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Algebra

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Algebra

Algebra is the branch of mathematics in which we perform mathematical operations with the help of numbers (constants) and alphabets (variables).

Algebraic Expression

The combination of constants, variables and elementary arithmetic operations (+, -, ×, ÷) is called algebraic expression.

Eg: 3x + 5y, 4x3 -5y

1/3, 5x – 6y + 7z ÷ 8y

1/3 etc.

On the basis of number of terms

There are following types of algebraic expressions on the basis of number of terms:

1. Monomial:

The algebraic expression that consists of only one term is called monomial.

Eg: 4x, 4y/3, -6.33z etc.

2. Binomial:

The algebraic expression that consists of two terms is called binomial.

Eg: 2x + 3y, 4x + x2 etc.

3. Trinomial:

The algebraic expression that consists of three terms is called trinomial.

Eg: 4xy + 3x + 5y2

4. Polynomial:

The algebraic expression that consists of two or more than two terms is called polynomial.

The general form of polynomial is given by P(x) = a0 + a1x + a2x2 + … + an x

n

Where, a0, a1, a2, a3 … and an, are real numbers and n is a whole number (none negative integer).

Eg: 1 + 2x – 3x2 + 5x

4, 6x

3 – 2x

2 + 4 etc.

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Definition : Coefficient

The number or the fixed value multiplied to the variable in a term of an algebraic expres-sion is called the coefficient of the term.

Eg: In polynomial 4x3 + 5x

2 – 7x – 2, the coefficient of x

3 is 4, coefficient of x

2 is 5, coef-

ficient of x is -7 and the constant term is -2.

Definition : Degree of Polynomial

The highest power of the variables in a polynomial is called the degree of polynomial.

Eg: The degree of polynomial 4x3 + 5x

2 – 7x – 2 is 3, the degree of polynomial 1 + 2x –

3x2 + 5x

4 is 4.

On the basis of Degree of polynomial

There are following types of polynomial on the basis of degree of polynomial:

1. Linear Polynomial:

The polynomial in which the highest power of the variables is one or, the polynomial of degree 1 is called linear polynomial.

Eg: 2x + 3

2. Quadratic Polynomial:

The polynomial of degree two is called quadratic polynomial.

Eg: x2 – 4x + 5, x

2 – 5 etc.

3. Cubic Polynomial:

The polynomial of degree three is called cubic polynomial.

Eg: x3 + 4x

2 – 3x + 4, x

3 – 4x – 3 etc.

4. Quartic Polynomial:

The polynomial of degree four is called quartic equation.

Eg: 5x4 – 4x + 2, 5x

4 + 4x

3 – 3x etc.




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Points to Remember

Zeroes of Polynomial: Those value(s) of x(variable) which satisfy the polynomial is/are called the zero/zeros of polynomial.

Eg: since x = 2 satisfy the polynomial x2 – 5x + 6, so we can say that x = 2 is a

zero of polynomial.

Remainder Theorem: When P(x) is divided by (x – a), then remainder will be P(a).

Eg: when P(x) = x2 -3x + 6 is divided by (x – 2), then remainder will be P(2)

⇒ P(2) = 22 – 3 × 2 + 6 = 4

Factor Theorem: If (x – a) is a factor of P(x), then remainder will be zero i.e. P(x) = 0

Eg: Find the value of k if x – 2 is a factor of x2 – k x + 6

Solution: according to question P(2) = 0

⇒ 22 – k × 2 + 6 = 0

Hence, k = 5

Equations

When we equate an expression or polynomial with a number, then the resultant is called an equation.

There are following types of equations:

Linear Equation in One Variable

The equation having only one variable with highest power 1 is called a linear equation in one variable.

The general form of linear equation in one variable is ax + b = 0, where a and b are real constants and a ≠ 0

Eg: 23 + x = 30




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Linear Equation in Two Variable

The equation having two variables with the highest power of the variable 1 is called linear equation in two variables.

The general form of the linear equation in two variable is given by ax + by + c = 0, where a, b and c are real numbers and (a, b) ≠ (0, 0).

Eg: 12x + 23y = 30

The value of the variables that satisfies the equation is called the solution of the equation.

The graph of the linear equation ax + by + c = 0 is a straight line.

Eg: since, x = 2 and y = 1 satisfy the linear equation in two variable x – 4y + 2 = 0, so we can say that x = 2 and y = 1 is a solution of the above equation.

Consistency and Inconsistency of Linear Equation in Two Variables

A system of a pair of linear equations in two variables is said to be consistent if it has at least one solution.

A system of a pair of linear equations in two variables is said to be inconsistent if it has no solution.

