5)a²a⁻³b³b⁻²
= (2/5)a⁽²⁻³⁾b⁽³⁻²⁾
= (.4)a⁻¹b¹

Operations in algebra (Addition and Subtraction)

There are four operations which can be performed on variables. These are addition, subtraction, multiplication and division. Repeated multiplication and division are parts of exponents.

Let us consider three arithmetic expressions.
1×3+2×3+3×3+4×3 = 3 + 6 + 9 + 12 = 30
1×5+2×5+3×5+4×5 = 5 + 10 + 15 + 20 = 50
1×8+2×8+3×8+4×8 = 8 + 16 + 24 + 32 = 80
When we look at the above expressions we find that 3 changes to 5 and 5 changes to 8.

As the value changes we can put a symbolic constant in place of it. Let it be a. Then the expression turns to
1×a+2×a+3×a+4×a
We can write
1a+2a+3a+4a
as the symbol of product in algebra can be replaced by nothing.

The numbers occurring in front of the symbolic constants are called the coefficients. They are also called coefficients if the symbolic constants are replaced by variables.

To perform the operation of addition simply add the coefficients of the symbolic constants.

Here we get (1+2+3+4)a = 10a

When we put different values of a, we get
a=3, 10a = 30
a=5, 10a = 50
a=8, 10a = 80.
Hence we have verified the operation.

Subtraction
To perform subtraction simply do subtraction for the coefficients.
23a − 7a = (23 − 7)a = 16a

Like and unlike terms

When the symbol for the symbolic constant or the variables are same for all the terms then the terms are called the like terms. Else they are unlike terms.

3a and 6a are like terms. Some examples of like terms are
3d and 7d, 23ad and 67ad, 3x and 7x, 4xy and 5xy, etc.
When the variables or symbolic constants are different for the considering terms then the terms are called unlike terms.Some examples of unlike terms are
3x and 6y, 56df and 78hu, etc.

Rules for addition and subtraction

1. Group the like terms.
2. Add or subtract the coefficients.
3. Only like terms can be added to or subtracted from each other.

Example 1:
Add 2x + 3y + 4z and 3x + 7x + 7y + 6z
2x + 3y + 4z + 3x + 7x + 7y + 6z
= 2x + 3x + 7x + 3y + 7y + 4z + 6z
= (2+3+7)x + (3+7)y + (4+6)z
= 12x + 10y + 10z

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.

Method of differences

There are many occurrences when we have certain numbers and we have to find relation between them. The relation is usually in form of expressions. The expression can contain log function, exponential function and algebraic functions.The simplest among them is to find polynomial expressions.In this post I will describe how to find functions when we know a series which has its terms separated by certain constant. The method is similar to differentiation but a lot different from them.

Suppose we have the numbers (from a function) like given in the following table

1	6	15	28	45	66	91
We find the difference of successive term
5		9	13	17	21	25
We again find the difference of successive term
4			4	4	4	4
We find the differences till we get a constant term for all

As we got the constant term so we can create a table of possible polynomials for all. We start from the bottom.

c''	4			4	4	4	4
a'x+b'	5		9	13	17	21	25
ax2+bx+c	1	6	15	28	45	66	91

We got a quadratic because we had three rows in the table.
Now we have
f(x) =ax²+bx+c
f(0) = c = 1
f(1) = a + b + c = 6
f(2) = 4a + 2b + c = 15
from f(0): c =1
from f(1): a + b = 5
from f(2): 4a + 2b = 14
so solving the simultaneous equations a = 2 and b = 3 hence the function is f(x) = 2x² + 3x + 1.
Above it must be known from where the values start for x and what is their differences in each step.

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.

Probability and Statistics

Probability and Statistics is one of the main branch of mathematics. It deals with population and chance. In statistics we read about stats of different kinds their arrangement, collection, classification, analysis and interpretation. In probability we deal with the study of chances and their ways. Probability mainly consists of permutations and combinations. There are many uses of probability in statistics. Let us have a look at what we study in probability and statistics.

