Class 10 Mathematics

Quadratic Equation

Exercise 4.3 (NCERT) Part 3

Question 3: Find the roots of the following equations:

Question (i): `x-1/x=3: x≠0`

Solution:

`x-1/x=3`

Or, `(x^2-1)/(x)=3`

Or, `x^2-3x-1=0`


We have; a = 1, b = - 3 and c = - 1

Root can be calculated as follows:

`text(Root)=(-b±sqrt(b^2-4ac))/(2a)`

`=(3±sqrt((-3)^2-4xx1xx-1))/(2xx1)`

`=(3±sqrt(13))/(2)`

Hence, `α=(3+sqrt(13))/(2)`

And, `β=(3-sqrt(13))/(2)`


Question (ii): `(1)/(x+4)-(1)/(x-7)=(11)/(30) x≠-4 or 7`

Solution:

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

Or, `(x-7-x-4)/((x+4)(x-7))=(11)/(30)`

Or, `(x^2-7x+4x-28)/(30)=-1`

Or, `x^2-3x-28=-30`

Or, `x^2-3x-28+30=0`

Or, `x^2-3x+2=0`

Or, `x^2-2x-x+2=0`

Or, `x(x-2)-1(x-2)=0`

Or, `(x-1)(x-2)=0`

Hence,roots are; 1 and 2


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

Solution: Let us assume, Rehman’s present age = x

Therefore, 3 years ago, Rehman’s age = x – 3

And, 5 years from now, Rehman’s age = x + 5

As per question;

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

Or, `(x+5+x-3)/((x-3)(x+5))=1/3`

Or, `(2x+2)/(x^2+5x-3x-15)=1/3`

Or, `6x+6=x^2+2x-15`

Or, `x^2+2x-15-6x-6=0`

Or, `x^2-4x-21=0`

Or, `x^2-7x+3x-21=0`

Or, `x(x-7)+3(x-7)=0`

Or, `(x+3)(x-7)`=0

Ruling out the negative value; Rehman’s Age = 7 year


Prev    Next

Exercise 4.1(1)

Exercise 4.1(2)

Exercise 4.2(1)

Exercise 4.2(2)

Exercise 4.2(3)

Exercise 4.3(part1)

Exercise 4.3(part2)

Exercise 4.3(part4)

Exercise 4.3(part5)

Exercise 4.4(part1)

Exercise 4.4(part2)