The system of a pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has:

I. A unique solution (i.e. consistent) if:

In this case the graphs of the pair of linear equations intersect at only one point and the co-ordinate of points is the solution of the pair of liner equation.

II. No solution (i.e. inconsistent) if:

In this case the graphs of the linear equations are parallel to each other that is the lines do not intersect each other.

III. An infinite number of solution (i.e. consistent) if:

In this case the graphs of the linear equations coincide each other that is the lines overlap each other.

1 1

2 2

a b

a b

1 1 1

2 2 2

a b c

a b c

1 1 1

2 2 2

a b c

a b c




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Quadratic Equation

An equation of the form ax2 + bx + c = 0 is called a quadratic equation.

Eg: 4x2 + 2x + 1 = 0, 5x

2 – 2x + 4 = 0 etc.

The standard form of the quadratic equation is ax2 + bx + c = 0, where a, b and c are

real numbers and a ≠ 0

Points to Remember

Root of Equation: The value(s) of x which satisfy the equation is/are called root of equation.

Eg: x = 3, satisfy the equation x2 – 6x + 9 = 0, so we can say that x = 3 is a root

of equation.

Maximum number of real roots of a linear, quadratic, cubic and a quartic equation are 1, 2, 3 and 4 respectively.

Finding roots of a quadratic equation:

We can find the roots of quadratic equation by the help of factorization method, and quadratic formula method.

Rule #1: Factorization Method

x2 – 8x + 12 = 0

⇒ x2 – 6x – 2x + 12 = 0

⇒ x (x – 6) – 2(x - 6) = 0

⇒ (x - 6) (x – 2) = 0

So, x = 6 or x = 2

Rule #2: Quadratic Formula method

x2 – 8x + 12 = 0

2 8 8 4 1 12

2 1

x




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So, x = 6 or x = 2

8 4

2x

Points to Remember

To solve a quadratic equation by the help of quadratic formula method, we first need to find the discriminant (D), where, D = b

2 – 4ac.

If D > 0, then the roots of the quadratic equation are real and distinct.

If D = 0, the roots of the quadratic equation are real and equal.

If D < 0, the roots of the quadratic equation are imaginary.

The roots of the quadratic equation are given by:

Usually we denote the roots of quadratic equation by α and β.

where α + β = -b/a and α × β = c/a

If it is given that α and β are the roots of the quadratic equation, then we can find the quadratic equation by the help of the formula [x

2 – (α + β)x + αβ = 0]

Eg: if x = 3 and x = 4 are the roots of a quadratic equation then we can find the equation by taking α = 2 and β = 4.

α + β = 7, α.β = 3 × 4 = 12

So, the required quadratic equation is x2 – 7x + 12 = 0

b Dx

2a

Cubic Equation

An equation of the form ax3 + bx

2 + cx + d = 0 is called a cubic equation.

Where a, b, c and d are real constants and a ≠ 0.




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Points to Remember

If α, β and δ are the roots of equation ax3 + bx

2 + cx + d = 0, then

α + β + δ = -b/a, αβ + βδ + δα = c/a and αβδ = -d/a

If α, β and δ are the roots of a cubic equation, then we can find the equation by using the formula

x3 – (α + β + δ)x

2 + (αβ + βδ + δα)x – αβδ = 0

Important Algebraic Identities

(a + b)2 = a

2 + 2ab + b

2

(a – b)2 = a

2 – 2ab + b

2

(a + b)2 = (a – b)

2 + 4ab

(a – b)2 = (a + b)

2 – 4ab

a2 + b

2 = (a + b)

2 – 2ab

a2 + b

2 = (a – b)

2 + 2ab

a2 – b

2 = (a + b) (a – b)

a3 + b

3 = (a + b) (a

2 – ab + b

2)

a3 – b

3 = (a – b) (a

2 + ab + b

2)

(a + b)3 = a

3 + b

3 + 3ab (a + b)

(a – b)3 = a

3 – b

3 – 3ab (a – b)

a3 + b

3 = (a + b)

3 – 3ab (a + b)

a3 – b

3 = (a – b)

3 + 3ab (a – b)




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(a + b + c)2 = a

2 + b

2 + c

2 + 2ab + 2bc + 2ca

a3 + b

3 + c

3 – 3abc = (a + b + c) (a

2 + b

2 + c

2 – ab – bc – ac) = ½

(a + b + c)[(a – b)2 + (b – c)

2 + (c – a)

2]

a3 + b

3 + c

3 = 3abc If (a + b + c) = 0

(a + b + c)3 = a

3 + b

3 + c

3 + 3(a + b)(b + c)(c + a)

a4 + b

4 + a

2b

2 = (a

2 – ab + b

2)(a

2 + ab + b

2)

Table-1 : Algebraic Identities (Most Important Formulae are highlighted in GREEN)

Important Special Cases

Type 1 : If x2 + y2 + z2 = 0, then x = y = z = 0

Q. If (a – 2)2 + (b + 3)

2 + (c – 5)

2 = 0, then find the value of a – b – c.

