Knowing the Algebraic formulae

Algebra is a branch of mathematics in which all unknown quantities are expressed as abstract symbols rather than specific numbers. Algebra is the branch of mathematics that helps in the representation of problems or situations in the form of mathematical expressions. By analyzing the mathematical expression one can understand various dependency of various factors associated with the problem.  Algebra is helpful to develop skill in problem solving . There ar many Algebraic formulae  Let us try to learn these algebraic formulae from fundamental concepts in mathematics.

a² – b² = (a + b) (a - b)

Adding  and subtracting a common  term ab with a² – b^2,  we get

a² + ab – ab - b² where a is commen in the first two terms andb is common in the last two terms.

a(a+b) – b(a+b), where (a+b) is common in both the terms, It is simplified into (a+b)    (a-b)

(a + b)² = a² + 2ab + b²

(a+b)²  = (a+b) (a+b) = a²  + ab + ab + b^2, = a²  + 2ab + b^2,. It gives a supplementary formula  a² + b² = (a + b)² – 2ab.

(a – b)² = a² – 2ab + b²

(a-b)² = (a-b)(a-b) = a²  - ab - ab + b^2, = a²  - 2ab + b^2. It gives a supplementary formula  a² + b² = (a - b)² + 2ab . It shows that (a-b)² , (a² + b²) , (a+b)²  are always in arithmetic progression with a common difference 2ab.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

(a+b+c) (a+b+c) = a²  ab + ac + ab + b² + bc + ac + bc + c² =  a² + b² + c² + 2ab + 2bc + 2ca.

(a – b – c)² = (a-b-c) (a-b-c) =  a² + b² + c² – 2ab + 2bc – 2ca.

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

(a+b)³  = (a+b) (a+b) (a+b) = (a² +2ab + b²⁾)(a+b) = (a³ + 2a² b +ab² ) + (a²b + 2ab² + b³) = a³ + 3a² b + 3ab² + b³ . The middle two terms have a common factor 3ab which reduces the formula into (a + b)³  =  a³ + b³ + 3ab(a + b).It provides another supplementary formula a³ + b³  =(a + b)³  - 3ab(a+b). Since (a+b) is common it is changed into  a³ + b³  =(a + b)³  - 3ab(a+b) = (a+b) [(a+b)² – 3ab] =(a+b)(a² –ab + b² ]

The same formula can be obtained directly from a³ + b³  by imply adding and subtracting  common and conjugate pairs a²b + ab²

a³ + b³ = a³ + a²b + b³ + b²a - a²b - ab²

The first two terms have a common factor a², the middle two terms have a common factor b², while the last two terms have a common factor ab..It reduces into

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

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

(a-b)³  = (a-b) (a-b) (a-b) = (a² -2ab + b²⁾)(a-b) = (a³ - 2a² b +ab² ) + (-a²b + 2ab² - b³) = a³ - 3a² b + 3ab² - b³ . The middle two terms have a common factor 3ab which reduces the formula into (a - b)³  =  a³ - b³ - 3ab(a - b).It provides another supplementary formula a³ - b³  =(a - b)³ + 3ab(a-b). Since (a-b) is common it is changed into  a³ -  b³  =(a - b)³  -+ 3ab(a- b) = (a-b) [(a-b)² + 3ab] =(a-b)(a² + ab + b² ]

The same formula can be obtained directly from a³ - b³  by imply adding and subtracting  common and conjugate pairs a²b + ab²

a³ - b³ = a³ - a²b - b³ + b²a + a²b - ab²

The first two terms have a common factor a², the middle two terms have a common factor b², while the last two terms have a common factor ab..It reduces into

a²(a-b) + b² (a-b) –+ab(a-b) = (a-b) (a² + ab + b² )

By following  a similar procedure one can establish algebraic formula for a^{n –}  bⁿ   where the exponent may have any integral value.. For example at first let us consider a⁴ – b^{4 .}.In the first step add and subtract  ( a³ b – ab³) ,a conjugate pair with power 4

a⁴ – b⁴  = (a+b) ( a³ – a²b + ab² – b³ )                                                                                             a⁴ – b⁴  = (a - b) ( a³ + a²b + ab² + b³ )

a⁴ – b⁴  = a⁴ – b⁴   + ( a³ b – ab³) - ( a³ b – ab³)

            = a³ (a+b) – b³ (a+b)  - ( a³ b – ab³)  

In the second step add and subtract a² b²  a conjugate pair  with power 4

a⁴ – b⁴  = a³ (a+b) – b³ (a+b)  - ( a³ b – ab³) - a² b²  + a² b²

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

(a+b) is common in all the terms which reduces the relation

       a⁴ – b⁴  = (a+b) ( a³ – a²b + ab² – b³ )

a⁴ – b⁴   can be expressed in another way.In the first step subtract and add (-a³ b + ab³), the conjugate pair with power 4.

a⁴ – b⁴  = a⁴ – b⁴   + ( -a³ b + ab³) - (- a³ b + ab³)

            = a³ (a-b) + b³ (a-b)  + ( a³ b – ab³)

In the second step add and subtract a² b²  a conjugate pair  with power 4

a⁴ – b⁴  = a³ (a-b) + b³ (a-b)  + ( a³ b – ab³) + a² b²  - a² b²

           = a³ (a-b) – b³ (a-b) +a²b (a- b) + ab² (a-b)

