important concepts and formulas - algebra

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Home > Quantitative aptitude questions and answers with explanation > Numbers > Important Concepts and Formulas - Algebra

Important Concepts and

Formulas - Numbers


Formulas - Algebra


Formulas - Sequence and

Series


Formulas - Logarithms


Formulas - Trigonometry


Formulas - Complex

Numbers

Solved Examples - Set 1

Solved Examples - Set 2Solved Examples - Set 3

Important Concepts and Formulas - Algebra

I. Basic Algebraic Formulas

1. (Distributive Law)

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

01.2kLike Share

Java RDBMS Quantitative Aptitude Online Tests Aptitude Q&A Speed Mathematics Scientif ic Cal culator Miscellaneous

a ( b + c ) = a b

+ a c

( a + b = + 2 a b

+

)

2

a

2

b

2

( a − b = − 2 a b

+

)

2

a

2

b

2

( a + b + (

a − b = 2 ( + )

)

2

)

2

a

2

b

2

( a + b = + 3

b + 3

a + = + 3 a b

( a + b ) +

)

3

a

3

a

2

b

2

b

3

a

3

b

3

( a − b = − 3

b + 3

a − = − 3 a b

( a − b ) −

)

3

a

3

a

2

b

2

b

3

a

3

b

3

− = ( a − b

) ( a + b )

a

2

b

2

+ = ( a + b

) ( − a b + )

a

3

b

3

a

2

b

2

− = ( a − b

) ( + a b + )

a

3

b

3

a

2

b

2

− = ( a − b

) ( + b

+ + . . . + )

a

b

a

− 1

a

− 2

a

− 3

b

2

b

− 1

( a + b + c = + + + 2 ( a b

+ b c

+ c a

)

)

2

a

2

b

2

c

2

( a + b ) (

a + c ) = + (

b + c ) a + b c

a

2

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1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

1.

2.

3.

4.

5.

6.

7.

8.

= a . a . a . . . ( n t i m e s )

a

. . . . = a

a

a

a

+ + . . . +

. = a

a

a

+

=

a

a

a

−

= = ( ) a

a

( ) a

= ( a b

)

a

b

=

a

b

a

b

= a

−

1

a

= a

1

a

−

= 1 w h e r e a

∈ R , a ≠ 0 a

0

= a

/

a

− −

√

i f = w h e r e a

≠ 0 a n d a

≠ ± 1 , t h e n = a

a

i f = w h e r e

≠ 0 , t h e n a

= ± b a

b

(

+ a = + 2

a

+ )

2

2

a

2

( − a = − 2 a

+ )

2

2

a

2

( + a ) (

+ b ) = + (

a + b ) + a b

2

− = ( − a

) ( + a )

2

a

2

( + 1 ) i s c o m p l e t e l y d i v i s i b l e b y ( x + 1 ) w h e n n i s o d d

( + ) i s c o m p l e t e l y d i v i s i b l e b y ( x + a ) w h e n n i s o d d

a

( − ) i s c o m p l e t e l y d i v i s i b l e b y ( − a

) f o r e v e r y n a t u r a l n u m b e r n

a

( − ) i s c o m p l e t e l y d i v i s i b l e b y ( + a

) w h e n n i s e v e n

a


3/11



Binomial Theorem

If a and b are any real numbers and n is a positive integer,

Binomial Coefficient also occurs in many other mathematical areas than algebra, especially in combinatorics wher

represents the number of combinations of n distinct things taking r at a time and is denoted by nCr or C(n,r).

The properties of binomial coefficients have led to extending its meaning beyond the basic case where n and r are nonn

integers with r ≤ n; such expressions are also called binomial coefficients.

Binomial Expansions

... and so on

Binomial Series

One of Newton's achievement was to extend the Binomial Theorem to the case where

n can be any real number.

If n is any real number and -1 < x < 1, then

II. Surds and Important Properties

i. Surds

Let be a rational number and be a positive integer such that is irrational.

