Quadratic Equation
NCERT Exercise 4.3
Part 3
Question 3: Find the roots of the following equations:
Question (i): `x-1/x=3: x≠0`
Answer:
`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`
Answer:
`(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.
Answer: 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