Ag 02 || QUADRATIC POLYNOMIALS AND EQUATIONS (Type 02) a² + b² + c² + 2ab + 2bc + 2ca = 0

UADRATIC POLYNOMIALS AND EQUATIONS (Type 02) 

ax² + bx + c type

By

Om Jitender Singh Tomar 

(ॐ जितेन्द्र सिंह तोमर)

M.A., B.Ed., DNYS, MASSCOM

21/2/21/12/2021

QUADRATIC EQUATIONS

Today, we will discuss with quadratic polynomial, quadratic polynomial are polynomial of the second degree with at least one term.

The general form of a quadratic polynomial is ax² + bx + c. 

QUADRATIC EQUATIONS

Now, we will discuss with quadratic equations, quadratic equations are polynomial of the second degree with at least one term.

The general form of a quadratic equations is ax² + bx + c = 0. 

कोई भी द्विघात समीकरण ax² + bx + c = 0. के रूप में होती है। 

जहां 

a =  x² का गुणज

b =  x का गुणज

c =  अचर पद

ax² + bx + c = 0

ax² + mx + nx + c = 0

FACTOR BY FACTORIZATION METHOD

x² + 8x + 12 = 0

x² + 6x + 2x + 12 = 0

x( x + 6) + 2(x + 6) = 0

( x + 6) (x + 2) = 0

( x + 6) = 0     or   (x + 2) = 0

 x + 6 = 0        or    x + 2 = 0

 x = – 6          or    x = – 2 

 x = – 6,  – 2 

Mind something that we will get four types of the equations here.

(01)

              ax² + bx + c = 0

              ax² – bx + c = 0

              ax² + bx – c = 0

               ax² – bx – c = 0

(02)

In Type 1 and 2, the product of a and c is positive (+) so that we will take such type of two parts of b, whose sum and product is equal to ac.

Condition 1

              ax² + bx + c = 0

              a × c = b (pair with sum b)

If +b = +m & +n (both are Positive)

              ax² + mx + nx + c = 0

Condition 2

              ax² – bx + c = 0

              a × c = b (pair with sum b)

If –b = –m & –n (both are Negative)

              ax² + mx + nx + c = 0

In Type 3 and 4, the product of a and c is Negative (–) so that we will take such type of two parts of b, whose difference and product is equal to ac.

Condition 3

              ax² + bx – c = 0

              a × c = b (pair with dif. b)

If +b = +m & –n 

(Sign of both parts are opposite and greater part will bear the sign of b) 

              ax² + mx – nx – c = 0

Condition 4

              ax² – bx – c = 0

              a × c = b (pair with dif. b)

If –b = –m & +n 

(Sign of both parts are opposite and greater part will bear the sign of b) 

              ax² – mx + nx – c = 0

FACTORIZATION BY THE VEDIC MATHS

Factorization using Vedic Mathematics is done by using two Sutras.

Combo Rule (Perfect Quadratic Expression):

We use combination of 2 sutras.

1. Anurupyena  (Proportionalty)

2. Adyamadyenantyamantya (1st by 1st and last by last) (explained below):

We are explaining below:

In Anurupyena, we split the middle term (coefficient of x) of quadratic equation in 2 terms such that Proportion/Ratio of coeff of x² term to 1st coeff of x term = Ratio of 2nd coeff of x term to constant term. That ratio of the 1st 2 coeff is one of the root of equation.

1. Adyamadyenantyamantya

In Adyamadyenantyamantya (Commonly called as Adyamadyena), we divide the first term’s coeff of eq with 1st term of factor obtained above and last term of eq with the last term of the same factor.

Sanskrit Name (For Adyamadyenantyamantya):

आद्यमाद्ये नान्त्यमन्त्येन

English Translation (For Adyamadyenantyamantya):

1st by 1st and last by last

TYPE 01

ax² + bx + c = 0 Type quadratic equation.

हम जानते हैं कि कोई भी द्विघात समीकरण ax² + bx + c = 0. के रूप में होती है। 

जहां 

a =  x² का गुणज

b =  x का गुणज

c =  अचर पद

ax² + bx + c = 0

ax² + mx + nx + c = 0

a c = m n

a         n

__  =  ___

m        c

a (पहला गुणांक)            n (b का दूसरा भाग)

________________ =    ______________

m (b का दूसरा भाग)       c (अन्तिम गुणांक)

a c = m n

a         a₁

__  =  ___      (Final rato of a/m)

m        m₁

First Factor

  x¹          x⁰

m₁             a₁          

( a₁x¹ + m₁ x⁰ ) = ( a₁x + m₁)

Second Factor

 x¹          x⁰

a/a₁             c/m₁          

( a₁x¹ + m₁ x⁰ ) 

= ( a/a₁x + c/m₁)

Examples:

1x² + 8x +12 = 0

 Anurupyena: Split middle terms coefficient of x, +5 in 2 parts such that product of coefficient of x² and constant term is equal to the product of parts of b.

1st coeff of x term = Ratio of 2nd coeff of x term to constant term.

