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Basic Formulae

Algebra requires dealing with unknown quantities. We use symbols to get general results. Hence it is useful to know the basic formulae.

 

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

II.         (a – b)2 = a2 + b2 – 2ab.

III.      (a2 – b2) = (a + b) (a – b).

IV.       (a + b)2  - (a-b)2 = 4ab.

V.         (a + b)2 + (a-b)2 = 2(a2 + b2).                                

 

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

VII.     (a – b)3 = a3 - b3 - 3ab(a – b).

VIII.  (a3 + b3) = (a + b)(a2 + b2 – ab).

IX.       (a3 – b3) = (a – b) (a2 + b2 + ab).

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

XI.       (a + b + c + d)2 = [a2 + b2 + c2 + d2+ 2a(b + c + d) + 2b(c + d) + 2cd].

XII.     (a3 + b3 + c3 –3abc) = (a + b + c) (a2 + b2 + c2 – ab – bc – ca). If a + b + c = 0 a3 + b3 + c3 = 3abc.

XIII.  (x + a)(x + b) = x2+ (a + b)x + ab.

XIV.   (x + a)(x + b)(x + c) = x3 + (a + b + c) x2 + (ab + bc + ca) x + abc.

 

Illustrations

Illustration 1:

If x + 1/x = 7, what is the value of x2 + 1/x2?

Solution

 x + 1/x = 7. Squaring both sides, we get x2 + 1/x2 + 2x(1/x) = 49;
 hence       x2 + 1/x2 = 49 – 2 = 47.


Illustration 2:

Given x2 + 1/x2 = 27, find the value of x – 1/x.

Solution

(x – 1/x)2 = x2 + 1/x2 – 2x(1/x) = x2 + 1/x2 – 2.
Hence x2 + 1/x2 = 27 – 2; (x – 1/x)2 = 25 ⇒ x – 1/x = √25 = ± 5.

  

 

Illustration 3:

Find the value of following expression:
     6.79 × 6.79 × 6.79 + 3.21 × 3.21 × 3.21
     6.79 × 6.79 – 6.79 × 3.21 + 3.21 × 3.21

Solution

The expression is of the form (a3 – b3)/(a2 – ab + b2). By using the above formulae, we can cancel out the terms leaving (a + b). Substitute ‘a’ = 6.79 and ‘b’ = 3.21, hence a + b = 6.79 + 3.21 = 10.

 

Basics of Algebra

Forming equations. One problem in algebra is how to formulate equations.

Letters such as x or n are used to represent unknown quantities.  For example, if Anil has 5 more pencils than Bela, then, if x represents the number of pencils that Bela has, then the number of pencils that Anil has is x + 5.

 

Or, if Subhash’s present salary y is increased by 7%, then his new salary is 1.07y.

A combination of letters and arithmetic operations, such as x + 5, 3x2/ (2x – 5) and 19x2 - 6x + 3, are called algebraic expressions.

 

Coefficients. The expression 19x2 - 6x + 3 consists of the terms 19x2, -6x, and 3, where 19 is the coefficient of x2, - 6 is the coefficient of x, and 3 is a constant term (or coefficient of x0 = 1). Such an expression is called a second degree (or quadratic) polynomial in x since the highest power of x is 2. The expression x + 5 is a first degree (or linear) polynomial in x since the highest power of variable x is 1.

 

Simplification. Algebraic expressions can be simplified by factoring or combining like terms.  For example, 6x + 5x is equivalent to (6 + 5)x. The expression 9x - 3y can be written as = 3(3x - y).  In the expression 5x2 + 6y, there are no like terms and no common factors.

 

Multiplication. To multiply two algebraic expressions, each term of one expression is multiplied by each term of the other expression. 

For example: (3x - 4) (9y + x) = 3x (9y + x) – 4 (9y + x) =

 (3x) (9y) + (3x) (x) + (-4) (9y) + (-4) (x) = 27xy + 3x2 - 36y - 4x.

 





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