Then is called a surd of order n.

( a + b )

= + b + + + ⋯ + + ⋯a

a

− 1

( − 1 )

2 !

a

− 2

b

2

( − 1 ) (

− 2 )

3 !

a

− 3

b

3

( − 1 ) (

− 2 ) ⋯ (

− + 1 )

!

a

−

b

= + b + + + ⋯ +

+ ⋯ +

0

a

1

a

− 1

2

a

− 2

b

2

3

a

− 3

b

3

a

−

b

b

= ∑

= 0

a

−

b

w h e r e

i s k n o w n a s B i n o m i a l C o e f f i c i e n t a n d i s d e f i n e d b y

= = w h e r e r = 1 , 2 , . . . , n a n d = 1

!

( ! ) (

− ) !

( − 1 ) (

− 2 ) ⋯ (

− + 1 )

!

0

( a + b = 1

)

0

( a + b = a + b )

1

( a + b = + 2 a b

+ )

2

a

2

b

2

( a + b = + 3

b + 3

a + )

3

a

3

a

2

b

2

b

3

( a + b = + 4

b + 6 + 4

a + )

4

a

4

a

3

a

2

b

2

b

3

b

4

( 1 + )

= 1 + + + + ⋯

(

− 1 )

2 !

2

(

− 1 ) (

− 2 )

3 !

3

= +

+

+

+ ⋯

0

0

1

1

2

2

3

3

= ∑

= 0

∞

w h e r e

= w h e r e r ≥ 1 a n d

= 1

( − 1 ) (

− 2 ) ⋯ (

− + 1 )

!

0

a a

√

a

√

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4/11



Examples :

is a surd of order 2



Please note that numbers like , etc are not surds because they are not

irrational numbers

Every surd is an irrational number. But every irrational number is not a surd. (eg

: , etc are not surds though they are irrational numbers.)

ii. Quadratic Surds

A surd of order 2 is called a quadratic surd

Examples :

iii. Rules of Surds

1.

2.

3.

4.

5. if

1. The conjugate surd of

2. if and are quadratic surds and if ,then and

i.e., rational part on the lef t side = rational part on the right side.

and irrational part on the left side = irrational part on the right side.

3. if and are quadratic surds and if ,

then and

III. Quadratic Equations and How to Solve Quadratic Equations

A. Quadratic Equations

A quadratic equation is a second degree univariate polynomial equation.

A quadratic equation can be written as (general form or standard form of quadratic

equation)

where x is a variable, a, b and c are constants and

Example :

(Please note that if a=0, equation becomes a linear equation)

The solutions of a quadratic equation are called its roots.

B. How to Solve Quadratic Equations

There are many methods to solve Quadratic equations. Quadratic equations can be

solved by factoring, completing the square, graphing, Newton's method, and using the

quadratic formula. We can go through some of the popular methods for solving

quadratic equations here.

3 √

7 √

3

4 3 √

9 √ 2 7

− −

√

3

π

, , ( 3 + ) , e t c . 2

√

3 √ 5

√

× =

− −

√

√

− − −

√

=

− −

√

√

− − −

√

= a . a . a ⋯ n t i m e s

− − − − − − − − − −

√

− − − − − − − − − − − − −

√

− − − − − − − − − − − − − − − − −

√

a

1 −

1

2

= a a

. a

. a ⋯ ∞

− − − − − −

√

− − − − − − − − −

√

− − − − − − − − − − − − −

√

= , t h e n

( − 1 ) = a a

+ a

+ a ⋯ ∞

− − − − − −

√

− − − − − − − − − − −

√

− − − − − − − − − − − − − − − −

√

+ = ± ( − ) a

√

b

√

a

√

b

√

b

√

√

a + = c +

b

√

√

a =

c b

=

b

√

√

a + =

b

√

√

a = 0

b =

a + b

+ c = 0

2

a ≠ 0

+ 5 + 6 = 0

2


5/11



1. How to Solve Quadratic Equations By Factoring

Step 1: Write the quadratic equation in standard form.

i.e. , bring all terms on the left side of the equal sign and 0 on the right side of

the equal sign.