Statistics has two different meanings. One is the data and other is a discipline. The data is facts and figures which we come across different conditions in our life and discipline refers to the collection, analysis and interpretation of those data. In the beginning of the study of statistics we deal with its different types of distributions. Then we deal with measures of central tendency. They are mean, median and mode. After it we deal with measures of dispersion. They are range, mean deviation and standard deviation. This study gives us a broad view of the data. The mean is the value which is the most probable when the whole data is taken into account. Median tells us about the middle, quartile, decile and percentile. Mode tells us which value occurs the most. The measure of dispersion tells us about the trend of the data, how much it is away with respect to the mean and the square of the mean. Then we deal with skewness and kurtosis. A measure of skewness would tell us how far the frequency curve of the given frequency distribution deviates from a symmetric one. On the other hand, a measure of kurtosis gives us some information about the degree of flatness (or peakedness) of the frequency curve. Then we deal with how one type of data is related to other type of data. Such study is dealt in correlation and regression. These things deals with statistics. To study more in statistics we need knowledge of probability.

In probability we deal with Random Experiment, sample space, event and algebra of events. Then we deal with probability of discrete sample space. After it we study about probability distributions. These are Bernoulli, Binomial, Multinomial, Hypergeometric, Geometric, Poisson and Negative Binomial. These are standard probability distributions. Then we deal with univariate distributions. After it we deal with standard continuous distributions. They are Normal, Gamma or Exponential, Beta Distributions. Then we deal with bivariate distributions. Then comes functions of random variables chi-square distribution, t-distribution and F-distribution. Then we end the topic by studying the methods used in factories.

Complex numbers and its origin

When we hear of complex numbers, one thing comes to our mind and it is something complex. One thing I would like to tell that complex numbers are not as complex as it seems to be. Even Euler wanted it to be taught in schools. It is a combination of symbols, idea and presentation. the evolution of complex numbers is attributed to the equation x² + 1 = 0. As we can see from the above equation that the square of a number is negative. And we know that the square of any integer cannot be negative. To solve such type of equations we use imaginary numbers. The basic of imaginary numbers is √-1. We represent it by i. One of the most interesting thing in the complex numbers is the product of two complex number i. i×i = √-1 × √-1 is not 1 as can be thought. The reasoning of the above is √-1 × √-1 = √(-1×-1) = √1 = 1. But we must keep in mind that now the number can not be thought as per the rules of real algebra. Any negative number n under the root sign can not be thought as a combination of a number, sign and a root. But it is to be interpreted as a sign i√n. Where the number n denotes the magnitude of the negative number. So the new reasoning is i×i = i² = -1.

A complete imaginary number consists of real as well as imaginary parts.But one thing which comes to our mind is what is the representation of these numbers. To show such kind of numbers we use Argand plane or complex plane. A complex number has two parts the real part and the imaginary part. The real part is denoted on the real axis and imaginary part is denoted on the imaginary axis. The real part is equivalent to the x-axis and the imaginary part is equivalent to the y-axis of the Cartesian Coordinates.

Euler established a relationship between e, i, cos θ and sin θ. He found that they form a definition and it is
e^iθ = cos θ + i sin θMuch of the mathematics of complex numbers is result of identity. We find that e^iθ = cos θ + i sin θhas no simple proof but as we expand e^iθ we find that its representation is like this
e^iθ = 1 + iθ + i²θ² + i³θ³ + i⁴θ⁴ + ...
= 1 + iθ − θ² − iθ³ + θ⁴ + ...
= (1 − θ² + θ⁴ ...) + i(θ − θ³ + ...)
We know that (1 − θ² + θ⁴ ...) is expansion of cos θ and (θ − θ³ + ...) is the expansion of sin θ
= cos θ + i sin θ

Similarly by induction we can prove that (cos nθ + i sin nθ) = (cos θ + i sin θ)ⁿ.

There are many such results which form the basis of complex numbers. But the origin of complex numbers if the result of deep knowledge of mathematics as it takes ideas from different fields.

Solution of Quadratic by graph

The solution of a quadratic equation, cubic equation is possible with graph. I have found a method by which we can find the solution of a quadratic, cubic, etc, equations with the help of graph when one of the solution is known. The solution consists of finding a equation which relates the roots of the equation. It is very easy to find such a solution. Although the equation consists of only two roots but when we examine the roots then we can find all the roots of the equation. We will find the expression for quadratic equation in this post and discuss cubic equation in the next post. The method goes like this.