[a × c = b ( m × n)]

(1 × 12) = 8 (2 × 6)

1/2  = 6/12

First Factor

  x¹         x⁰

 1              ₂        

  ( 1x + 2 ) 

Adyamadyenantyamantya: Divide the first term’s coeff (2) of eq by 1st term of factor(1) and divide last term of eq (–3) by 2st term of factor (3)So 2nd factor: 2x–1

Second Factor

 x¹                x⁰         

1/1            12/2           (1x + 2 ) 

( 1x + 6 ) 

By Factorization Method

2x² + 5x –3 

= 2x² + 6x –1x –3

= 2x(x + 3) –1(x + 3)

= (x + 3) (2x –1)

= 2(x + 3) –1(x + 3)

By Vedic maths Method

2x² + 5x –3

a = 2, b = 5 & c = –3

[a×c = 2 × –3 = –6

b = 5 = +2 × –3 = +5]

2 × –3 = –3, +2            a × c = m × n 

2/–3 = +2/–3                 a/m = n/c

First Factor          Second Factor

(2x –1)                    (2/2x +3/1)

(2x –1)                    (x +3)

TYPE 02

a² + b² + c² + 2ab + 2bc + 2ca = 0 Type quadratic equation.

Lopana Sthapanabhyam (Subsutra of Adyamadyenantyamantya):

Sanskrit Name:

लोपनस्तापनाभ्याम्

English Translation:

By Alternare Elimination and Retention.

Examples:

1. Factorize 

2x² + 6y² + 3z² + 7xy + 11yz + 7zx

We have 3 variables x, y, z.

Applying Lopanasthapana, remove any of the variable. Lets Eliminate z by putting z=0.
Hence the given expression

E = 2x² + 6y² + 7xy

= (x+2y) (2x+3y) … 

(Combination of Anurupyena & Adyamadyenantyamantya).
Similarly, if y=0, then
E = 2x² + 3z² + 7zx

= (x+3z) (2x+z)

As x and 2x are present separately and uniquely.Hence we cay map to get Factors.
E = (x+2y+3x) (2x+3y+z)

2. Factorize 

x² + xy -2y² + 2xz -5yz -3z²
Eliminating z and factorizing gives 

(x-y)(x+2y)

Eliminating y and factorizing gives 

(x-z)(x+3z)

But as we see x term is present in all factors so we cannot map with correct terms to get factors. So we need the eliminate 3rd variable as well

Eliminating x and factorizing gives 

(-y-z)(2y+3z)

Now very easily we can do mapping and get E = (x-y-z)(x+2y+3z)

Note:

• We can eliminate 2 variables at a time as well. Thus we will have to map on the basis of Single independent term. But it will have few steps more.
• Lopanasthapana method can also be used for calculating HCF of expressions.
• We can also apply Gunitasamuccaya for checking.

Cubic Polynomials

Factorization of Cubic Polynomials:

Cubic Polynomial: ax³ + bx² + cx + d
Initial Step:

• Check a+b+c+d = 0, If sum of all coefficients=0 then directly we say (x-1) is 1 factor. OR
• Check a+c= b+d , If sum of odd terms = sum of even terms then (x+1) is 1 factor

and then apply Paravartya Sutra rule to get a quadratic Equation and apply usual Combo rule of Adyamadyena and Adyamadyena for solving quadratic equation.

Examples:

Solve x³ – 6x² + 11x -6

1. Sum of all coefficients=0 so    (x-1) is 1 factor.
2. Using Paravatya rule (x³ – 6x² + 11x -6)/(x-1) gives x² – 5x + 6
3. Solve quadratic equation using Combo Rule to get (x-2)(x-3)
4. E = (x-1)(x-2)(x-3).

Solve x³ + 6x² + 11x +6

1. Sum of odd terms = sum of even terms= 12, so (x+1) is 1 factor
2. Using Paravatya rule (X³ + 6x² + 11x +6)/(x+1) gives x² + 5x + 6
3. Solve quadratic equation using Combo Rule to get (x+2)(x+3).
4. E = (x+1)(x+2)(x+3).

If none of the Initial steps are satisfied then we apply following approach.

Solve x³ + 12x² + 44x + 48

1. None of the above Initial Condition is satisfied.
2. Last term(d) = 48. Factors of 48 are 1,2,3,4,6,8,12,16,24 and 48.
3. We need to find 3 numbers of these factors such that their sum = b(12) and product = d(44).
4. By trial and error method we observe 2+ 4 + 6 and 2 x 4 x 6 = 48
5. So the factors are (x+2)(x+4)(x+6)

 

Solve x³ -2x² -23x + 60

1. None of the above Initial Condition is satisfied.
2. Last term(d) = 60. Factors of 60 are 1,2,3,4,5,6,10,12,15,20,30 and 60.
3. We need to find 3 numbers of these factors such that their sum = b(-2) and product = d(60).
4. By trial and error method we observe -3 -4 + 5 = -2 and -3 x -4 x 5 = 60.
5. So the factors are (x-3)(x-4)(x+5).