Step 2: Factor the left side of the equation

Step 3: Equate each factor to 0 and solve the equations

Factoring is one of the fastest ways for solving quadratic equations. The concept

will be clear from the following examples.

Example1 : Solve the Quadratic Equation

Step 1 : Write the quadratic equation in standard form.

Step 2 : Factor the left side of the equation.

We know that . We will use the same

concept for factoring.

Assume

We need to find out a and b such that a + b = 3 and ab = -10

a = +5 and b = -2 satisfies the above condition.

Hence

Step 3 :Equate each factor to 0 and solve the equations

(x + 5)(x - 2) = 0

=> (x + 5) = 0 or (x - 2) = 0

=> x = -5 or 2

Hence, the solutions of the quadratic equation are x = -5

and x = 2.

(In other words, x = -5 and x = 2 are the roots of the quadratic equation


This equation is already in the standard form. Hence let's go to step 2


Here we need to find out a and b such that a + b = -7 and ab = +10

a = -5 and b = -2 satisfies the above condition.

Hence

Hence,


(x - 5)(x - 2) = 0

=> (x - 5) = 0 or (x - 2) = 0

=> x = 5 or 2

Hence, the solutions of the quadratic equation are x = 5 and x

+ 4

− 1 1 =

− 1

2

+ 4 − 1 1 = − 1

2

⇒ + 4

− 1 1 −

+ 1 = 0

2

⇒ + 3

− 1 0 = 0

2

+ ( a + b ) +

a b = ( + a

) ( + b )

2

+ 3

− 1 0 = + ( a + b ) +

a b

2

2

+ 3

− 1 0 = (

+ 5 ) (

− 2 )

2

+ 3

− 1 0 = 0

2

⇒ (

+ 5 ) (

− 2 ) = 0

+ 4

− 1 1 =

− 1

2

+ 4

− 1 1 =

− 1

2

− 7

+ 1 0 = 0

2

− 7

+ 1 0 = (

− 5 ) (

− 2 )

2

+ 3

− 1 0 = 0

2

⇒ (

− 5 ) (

− 2 ) = 0

− 7

+ 1 0 = 0

2


6/11



= 2.

(In other words, x = 5 and x = 2 are the roots of the quadratic equation

)




Here we need to follow a slightly different approach for factoring because here

the coefficient of x2 ≠ 1 (whereas In example 1 and example 2, the coefficient of

x2 was 1)

Hence, to factor , we need to do the followings

1. Product of the second degree term and the constant. i.e.,

2. We got product as 24x2 and have middle term as -14x. From this , it is clear

that we can take -12x and -2x such that their sum is -14x and product is 24x 2

Hence, =

Hence,


(x - 5)(x - 2) = 0

=> (x - 4) = 0 or (3x - 2) = 0

=> x = or .

Hence, the solutions of the quadratic equation are x = and

x =

(In other words, x = and x = are the roots of the quadratic equation




coefficient of x2 ≠ 1

Hence, to factor , we need to do the followings

1. Product of the second degree term and the constant. i.e.,

2. We got product as 72x2 and have middle term as -17x. From this , it is clear

that we can take -8x and -9x such that their sum is -17x and product is 72x 2

Hence,

Hence,


(3x - 4)(2x - 3) = 0

=> (3x - 4) = 0 or (2x - 3) = 0

=> x = or .

Hence, the solutions of the quadratic equation are x =

and x = .

