{"id":348,"date":"2025-01-08T13:09:40","date_gmt":"2025-01-08T07:39:40","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=348"},"modified":"2025-01-09T10:07:34","modified_gmt":"2025-01-09T04:37:34","slug":"samacheer-kalvi-9th-maths-guide-chapter-3-ex-3-7","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-9th-maths-guide-chapter-3-ex-3-7\/","title":{"rendered":"Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7"},"content":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.7<\/h2>\n

Question 1.
\nFind the quotient and remainder of the following.
\n(i) 4x3<\/sup> + 6x2<\/sup> – 23x + 18) \u00f7 (x + 3)
\nSolution:
\n\"Samacheer
\n\u2234 The quotient = 4x2<\/sup> – 6x – 5
\nThe remainder = 33<\/p>\n

(ii) (8y3<\/sup> – 16y2<\/sup> + 16y – 15) \u00f7 (2y – 1)
\nSolution:
\n\"Samacheer
\n\u2234 The quotient = 4y2<\/sup> – 6y + 5
\nThe remainder = -10<\/p>\n

\"Samacheer<\/p>\n

(iii) (8x3<\/sup> – 1) \u00f7 (2x – 1)
\nSolution:
\n\"Samacheer
\n\u2234 The quotient = 4x2<\/sup> + 2x + 1
\nThe remainder = 0<\/p>\n

(iv) (-18z + 14z2<\/sup> + 24z3<\/sup> + 18) \u00f7 (3z + 4)
\nSolution:
\n\"Samacheer
\n\u2234The quotient = 8z2<\/sup> – 6z + 2
\nThe remainder = 10<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nThe area of a rectangle is x2<\/sup> + 7x + 12. If its breadth is (x + 3) then find its length.
\nSolution:
\nLet the length of the rectangle be “l”
\nThe breadth of the rectangle = x + 3
\nArea of the rectangle = length \u00d7 breadth
\nx2<\/sup> + 7x + 12 = l(x + 3)
\n\"Samacheer
\nLength of the rectangle = x + 4
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nThe base of a parallelogram is (5x + 4). Find its height if the area is 25x2<\/sup> – 16.
\nSolution:
\nLet the height of the parallelogram be “h”.
\nBase of the parallelogram = 5x + 4
\nArea of a parallelogram = 25x2<\/sup> – 16
\n\u2234 Base x Height = 25x2<\/sup> – 16
\n(5x + 4) \u00d7 h = 25x2<\/sup>– 16
\n\"Samacheer
\nHeight of the parallelogram = 5x – 4
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nThe sum of (x + 5) observations is (x3<\/sup> + 125). Find the mean of the observations.
\nSolution:
\nSum of the observation = x3<\/sup> + 125
\nNumber of observation = x + 5
\n\"Samacheer
\nMean = x2<\/sup> – 5x + 25
\n\"Samacheer<\/p>\n

Question 5.
\nFind the quotient and remainder for the following using synthetic division:
\n(i) (x3<\/sup> + x2<\/sup> – 7x – 3) \u00f7 (x – 3)
\nSolution:
\np(x) = x3<\/sup> + x2<\/sup> – 7x – 3
\nd(x) = x – 3 [p(x) = d(x) \u00d7 q(x) + r]
\n\"Samacheer
\nx – 3 = 0
\nx = 3
\nHence the quotient = x2<\/sup> + 4x + 5
\nRemainder = 12<\/p>\n

\"Samacheer<\/p>\n

(ii) (x3<\/sup> + 2x2<\/sup> – x – 4) \u00f7 (x + 2)
\nSolution:
\np(x) = x3<\/sup> + 2x2<\/sup> -x – 4
\nd(x) = x + 2
\n\"Samacheer
\nx + 2 = 0
\nx = -2
\nThe quotient = x2<\/sup> – 1
\nRemainder = -2<\/p>\n

(iii) (3x3<\/sup> – 2x2<\/sup> + 7x – 5) \u00f7 (x + 3)
\nSolution:
\np(x) = 3x3<\/sup> – 2x2<\/sup> + 7x – 5
\nd(x) = x + 3
\n\"Samacheer
\nx + 3 = 0
\nx = -3
\nThe quotient = 3x2<\/sup> – 11x + 40
\nRemainder = -125<\/p>\n

(iv) (8x4<\/sup> – 2x2<\/sup> + 6x + 5) \u00f7 (4x + 1)
\nSolution:
\np(x) = 8x4<\/sup> – 2x2<\/sup> + 6x + 5
\nd(x) = 4x + 1
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nIf the quotient obtained on dividing (8x4<\/sup> – 2x2<\/sup> + 6x – 7) by (2x + 1) is (4x3<\/sup> + px2<\/sup> -qx + 3), then find p, q and also the remainder.
\nSolution:
\np(x) = 8x4<\/sup> – 2x2<\/sup> + 6x – 7
\nd(x) = 2x + 1
\n\"Samacheer
\n2x + 1 = 0
\n2x = -1
\nx = –\\(\\frac{1}{2}\\)
\nThe quotient = \\(\\frac{1}{2}\\) [8x3<\/sup> – 4x2<\/sup> + 6]
\n= 4x3<\/sup> – 2x2<\/sup> + 3
\n= 4x3<\/sup> – 2x2<\/sup> + 0x + 3
\nThe given quotient is = 4x3<\/sup> + px2<\/sup> – qx + 3
\n(compared with the given quotient)
\nThe value of p = -2 and q = 0
\nRemainder = -10<\/p>\n

\"Samacheer<\/strong><\/p>\n

Question 7.
\nIf the quotient obtained on dividing 3x3<\/sup> + 11x2<\/sup> + 34x + 106 by x – 3 is 3x2<\/sup> + ax + b, then find a, b and also the remainder.
\nSolution:
\np(x) = 3x3<\/sup> + 11x2<\/sup> + 34x + 106
\nd(x) = x – 3
\n\"Samacheer
\nx – 3 = 0
\nx = 3
\nThe quotient is = 3x2<\/sup> + 20x + 94
\nThe given quotient is = 3x2<\/sup> + ax + b
\nCompared with the given quotient
\nThe value of a = 20 and b = 94
\nThe remainder = 388<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.7 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[3],"tags":[],"class_list":["post-348","post","type-post","status-publish","format-standard","hentry","category-class-9"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/348"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=348"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/348\/revisions"}],"predecessor-version":[{"id":39814,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/348\/revisions\/39814"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}