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

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Question 4 (i) Simplify the following: (i) (4𝑥^2 + 4𝑥 + 1)/(4𝑥^2 − 1) Factorising numerator and denominator separately Factorising Numerator We have to factorise 4𝑥^2 + 4𝑥 + 1 This looks like (𝒂+𝒃)^𝟐 formula Now, 4𝑥^2 + 4𝑥 + 1 =" " 4𝑥^2 +1+4𝑥 =(2𝑥)^2+1^2+2 × 2𝑥 × 1 Using (𝑎+𝑏)^2 = 𝑎^2 + 𝑏^2 + 2ab Where 𝑎 = 2𝑥, b = 1 =(𝟐𝒙+𝟏)^𝟐 Factorising Denominator We have to factorise 4𝑥^2 − 1 This looks like 𝒂^𝟐−𝒃^𝟐 formula Now, 4𝑥^2 − 1 =(𝟐𝒙)^𝟐−(𝟏)^𝟐 Using 𝑎^2−𝑏^2=(𝑎+𝑏)(𝑎−𝑏) Where 𝑎 = 2𝑥, b = 1 =(𝟐𝒙+𝟏)(𝟐𝒙−𝟏) Thus, our rational expression becomes (𝟒𝒙^𝟐 + 𝟒𝒙 + 𝟏)/(𝟒𝒙^𝟐 − 𝟏)=(2𝑥 + 1)^2/((2𝑥 + 1)(2𝑥 − 1)) =(2𝑥 + 1)(2𝑥 + 1)/((2𝑥 + 1)(2𝑥 − 1)) =((𝟐𝒙 + 𝟏))/( (𝟐𝒙 − 𝟏)) Question 4 (ii) Simplify the following: (ii) 9(3𝑎^3 − 24𝑏^3 )/(9𝑎^2 − 36𝑏^2 ) Factorising numerator and denominator separately Factorising Numerator We have to factorise 9(3𝑎^3 − 24𝑏^3 ) This looks like 𝒂^𝟑−𝒃^𝟑 formula Now, 9(3𝑎^3 − 24𝑏^3 ) Taking 3 common from terms inside brackets =9 × 3(3/3 a^3−24/3 b^3 ) =9 × 3(a^3−8b^3 ) =27(a^3−8b^3 ) =𝟐𝟕(𝒂^𝟑−(𝟐𝒃)^𝟑 ) Using 𝑎^3−𝑏^3=(𝑎−𝑏)(𝑎^2+𝑎𝑏+𝑏^2) Where 𝑎 = 𝑎, b = 2𝑏 Factorising Denominator We have to factorise 9𝑎^2 − 36𝑏^2 This looks like 𝒂^𝟐−𝒃^𝟐 formula Now, 9𝑎^2 − 36𝑏^2 =(𝟑𝒂)^𝟐−(𝟔𝒃)^𝟐 =(𝟑𝒂+𝟔𝒃)(𝟑𝒂−𝟔𝒃) Taking 3 common from both brackets =3(𝑎+2𝑏) × 3 (𝑎−2𝑏) =𝟗(𝒂+𝟐𝒃) (𝒂−𝟐𝒃) Using 𝑎^2−𝑏^2=(𝑎+𝑏)(𝑎−𝑏) Where 𝑎 = 3𝑎, b = 6𝑏 Thus, our rational expression becomes 9(3𝑎^3 − 24𝑏^3 )/(9𝑎^2 − 36𝑏^2 )=(27(𝑎 − 2𝑏) (𝑎^2 + 2𝑎𝑏 + 4𝑏^2 ))/(9(𝑎 + 2𝑏) (𝑎 − 2𝑏)) =𝟑(𝒂^𝟐 + 𝟐𝒂𝒃 + 𝟒𝒃^𝟐 )/((𝒂 + 𝟐𝒃) ) Question 4 (iii) Simplify the following: (iii) (𝑠^3 + 125𝑡^3)/(𝑠^2 − 2𝑠𝑡 − 35𝑡^2 ) Factorising numerator and denominator separately Factorising Numerator We have to factorise 𝒔^𝟑+𝟏𝟐𝟓𝒕^𝟑 This looks like 𝒂^𝟑+𝒃^𝟑 formula Now, 𝑠^3 + 125𝑡^3 =𝒔^𝟑+(𝟓𝒕)^𝟑 Using 𝑎^3+𝑏^3=(𝑎+𝑏)(𝑎^2−𝑎𝑏+𝑏^2) Where 𝑎 = 𝑠, b = 5𝑡 = (𝑠+5𝑡) (𝑠^2−𝑠 × 5𝑡+(5𝑡)^2 ) =(𝒔+𝟓𝒕)(𝒔^𝟐−𝟓𝒔𝒕+𝟐𝟓𝒕^𝟐 ) Factorising Denominator We have to factorise 𝑠^2−2𝑠𝑡−35𝑡^2 Now, 𝑠^2−2𝑠𝑡−35𝑡^2 = 𝒔^𝟐−𝟐𝒕𝒔−𝟑𝟓𝒕^𝟐 Taking s as main variable, and t as constant Factorising by Splitting the middle term Splitting the middle term method We need to find two numbers whose Sum = –2t Product = 1 × –35t2 = –35t2 Since both sum and product are negative. Thus, one number is positive, one is negative. And bigger is negative = 𝑠^2−𝟕𝒕𝒔+𝟓𝒕𝒔−35𝑡^2 = 𝑠(𝑠−7)+5𝑡(𝑠−7) = (𝒔+𝟓𝒕)(𝒔−𝟕) Thus, our rational expression becomes (𝑠^3 + 125𝑡^3)/(𝑠^2 − 2𝑠𝑡 − 35𝑡^2 )=(𝑠 + 5𝑡)(𝑠^2 − 5𝑠𝑡 + 25𝑡^2 )/((𝑠 + 5𝑡)(𝑠 − 7) ) =((𝒔^𝟐 − 𝟓𝒔𝒕 + 𝟐𝟓𝒕^𝟐 ))/((𝒔 − 𝟕) )