− 7

+ 1 0 = 0

2

3 − 1 4 + 8 = 0

2

3 − 1 4

+ 8

2

3 × 8 = 2 4

2

2

3 − 1 4

+ 8 = 0

2

3 − 1 2

− 2

+ 8 = 3 (

− 4 ) − 2 (

− 4 ) = (

− 4 ) ( 3

− 2 )

2

3 − 1 4

+ 8 = 0

2

⇒ (

− 4 ) ( 3

− 2 ) = 0

4

2

3

3 − 1 4

+ 8 = 0

2

4

2

3

4

2

3

3 − 1 4

+ 8 = 0

2

6 − 1 7

+ 1 2 = 0

2

6 − 1 7

+ 1 2

2

6 × 1 2 = 7 2

2

2

6 − 1 7

+ 1 2 = 6 − 8

− 9

+ 1 2 = 2

( 3

− 4 ) − 3 ( 3

− 4 ) = ( 3

− 4 ) ( 2

− 3 )

2

2

6 − 1 7

+ 1 2 = 0

2

⇒ ( 3

− 4 ) ( 2

− 3 ) = 0

4

3

3

2

6 − 1 7

+ 1 2 = 0

2

4

3

3

2

4 3


7/11




(Special Factorization Examples)




Here, is in the form where a = x and b = 3

We know that We will use the same concept for

factoring this.

Hence,


(x - 3)(x + 3) = 0

=> x = 3 or -3

Hence, the solutions of the quadratic equation are x = 3 and x = -3

(In other words, x = 3 and x = -3 are the roots of the quadratic equation




Hence,

Step 3 : Equate each factor to 0 and solve the equations

(5x - 4)(5x + 4) = 0

=>5x = 4 or -4

.

Hence, the solutions of the quadratic equation are x = and x =







factoring this.

2

4

3

3

2

6 − 1 7

+ 1 2 = 0

2

− 9 = 0

2

− 9

2

− a

2

b

2

− = ( a − b

) ( a + b ) a

2

b

2

− 9 = − = (

− 3 ) (

+ 3 )

2

2

3

2

− 9 = 0

2

⇒ (

− 3 ) (

+ 3 ) = 0

− 9 = 0

2

− 9 = 0

2

2 5 − 1 6 = 0

2

2 5 − 1 6 = ( 5

− = ( 5

− 4 ) ( 5

+ 4 )

2

)

2

4

2

2 5 − 1 6 = 0

2

⇒ ( 5

− 4 ) ( 5

+ 4 ) = 0

= o r

4

5

− 4

5

2 5 − 1 6 = 0

2

4

5

− 4

5

4

5

− 4

5

2 5 − 1 6 = 0

2

+ 6

+ 9 = 0

2

+ 6

+ 9

2

+ 2 a b + a

2

b

2

+ 2 a b + = ( a + b a

2

b

2

)

2

+ 6

+ 9 = + 2 ×

× 3 + = (

+ 3

2

2

3

2

)

2

+ 6

+ 9 = 0

2


8/11



Hence,

Step 3 :

=> x + 3 = 0

=> x = -3

Hence, the solution of the quadratic equation is x = -3

(In other words, x = -3 is the root of the quadratic equation






factoring this.

Hence,

Step 3 :

=> x - 3 = 0

=> x = 3

Hence, the solution of the quadratic equation is x = 3

(In other words, x = 3 is the root of the quadratic equation

2. How to Solve Quadratic Equations By Quadratic Formula

Quadratic Formula

Consider a quadratic equation where

Its solutions can be given by


Hence, the solutions of the quadratic equation are x = 2

and x = -5


+ 6

+ 9 = 0

2

⇒ (

+ 3 = 0 )

2

( + 3 = 0

)

2

+ 6

+ 9 = 0

2

+ 6

+ 9 = 0

2

− 6

+ 9 = 0

2

− 6

+ 9

2

− 2 a b + a

2

b

2

− 2 a b + = ( a − b a

2

b

2

)

2

− 6

+ 9 = − 2 ×

× 3 + = (

− 3

2

2

3

2

)

2

+ 6

+ 9 = 0

2

⇒ (

− 3 = 0 )

2

( − 3 = 0

)