1. Let the equation be ax² + bx + c = 0
2. Take two roots of the equation. Let it be x₁ and x₂.
3. Form one equation with one root.
   ax₁² + bx₁ + c = 0 ---(i)
   Form one more equation with other root.
   ax₂² + bx₂ + c = 0 ---(ii)
   
4. Subtract (ii) from (i) and solve it
   (ax₁² − ax₂²) + (bx₁ − bx₂) + c − c = 0
   a(x₁² − x₂²) + b(x₁ − x₂) = 0
   (x₁ − x₂)[a(x₁ + x₂) + b] = 0
   Suppose both the roots are different, then
   [a(x₁ + x₂) + b] = 0
   (x₁ + x₂) = −b/a
   
You will get a relation in the roots.
5. Plot the relation on a graph by taking x₁ to be y and x₂ to be x.
   The relation becomes y = −x − (b/a)

Graphic Solution of two quadratic expressions y = x² + 5x + 6 and y = x² − 5x + 6 and their respective relation equation between roots are given. The roots can be found by checking where the quadratic graph cuts the x-axis.

The two straight lines gives the solution of the equations x² + 5x + c₁ = 0 and
x² − 5x + c₂ = 0. If we know one root of the equation then we can find the other root by looking at the point's corresponding coordinate point where one coordinate is equal. For example as one roots of x² + 5x + 6 = 0 is 2 so the corresponding point on the curve taking x = 2 is (2,3) so the other root is 3. We can also find the roots by looking at the point (a,b) where the area formed by the rectangle between x-axis, y-axis, x = a and y = b have area equal to the constant term. The condition for the roots for a particular equation is that the roots has product equal to c/a, where c is the constant term and a is coefficient of x².

Solution of a Cubic by graph

As we have seen in the post 'Solution of Quadratic by Graph' we can represent solution of a quadratic equation. Similarly we can represent solution of a cubic equation also by graph. The method is the same we represent a relation between two roots. The relation gives us a graph which has solution as the value of x coordinate and the corresponding y coordinate. In the case of cubic we get an ellipse. Let us first find the relation between two roots of a cubic equation.

1. Let the equation be ax³ + bx² + cx + d = 0
2. Take two roots of the equation. Let it be x₁ and x₂.
3. Form one equation with one root.
   ax₁³ + bx₁² + cx₁ + d = 0 ---(i)
   Form one more equation with other root.
   ax₂³ + bx₂² + cx₂ + d = 0 ---(i)
   
4. Subtract (ii) from (i) and solve it
   (ax₁³ − ax₂³) + (bx₁² − bx₂²) + (cx₁ − cx₂) + d − d = 0
   a(x₁³ − x₂³) + b(x₁² − x₂²) + c(x₁ − x₂) = 0
   (x₁ − x₂)[a(x₁² + x₂² + x₁x₂) + b(x₁ + x₂) + c] = 0
   Suppose both the roots are different, then
   [ax₁² + ax₂² + ax₁x₂ + bx₁ + bx₂ + c] = 0
   [ax₁² + ax₁x₂ + bx₁ + ax₂² + bx₂ + c] = 0
   [ax₁² + x₁(ax₂ + b) + (ax₂² + bx₂ + c)] = 0
   solving in x₁ we get,
   x₁ = (1/2a)( − (ax₂ + b)±√((ax₂ + b)² − 4a(ax₂² + bx₂ + c)))
   x₁ = (1/2a)( − (ax₂ + b)±√(−3a²x₂² − 2abx₂ + b² − 4ac))
   
You will get a relation in the roots.
5. Plot the relation on a graph by taking x₁ to be y and x₂ to be x.
   The relation becomes y = (1/2a)( − (ax + b)±√( − 3a²x² − 2abx + b² − 4ac))
   Divide the relation into two parts one with + sign and other with − sign. Plot them.

Solution of the cubic equation in the above graph is 1,2 and 3. As we can see from the graph that the solution exist where the line x=1 or x=2 or x=3 cuts the curve. When we look at the line x = 1, then we find that the line cuts the curve at y=2 and y=3. Similarly we see that the curve is cut by the line x=2 at y=1 and y=3. So is the case with x=3, where y=1 and y=2. So the solutions of the equation is the points satisfying the condition that the product of the roots is equal to the negative of the constant term divided by coefficient of x³.

Similarly on the guidelines for the previous graph we find that the solution is 1,2 and 4.

When the graph is plotted then the roots of the family of curves ax³ + bx² + cx + k = 0, where k is parameter is given by the curve which is like the ellipse. If one root is known to be α. Then other two roots can be found by looking at the points where the line x = α intersect the ellipse. By noting down the two values of y where the line x = α intersect we get the three roots. The two values of y and one value of x or two values of x and one value of y are the three roots of the curve.