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Basic Theory for Algebraic Expressions for Upcoming Bank PO and SSC Exam

Posted on - 05-02-2017

QA Math Algebra

Bank PO

In arithmetic we express 1+ 1+ 1= 3, 2 x 2 x 2 = 8, 3 – 2 = 1 etc., Here 1, 2, 3, 8 are known numbers. In algebra we deal with symbols like x, y, z, a, b, c, p etc., which represent unknown numbers. .

x + x + x = 3x, a x a x a = a3, 3x - 2x = x are some examples of algebraic statements. In the above examples 1, 2, 3 and 8 are known and x, y, z, c, b, c and p are unknowns. .

When x + y = 10, the value of x depends on the value of y and vice-versa. Here 10 is a constant and x, y are variables. .

We can define algebra as a branch of mathematics which deals with rules, procedures and operations involving wns and unknowns; and constants and variables.

Algebraic expressions are formed from variables and constants. 3x + 4y is an algebraic expression. The value of an expression depends on the values of the variables from which the expression is formed. .

Formulas in algebra are written in a general form using algebraic expressions. .

The following algebraic formulas should be memorized since they are very useful to solve certain problems. .

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

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

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

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

(a - b)3 = a3 – b3 - 3ab (a - b)

a3 + b3 = (a + b) (a2 - ab + b2)

a3 - b3 = (a - b) (a2 + ab + b2)

a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

if a + b + c = 0, then a3 + b3 + c3 = 3abc. .

Worked out examples

example.1

If a = 13, b = 12, then find the value of

Solution     

Given that a = 13, b = 12. .

= a + b

= = a + b

= 13 + 12 = 25

Example.2

x = 103, then find the value of

.

Solution

= =

(\ a3 - 3ba2 + 3ab2 – b3 = (a - b)3)

= x - 2 = 101

example.3    

If y + = 10, then find the value of .

Solution     

y + = 10

Multiplying by y on both sides,

y2 + 25 = 10y

y2 - 10y + 25 = 0

(y - 5)2 = 0

y - 5 = 0

y = 5

.

Example.4    

If 3y + = 3, then find the value of 64y3 + .

Solution     

3y + = 3

Multiplying both sides by 4y + = 4.

Cubing on both sides,

64y3 + + 3(4y)= 64

(\ (a + b)3 = a3 + b3 + 3ab (a + b))

64y3 + + 4 (4) = 64

64y3 + = 48

example.5

If y + , then find the value of

Solution

x =

 
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