{"id":22355,"date":"2020-02-04T14:42:50","date_gmt":"2020-02-04T09:12:50","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=22355"},"modified":"2021-07-02T15:25:50","modified_gmt":"2021-07-02T09:55:50","slug":"1st-puc-basic-maths-question-bank-chapter-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-question-bank-chapter-6\/","title":{"rendered":"1st PUC Basic Maths Question Bank Chapter 6 Theory of Equations"},"content":{"rendered":"

Students can Download Basic Maths Chapter 6 Theory of Equations Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Basic Maths Question Bank Chapter 6 Theory of Equations<\/h2>\n

Question 1.
\nFind the nature of roots of the equation 3x2<\/sup> – 4x + 9 = 0
\nAnswer:
\nNature of roots depends on b2<\/sup> – 4ac
\n\u21d2 a = 3, b = – 4, c = 9.
\n\u2234 b2<\/sup> – 4ac = (-4)2<\/sup> – 4 . (3) . 9 = 16 – 108 = – 92 < 0
\n\u2234 The roots are imaginary.<\/p>\n

Question 2.
\nIf a and B are the roots of 4x2<\/sup> + 3x – 5 = 0. Find the value of \\(\\frac{\\alpha^{2}}{\\beta}+\\frac{\\beta^{2}}{\\alpha}\\)
\nAnswer:
\na = 4, b = 3, c = – 5
\nSum of roots = \u03b1 + \u03b2 = \\(\\frac{-b}{a}=\\frac{-3}{4}\\)
\nProduct of roots = \u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{-5}{4}\\)
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 3.
\nFind three consecutive even natural numbers, such that the sum of their squares is 200.
\nAnswer:
\nLet the required even numbers be x, x + 2, x + 4
\nBy data x2<\/sup> + (x + 2)2<\/sup> + (x + 4)2<\/sup> = 200
\n\u21d2 x2<\/sup> + x2<\/sup> + 4x + 4 + x2<\/sup> + 8x + 16 – 200 = 0
\n3x2<\/sup> + 12x – 180 = 0 \u21d2 x2<\/sup> + 4x – 60 = 0
\n\u21d2 (x + 10)(x – 6) =0
\nwe require +ve the integers \u2234 x = 6
\n\u21d2 6,8, 10 be the numbers.<\/p>\n

Question 4.
\nA piece of cloth costs Rs. 200. If the piece were 5mts longer and each meter of cloth costed Rs. 2 less, the cost of piece would have remained unchanged. How long is the piece and what is its original rate per meter.
\nAnswer:
\nLet x be the length of the cloth, its cost = 200
\n\u2234 Rate of cloth per meter = \\(\\frac{200}{x}\\)
\nIncreased length of piece of cloth = (x + 5) mts
\nNow rate of cloth per meter = \\(\\frac{200}{21+5}\\)
\nBy date, old rate cloth – new rate of cloth = Rs 2.
\n\"1st
\n\"1st
\n\"1st
\n\u21d2 x2<\/sup> + 5x – 500 = 0
\n\u21d2 x2<\/sup> + 25x – 20x – 500 = 0
\n\u21d2 x(x + 25) – 20(x + 25) = 0
\n\u21d2 x = 20 or -25 (neglected) [\u2234 x = 20mts]
\nlength of cloth = 20 mts, rate per metre = Rs. 10.<\/p>\n

Question 5.
\nSolve : x3<\/sup> – 6x2<\/sup> + 11x – 6 = 0. since that ratio of two roots is 2 : 3
\nAnswer:
\nBy data \u03b1 : \u03b2 = 2 : 3
\n\u21d2 \\(\\frac{\\alpha}{\\beta}=\\frac{2}{3}\\) = \u03b1 = 2\u03b2
\nfrom the given equation,
\n\"1st
\n\u21d2 5\u03b13<\/sup> – 12\u03b12<\/sup> + 8 = 0
\nsolving \u03b1 = 2, \u2234 \u03b1 = 1
\n\u2234 The roots are 1,2,3<\/p>\n

\"1st<\/p>\n

Question 6.
\nSolve by synthetic division that it has atleast one integral root between -3 and 3 for the equation x4<\/sup> – 9x2<\/sup> + 4x + 12 = 0
\nAnswer:
\nLet f(x) = x4<\/sup> – 9X2<\/sup> + 4x + 12 = 0
\nBu inspection x = \u00b11, \u00b12, \u00b13 are the roots or not
\nf(-3) = 0, f(-1) = 0, f(2) = 0
\n\u2234 -3, -1, 2 are the roots of the given equation
\nand
\n\"1stq
\n\u2234 x = – 3 is a root and x = – 2 is not a root
\nTry for x = – 1, x = -1 is a root
\nNow the quotient is x2<\/sup> – 4x + 4 = 0
\n\u21d2 (x – 2)2<\/sup> = 0 = x = 2, 2
\nHence the form root are -3, -1,2, 2<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Chapter 6 Theory of Equations Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Basic Maths Question Bank Chapter 6 Theory of Equations Question 1. Find the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22355"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=22355"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22355\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=22355"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=22355"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=22355"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}