If both x – 2 and x – 1/2 are factors of px2 + 5x + r, show that p = r - End-of-Chapter Exercises

part 2 - Question 10 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I) - Class 9
part 3 - Question 10 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I) - Class 9

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Question 10 If both š‘„āˆ’2 and š‘„āˆ’1/2 are factors of š‘š‘„^2+5š‘„+š‘Ÿ, show that š‘=š‘Ÿ. If both š‘„āˆ’2 and š‘„āˆ’1/2 are factors of š‘š‘„^2+5š‘„+š‘Ÿ, Then we can write š’Œ(š’™āˆ’šŸ)(š’™āˆ’šŸ/šŸ)= š’‘š’™^šŸ+šŸ“š’™+š’“ Where k is any constant Now, solving our equation š’Œ(š’™āˆ’šŸ)(š’™āˆ’šŸ/šŸ)= š’‘š’™^šŸ+šŸ“š’™+š’“ š‘˜[š‘„(š‘„āˆ’1/2)āˆ’2(š‘„āˆ’1/2)]= š‘š‘„^2+5š‘„+š‘Ÿ š‘˜[š‘„^2āˆ’1/2 š‘„āˆ’2š‘„+2 Ɨ1/2]= š‘š‘„^2+5š‘„+š‘Ÿ š‘˜[š‘„^2āˆ’1/2 š‘„āˆ’2š‘„+1]= š‘š‘„^2+5š‘„+š‘Ÿ š‘˜[š‘„^2āˆ’š‘„(1/2+2) +1]= š‘š‘„^2+5š‘„+š‘Ÿ š‘˜[š‘„^2āˆ’š‘„((1 + 2 Ɨ 2)/2) +1]= š‘š‘„^2+5š‘„+š‘Ÿ š‘˜[š‘„^2āˆ’š‘„((1 + 4)/2) +1]= š‘š‘„^2+5š‘„+š‘Ÿ š‘˜[š‘„^2āˆ’5/2 š‘„ +1]= š‘š‘„^2+5š‘„+š‘Ÿ ć€–š’Œš’™ć€—^šŸāˆ’šŸ“š’Œ/šŸ š’™ +š’Œ = š’‘š’™^šŸ+šŸ“š’™+š’“ Comparing x2 terms š’Œ=š’‘ Comparing constant terms š’Œ=š’“ Thus, š’‘=š’“ Hence proved Comparing constant terms š’Œ=š’“ Thus, š’‘=š’“ Hence proved Note: By comparing x terms, we can find value of k. And hence find value of p & r

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