A. Given: (a – 2)2 + (b + 3)

2 + (c – 5)

2 = 0

So, a – 2 = 0

⇒ a = 2

And, b + 3 = 0

⇒ b = -3

And, c – 5 = 0

⇒ c = 5

Therefore, a – b – c = 2 – (-3) – 5 = 0

Type 2 : If a3 + b3 + c3 = 3abc Then (a + b + c) = 0

Q. If (2x + 3)3 + (x - 8)

3 + (x + 13)

3 = (2x + 3) (3x - 24) (x + 13), then what is the value

of x?

A. As we know,

If x3 + y

3 + z

3 - 3xyz = 0, then x + y + z = 0

⇒ (2x + 3)3 + (x - 8)3 + (x + 13)3 = (2x + 3) (3x - 24) (x + 13)

⇒ (2x + 3)3 + (x - 8)3 + (x + 13)3 - 3 (2x + 3) (x - 8) (x + 13) = 0




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⇒ (2x + 3) + (x - 8) + (x + 13) = 0

⇒ 4x + 8 = 0

⇒ 4x = -8

⇒ x = -8/4 = -2

Testbook Trick

1

2, 1If x then xx

1

2, 1If x then xx

21 1

, 4If x a then x ax x

21 1

, 4If x a then x ax x

2 22

1 1 , 2If x a then x a

x x

2 22

1 1 , 2If x a then x a

x x

3 33

1 1 , 3If x a then x a a

x x

31

1, 1If x then xx

31

1, 1If x then xx

61

3, 1If x then xx

Here are some algebraic expressions that are normally encountered in the questions.

Thus the following shortcuts might come handy while solving questions:




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Q. If then find the value of

A. Given:

So, x = 1

1

2xx

55

1 xx

1

2, xx

55

1 1So,x 1 2

x 1

Q.

A. Given:

So, x = -1

1

2, xx

5 45 4

1 1 1 2, If x then find the value ofx andx

x x x

45 4

45 4

1 1 1 1So,x 1 2 and x 1 2

x 1 x 1

Q.

A.

21

x 68 4x

1 1

68, If x thenfind xx x

1

x 64x

1

Hence, x 8 x

Q.

A.

21

x 60 4x

1 1

60, If x thenfind xx x

1

x 64x

1

Hence, x + 8 x




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Q.

A.

2 22

1x 2 2 2

x

22

1 1 2, If x thenfindthevalueofx

x x

Q.

A. 2

2

2

1x 2 2 4

x

22

1 1 2, If x thenfindthevalueof x

x x

Q.

3 33

1x 5 3 5 110

x

33


x x

A.

Q.

3 33

1x 5 3 5 140

x

33


x x

A.

Q.

1

x 1x

2021


x x

A. Given:

So, x = -1

2020

2121

1 1So, x 1 1 1 2

x 1

Q. 20

21


x x

1

x 1x

A. Given:

So, x = -1

2020

2121

1 1So, x 1 1 1 2

x 1




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So, x = +1

2020

2121

1 1So, x 1 1 1 0

x 1

Q.

1

x 3x

2442

1 1 3, If x thenfindthevalueofx

x x

A. Given:

So,

6x 1

2442

1x

x

4

6

76

1x

x

4

7

11 1 1 2

1

Testbook Trick

x x x . x

n

n

2 1

2x x x . ..n x

1 1 4xx x x .

2

1 1 4xx x x .

2

For the two forms mentioned below, if the difference between the factors of x is 1, then

the factor which is greater in value will be the answer of the given question.




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Q.

A. 3 3 3 . 3

3 3 3 . Findthevalue of

Q.

5

5

2 1

23

3 3 3 . 5 thFindthevalue of term

A.

Q.

1 1 20 45

2

20 20 20 .

A.

Q.

1 1 20 44

2

20 20 20 .

A.

Testbook Trick

1 1If xy 1, then 1

x 1 y 1

Q.

y 5 2 6

1 1 5 2 6 5 – 2 6,

1 1 If x andy thenfindthevalueof

x y

A.




PAGE 15

5 2 6 y 5 2 6

5 2 6

25 24y

5 2 6

1 y

5 2 6

15 6

5 2 6

xy 1

1 1Therefore, 1

x 1 y 1

Algebra - screativeworkscom.files.wordpress.com › 2020 › 04 › quant_al… · Algebra is the...

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