Exercise 4.2-Class 9th (P. Board)-Ytb

We will solve the problems of this exercise by using various Algebraic Formulae.  Different formulae used are listed below with examples thereof.

  1. (a + b)2 + (a – b)2 = 2(a2 + b2) and

(a + b)2 – (a – b)2 = 4ab

Example:

If a + b = 7 and a – b = 3 then find the value of a2 + b2 and ab

Now we know that

(a + b)2 + (a – b)2 = 2(a2 + b2)

Putting the values of (a + b) and (a – b)

= (7)2 + (3)2 = 2(a2 + b2)

= 49 + 9 = 2(a2 + b2) = 58

Hence a2 + b2 = 58/2 = 29

Similarly we know that

(a + b)2 – (a – b)2 = 4 ab

=(7)2 – (3)2 = 4 ab

= 49 – 9 = 4 ab = 40

Hence ab = 40/4 = 10

Answer: a2 + b2 = 29         and     ab = 10

  1. (a + b + c)2 = a2 + b2 + c2 + 2ab +2ac + 2bc

Example:

If a2 + b2 + c2 = 43 and ab + bc + ca = 3

Find the value of a + b + c

We have the formula

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

Putting the values given in the example

(a + b+ c)2 = (43) + 2(3) = 43 + 6 = 49

Taking the Square Root of both sides

(a + b + c) = 7          Answer

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

(a – b)3 = a3 – 3ab(a – b) – b3

Example:

If 2x – 3y = 10 and xy = 2 then find 8x3 – 27y3

We have the folrmula

(a – b)3 = a3 – 3ab(a – b) – b3

= (2x – 3y)3 = (2x)3 – 3 . 2x . 3y (10) – (3y)3 = (10)3

= 8x3 – 18xy (10)  – 27y3 = 1000 = 8x3 – 18 . 2 . 10 – 27 y3

= 8x3 – 27 y3 – 360 = 1000

= 8x3 – 27y3 = 1360           Answer

 

  1. a3 + b3 = (a + b) (a2 – ab + b2)

a3 – b3 = (a – b) (a2 + ab + b2)

 

The questions on these formulae may be solved in the same manner

QUESTIONS

1 2 3
4 5 6
7 8 9
10 11 12
13 14 15 (I)
15(II) 15(III) 15(IV)

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