2

− 6 + 9 = 0

2

− 6

+ 9 = 0

2

a + b

+ c = 0

2

a ≠ 0

=

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

+ 4

− 1 1 =

− 1

2

+ 4

− 1 1 =

− 1

2

⇒ + 4

− 1 1 −

+ 1 = 0

2

⇒ + 3

− 1 0 = 0

2

= = =

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

− 3 ± − 4 × 1 × ( − 1 0 ) 3

2

− − − − − − − − − − − − − − −

√

2 × 1

− 3 ± 9 + 4 0

− − − − −

√

2

=

− 3 ± 7

2

= o r

4

2

− 1 0

2

= 2 o r − 5

+ 4 − 1 1 = − 1

2

+ 4

− 1 1 =

− 1

2


9/11




Hence, the solutions of the quadratic equation are x = 5 and x

= 2.

(In other words, x = 5 and x = 2 are the roots of the quadratic equation


Hence, the solutions of the quadratic equation are x = and

x =



Hence, the solutions of the quadratic equation are x =

and x = .



Hence, the solutions of the quadratic equation are x =(-3 + i)

+ 4

− 1 1 =

− 1

− 7

+ 1 0 = 0

2

= = =

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

7 ± ( − 7 − 4 × 1 × 1 0

)

2

− − − − − − − − − − − − − − −

√

2 × 1

7 ± 4 9 − 4 0

− − − − − −

√

2

= =

7 ± 9 √

2

7 ± 3

2

= o r

1 0

2

4

2

= 5 o r 2

− 7

+ 1 0 = 0

2

− 7

+ 1 0 = 0 )

2

3 − 1 4

+ 8 = 0

2

= = =

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

1 4 ± ( − 1 4 − 4 × 3 × 8 )

2

− − − − − − − − − − − − − − −

√

2 × 3

1 4 ± 1 9 6 − 9 6

− − − − − − −

√

6

= =

1 4 ± 1 0 0

− − −

√

6

1 4 ± 1 0

6

= o r

2 4

6

4

6

= 4 o r

2

3

3 − 1 4

+ 8 = 0

2

4

2

3

4

2

3

3 − 1 4

+ 8 = 0

2

6 − 1 7

+ 1 2 = 0

2

= = =

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

1 7 ± ( − 1 7 − 4 × 6 × 1 2 )

2

− − − − − − − − − − − − − − − −

√

2 × 6

1 7 ± 2 8 9 − 2 8 8

− − − − − − − −

√

1 2

= =

1 7 ± 1

√

1 2

1 7 ± 1

1 2

= o r

1 8

1 2

1 6

1 2

= o r

3

2

4

3

6 − 1 7

+ 1 2 = 0

2

4

3

3

2

4

3

3

2

6 − 1 7

+ 1 2 = 0

2

+ 6 + 1 0 = 0

2

= = =

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

− 6 ± ( 6 − 4 × 1 × 1 0 )

2

− − − − − − − − − − − − − −

√

2 × 1

− 6 ± 3 6 − 4 0

− − − − − −

√

2

= =

− 6 ± − 4

− − −

√

2

− 6 ± 2

2

= − 3 ±

= ( − 3 +

) o r ( − 3 − )

+ 6

+ 1 0 = 0

2


10/11



and x = (-3 - i)

(In other words, x =(-3 + i) and x = (-3 - i) are the roots of the quadratic

equation


Hence, the solutions of the quadratic equation is

(In other words, is the roots of the quadratic equation


Hence, the solutions of the quadratic equation are x = 3 and x = -3


C. How to find out the number of solutions(roots) of a quadratic

equation quickly?

To find out the number of solutions (roots) of a quadratic equation quickly, we need to find out the discriminant, D. The

discriminant, D =

If the discriminant is positive, there will be two real solutions for the quadratic

equation.

If the discriminant is 0, there will be one real solution (it repeats itself) for the

quadratic equation.

If the discriminant is negative, there will be two complex solutions for the

quadratic equation.

