This page 10th cbse maths solution for exercise 4.3 part 2 is going to provide you solution for every problems that you find in the exercise no 4.3 part 2

10th CBSE maths solution for Exercise 4.3 part 2

(3) Find the roots of the following equations:

(i) x - (1/x) = 3, x ≠ 0

Solution:

(x² - 1)/x = 3

x² - 1 = 3 x

x² - 3 x - 1 = 0

a = 1 b = -3 c = -1

x = - b ±
√(b² - 4 a c)/2a

x = - (-3) ±
√((-3)² - 4 (1)(-1))/2(1)

x = 3 ±
√(9+4)/2

x = (3 ±
√13)/2

(ii) [1/(x + 4)] - [1/(x -7)] = 11/30

Solution:

[1/(x + 4)] - [1/(x -7)] = 11/30

[(x - 7) - (x + 4)]/(x + 4)(x - 7) = 11/30

(x - 7 - x - 4)/(x² + 4 x - 7 x - 28) = 11/30

-11/(x² - 3 x - 28) = 11/30

30 (-11) =11 (x² - 3 x - 28)

-30 = x² - 3 x - 28

x² - 3 x - 28 + 30 = 0

x² - 3 x + 2 = 0

x² - 1 x - 2 x + 2 = 0

x (x - 1) - 2 (x - 1) = 0

(x - 2) (x - 1) = 0

x - 2 = 0 x - 1 = 0

x = 2 x = 1

(4) The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

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Solution:

Let 'x' be Rehman's present age

Three years ago his age was 'x - 3'

5 years after his age will be 'x + 5'

Sum of the reciprocals of Rehman's ages = 1/3

[1/(x -3) + 1/(x + 5)] = 1/3

(x + 5 + x - 3)/(x-3)(x + 5) = 1/3

(2 x + 2)/(x² - 3 x + 5 x - 15) = 1/3

(2 x + 2)/(x² + 2 x - 15) = 1/3

3 (2 x + 2) = 1 (x² + 2 x - 15)

6 x + 6 = x² + 2 x - 15

x² + 2 x - 15 - 6 x - 6 = 0

x²- 4 x - 21 = 0

x² - 7 x + 3 x - 21 = 0

x (x - 7) + 3 (x - 7) = 0

(x + 3) (x - 7) = 0

x = 3 or x = 7

So his present age will be 7 years.

(5)
In a class test,the sum of Shefali's marks in Mathematics and English
is 30. Had she got 2 marks more in Mathematics and 3 marks less in
English, the product of their marks would have been 210. Find her marks
in the two subjects.

Solution:

Let 'x' be the mark got by Shefali in the subject mathematics

let '30 - x' be the mark got by Shefali in English