(a - b) is common in all the terms which reduces the relation

       a⁴ – b⁴  = (a - b) ( a³ + a²b + ab² + b³ )

This can be obtained directly  by factorizing the therm a⁴ – b⁴

 a ⁴ – b⁴ = (a² + b² )(a² - b²) = (a² + b² )(a+b) (a-b)

a⁴ – b⁴ =  (a+b) [(a-b) (a² + b² )] = (a+b)( a³ –a²b + ab² – b³ )

a⁴ – b⁴ =  (a-b) [(a+b) (a² + b² )] = (a-b)( a³ +a²b + ab² + b³ )

In the same way one can prove

a⁵ – b⁵ = (a-b) (a⁴ + a³b + a²b² + ab³ + b⁴)

a⁶  –  b⁶ = (a-b) (a⁵ + a⁴b + a³b² + a²b³ + ab⁴ + b⁵)

If n is a natural number aⁿ– bⁿ = (a-b) (a^n-1 + a^n-2b + a^n-3b² + …… ab^n-2+  b^n-1)

aⁿ – bⁿ is divisible by (a-b) what ever may be the value of the exponent. If the exponent is even it is divisible by both (a-b) and (a+b).

aⁿ – bⁿ  can also be expressed as a product with (a+b) for odd exponents. For example let us consider the case a⁵ – b^5.. In the first stage add and subtract a conjugate pair  for 5 th power a⁵ – b⁵   = a⁵ – b⁵  + a⁴b – b⁴a  + ( - a⁴b + b⁴a) =a⁴(a+b) – b⁴ (a+b) – a⁴b + b⁴a. in the second stage add and subtract –a³ b² + a² b³    we get a⁵ – b⁵   = (a+b) (a⁴ – b⁴) – a³b (a+b) + ab³ (a+b)  + a³ b² – a² b³ = (a+b) (a⁴ – a³b +  a³b -  b⁴)  + a² b² (a-b)

a³ – b³ = (a+b) { a² – b²) – ab(a-b)

a⁵ – b⁵   = (a+b) (a⁴ – a³b +  a³b -  b⁴)  + a² b² (a-b)

a⁷ – b^{7 =} (a+b)( a⁶ – a⁵b + a⁴b^2- a²b⁴ +a b⁵ – b⁶) – a³b³(a+b)

In general

a²ⁿ⁺¹ – b^{2n +1} = (a+b) [a²ⁿ – a^2n-1b + a^2n-2 b²

                                    -b²ⁿ + b^2n-1a – b^2n-2 a²]  + (-1)ⁿ aⁿbⁿ (a-b)

if n is even (n = 2k), aⁿ + bⁿ = (a + b)(a^n-1 – a^n-2b +…+ b^n-2a – b^n-1)

If n is odd (n = 2k + 1), aⁿ + bⁿ = (a + b)(a^n-1 – a^n-2b +a^n-3b²…- b^n-2a + b^n-1)

(a + b + c + d)² = a² + b² + c² + d² + 2ab + 2 ac + 2ad+2bc + 2bd + 2cd

(a + b + c + d)² = [ (a+b) + (c+d)]² = (a+b)²  + (c+d)² + 2(a+b)(c+d)

                                                        = a² + b² + 2ab + c² + d² + 2cd + 2ac + 2ad + 2bc + 2bd

                                                       = a² + b² + c² + d² + 2ab + 2ac + 2ad + +2bc + 2bd +2cd

Solution of Quadratic equation

The general form of  a quadratic equation is ax² + bx + c = 0 ,where a,b,c are some coefficients, invariables, while x has certain specific values which satisfy the equation, called solutions or roots.

Determination of roots of the quadratic equation

Step-1 : Divide by a, the coefficient of x^{2 ;}   x² + (b/a) x = - c/a

Step-2 : Make the terms in LHS to be a square .For that  the coefficient of x = b/a is halved and then squared .It gives b²/4a². And is added on both sides

     x² + (b/a) x +  b²/4a²= - c/a + b²/4a²

       (x + b/2a)² =  b²/4a²  - c/a

       (x + b/2a)  =  ±[ (b/2a) ²  - c/a]^1/2

       x = [ -b  ± (b² – 4ac)^1/2 ]/2a

Sum of two roots

Let x₁ and x₂ be two roots that satisfy the   quadratic equation  ax² + bx = -c

 ax₁ ² + bx₁ = -c and   ax₂ ² + bx₂ = -c

Subtracting one from the other

a( x₂² – x₁²) + b(x₂ – x₁) = (x₂ – x₁) [ a(x₂ + x₁) + b] = 0

If the two roots are different  a(x₂ + x₁) + b = 0 or x₁ + x_{2  = -} b/a

Product of two roots

ax₁² + bx₁ = - c and   ax₂² + bx₂ = - c Adding together we

a( x₂² + x₁²) + b(x₂ + x₁) = -2c_.

 Adding 2x₁ x₂  both side and reduce the equation  to

a( x₂² + x₁²)² + b(x₂ + x₁) = -2c_. +  2x₁ x₂

Substitute the value for sum of the two roots x₂ + x₁ = -b/a we have x₁x₂  = c/a