Example1 :

Find out the number of roots for

Discriminant, D =

D > 0. Hence will have two real roots.

Let's verify if this is correct by solving the quadratic equation.

Yes, two real roots we got.

Example2 :


Discriminant, D =

D = 0. Hence will have one real root (it repeats itself)


Yes, real root (it repeats itself).

+ 6

+ 1 0 = 0

2

+ 6 + 9 = 0

2

= = =

− b ± − 4

a c

b

2

− − − − − − −

√

2 a

− 6 ±

( − 6 − 4 × 1 × 9 )

2

− − − − − − − − − − − − − −

√

2 × 1

− 6 ±

( 3 6 − 3 6

− − − − − − −

√

2 × 1

= =

− 6 ± 0 √

2

− 6

2

= − 3

+ 6

+ 9 = 0

2

= − 3

= − 3 + 6

+ 9 = 0

2

− 9 = 0

2

= = =

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

0 ± ( 0 − 4 × 1 × ( − 9 ) )

2

− − − − − − − − − − − − − − − −

√

2 × 1

± 3 6

− −

√

2

= ±

6

2

= 3 o r − 3

− 9 = 0

2

− 9 = 0

2

a + b

+ c = 0

2

– 4 a c b

2

− 7

+ 1 2 = 0

2

– 4 a c = ( − ) − ( 4 × 1 × 1 2 ) = 4 9 − 4 8 = 1 . b

2

7

2

− 7

+ 1 2 = 0

2

− 7

+ 1 2 = 0 ⇒ (

− 3 ) (

− 4 ) = 0 ⇒

= 3 o r 4 .

2

− 4

+ 4 = 0

2

– 4 a c = ( − ) − ( 4 × 1 × 4 ) = 1 6 − 1 6 = 0 . b

2

4

2

− 4

+ 4 = 0

2

− 4

+ 4 = 0 ⇒ (

− 2 = 0 ⇒

= 2

2

)

2


11/11


Example3 :


Discriminant, D =

D < 0. Hence will have two complex roots


Yes, we got two complex roots, +i and -i

D. Sum and Products of the roots of a quadratic equation

Let and are the roots of a quadratic equation

Sum of the roots of the quadratic equation,

Product of the roots of the quadratic equation,

Example 1: Find the sum and product of the roots of the quadratic equation

sum of the roots

product of the roots

Example 2: Find the sum and product of the roots of the quadratic equation

sum of the roots

product of the roots

Comments(6) Newest

+ 1 = 0

2

– 4 a c = ( ) − ( 4 × 1 × 1 ) = 0 − 4 = − 4 b

2

0

2

+ 1 = 0

2

+ 1 = 0

2

= = = = = = ±

− b ± − 4 a c

b

2

− − − − − − −

√

2 a

0 ± − 4 × 1 × 1 0

2

− − − − − − − − − − − −

√

2 × 1

0 − 4

− − − − −

√

2

− 4

− − −

√

2

± 2

2

1

2

a + b

+ c = 0 w h e r e

a ≠ 0

2

+ =

1

2

− b

a

× =

1

2

c

a

− 7

+ 1 0 = 0

2

= = = 7

− b

a

7

1

= = = 1 0

c

a

1 0

1

2 + 3

+ 1 = 0

2

= =

− b

a

− 3

2

= =

c

a

1

2

| | | Li ke Di slike Rep ly Flag





SRK 30 Sep 2014 12:12 PM

PLEASE HELP ME IN FINDING HCF AND LCM OF POLYNOMIALS

Reshma 12 Jun 2014 11:11 AM

very helpful site for students...:)

catty 25 May 2014 5:09 PM

so helpful guys! Thanks :)

Parthiban 03 Mar 2014 6:06 PM

This is help to me pls can you send me all the formulas

vitthal 13 Feb 2014 1:26 PM

give me problems

Usama 28 Sep 2013 12:18 PM

Can u mail me the formulae?
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important concepts and formulas - algebra

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