Example 16 - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

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Example 16 Simplify the rational expression (𝑥^2 − 7𝑥 + 12)/(5𝑥^2 + 5𝑥 − 100), assuming that 5𝑥^2+5𝑥−100≠0. We have to simplify (𝑥^2 − 7𝑥 + 12)/(5𝑥^2 + 5𝑥 − 100) Here, we factorise both the numerator and denominator separately and then divide them Factorising Numerator We have to factorise 𝑥^2 − 7𝑥 + 12 x2 – 7x + 12 Factorising by splitting the middle term = x2 – 4x – 3x + 12 = x(x – 4) – 3(x – 4) = (x – 3) (x – 4) Factorising Denominator We have to factorise 5𝑥^2 + 5𝑥 − 100 Now, 𝟓𝒙^𝟐 + 𝟓𝒙 − 𝟏𝟎𝟎 Since 5 is multiplied in all terms, taking it common Splitting the middle term method We need to find two numbers whose Sum = –7 Product = 1 × 12 = 12 Since sum is negative but product is positive. Thus, both numbers are negative =5(5/5 𝑥^2 +5/5 𝑥 −100/5) =5(𝒙^𝟐 +𝒙 −𝟐𝟎) Factorising by splitting the middle term =5(𝑥^2 +𝟓𝒙−𝟒𝒙 −20) =5(𝑥(𝑥+5)−4(𝑥+5)) =𝟓(𝒙+𝟓)(𝒙−𝟒) Thus, our rational expression becomes (𝒙^𝟐 − 𝟕𝒙 + 𝟏𝟐)/(𝟓𝒙^𝟐 + 𝟓𝒙 − 𝟏𝟎𝟎)=((𝑥 − 3)(𝑥 − 4))/(5(𝑥 + 5)(𝑥 − 4)) Cancelling (𝑥 − 4) from numerator and denominator =((𝒙 − 𝟑) )/(𝟓(𝒙 + 𝟓) ) Splitting the middle term method We need to find two numbers whose Sum = 1 Product = 1 × –20 = –20 Since product is negative, one number is negative, one is positive. And, sum is positive: so it means bigger number is positive =((𝒙 − 𝟑) )/(𝟓(𝒙 + 𝟓) )