Question 8 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

Transcript

Question 8 If a number plus its reciprocal equals 10/3, find the number. Let Number = 𝒙 Thus, Reciprocal of our number = 𝟏/𝒙 Given that number plus its reciprocal equals 10/3 So, we can write Number + Reciprocal = 10/3 𝒙+𝟏/𝒙=𝟏𝟎/𝟑 (𝑥 × 𝑥 + 1)/𝑥=10/3 (𝑥^2 + 1)/𝑥=10/3 Cross multiplying 3(𝑥^2+1)=10𝑥 3𝑥^2+3=10𝑥 𝟑𝒙^𝟐−𝟏𝟎𝒙+𝟑=𝟎 Factorising by Splitting the middle term 3𝑥^2−𝟗𝒙−𝒙+3=0 3𝑥(𝑥−3)−1(𝑥−3)=0 (𝒙−𝟑)(𝟑𝒙−𝟏)=𝟎 Thus, 𝑥−3=0 𝒙=𝟑 And, 3𝑥−1=0 3𝑥=1 𝒙=𝟏/𝟑 So, our required numbers are 3 and 𝟏/𝟑 Splitting the middle term method We need to find two numbers whose Sum = –10 Product = 3 × 3 = 9 Since sum is negative but product is positive. Thus, both numbers are negative