{"id":25261,"date":"2020-09-15T16:50:15","date_gmt":"2020-09-15T11:20:15","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25261"},"modified":"2021-07-02T15:01:49","modified_gmt":"2021-07-02T09:31:49","slug":"kseeb-solutions-for-class-10-maths-chapter-9-ex-9-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-9-ex-9-4\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 9 Polynomials Ex 9.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.4<\/h2>\n

Question 1.
\nVerify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
\n(i) 2x3<\/sup> + x2<\/sup> – 5x + 2; \\(\\frac{1}{2}\\), 1, – 2
\n(ii) x3<\/sup> – 4x2<\/sup> + 5x – 2; 2, 1, 1
\nAnswer:
\nLet P(x) = 2x3<\/sup> + x2<\/sup> – 5x + 2 on comparing with general polynomial P(x) = ax3<\/sup> + bx2<\/sup> + cx + d we get
\na = 2, b = 1, c = – 5 and d = 2
\nGiven zeroes \\(\\frac{1}{2}\\), 1, – 2
\n\"KSEEB
\nP(1) = 2 (1)3<\/sup> + (1)2<\/sup> – 5(1) + 2
\n= 2 \u00d7 1 + 1 – 5 + 2
\n= 2 + 1 – 5 + 2 = 5 – 5 = 0
\nP(1) = 0
\nP(- 2) = 2(- 2)3<\/sup> + (-2)2<\/sup> – 5(- 2) + 2
\n= 2 \u00d7 – 8 + 4 + 10 + 2
\n= – 16 + 16
\nP(- 2) = 0
\nHence 1\/2, 1 & -2 are the zeroes of the given cubic polynomial
\nAgain consider \u03b1 = 1\/2, \u03b2 = 1 & \u03b3 = – 2
\n\"KSEEB<\/p>\n

ii) Let P(x) = x3<\/sup> – 4x2<\/sup> + 5x – 2 on comparing with general polynomial
\nP(x) = ax3<\/sup> + bx2<\/sup> + cx + d
\nwe get a = 1, b = – 4, c = 5 & d = 2.
\nGiven zeroes 2, 1, 1
\nP(2) = (2)3<\/sup> – 4(2)2<\/sup> + 5(2) – 2
\n= 8 – 4 \u00d7 4 + 10 – 2
\n= 8 – 16 + 10 – 2
\nP(2) = 18 – 18 = 0
\nP(2) = 0
\nP(1) = (1)3<\/sup> – 4(1)2<\/sup> + 5(1) – 2
\n= 1 – 4 + 5 – 2
\nP(1) = 6 – 6 = 0
\nP(1) = 0
\nHence, 2, 1 and 1 are the zeroes of the given cubic polynomial.
\nAgain, consider \u03b1 = 2, \u03b2 = 1 & \u03b3 = 1
\n\u03b1 + \u03b2 + \u03b3 = 2 + 1 + 1 = 4
\nand \u03b1 + \u03b2 + \u03b3 = \\(\\frac{-b}{a}=\\frac{-(-4)}{1}\\) = 4
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = 2(1) + (1) (1) + (1) \u00d7 (2)
\n= 2 + 1 + 2 = 5
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{c}{a}=\\frac{5}{l}\\) = 5
\n\u03b1\u03b2\u03b3 = 2 \u00d7 1 \u00d7 1 = 2
\n\u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}=\\frac{-(-2)}{1}=\\frac{2}{1}\\) = 2<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
\nAnswer:
\nLet the cubic polynomial be P(x) = ax3<\/sup> + bx2<\/sup> + cx + d
\nThen sum of zeroes = \\(\\frac{-b}{a}\\) = 2
\nSum of the product of zeroes taken two at a time = \\(\\frac{c}{a}\\) = – 7
\nProduct of zeroes = \\(\\frac{-d}{a}\\) = – 14
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 3.
\nIf the zeroes of the polynomial x3<\/sup> – 3x2<\/sup> + x + 1 are a – b, a, a + b, find a and b.
\nAnswer:
\nThe given Polynomial is x3<\/sup> – 3x2<\/sup> + x + 1
\ncomparing with Ax3<\/sup> + Bx2<\/sup> + Cx + D,
\nA = 1, B = – 3, C = 1 & D = 1
\nLet \u03b1 = a – b, \u03b2 = a & \u03b3 = a + b
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{B}{A}=\\frac{-(-3)}{1}\\) = 3
\na – b + a + a + b = 3
\n3a = 3
\na = \\(\\frac{3}{3}\\) = 1
\na = 1
\n\u03b1\u03b2\u03b3 = \\(\\frac{-D}{A}\\) = – 1
\n(a – b) a (a + b) = – 1 Put a = 1
\n(1 – b) 1 (1 + b) = – 1
\n1 – b2<\/sup> = – 1
\n1 + 1 = b2<\/sup>
\nb2<\/sup> = 2
\nb = \u00b1\\(\\sqrt{2}\\)
\n\u2234 a = 1 & b = \u00b1\\(\\sqrt{2}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIf two zeroes of the polynomial x4<\/sup> – 6x3<\/sup> – 26x2<\/sup> + 138x – 35 are 2 \u00b1\\(\\sqrt{3}\\). find other zeroes.
\nAnswer:
\nTwo zeroes of the given polynomial x4<\/sup> – 6x3<\/sup> – 26x2<\/sup> + 138x – 35 are 2 \u00b1 \\(\\sqrt{3}\\).
\n\u2234 x = 2 \u00b1 \\(\\sqrt{3}\\).
\nTherefore [(x – (2 + \\(\\sqrt{3}\\))] [x – (2 – \\(\\sqrt{3}\\))]
\n= (x – 2 – \\(\\sqrt{3}\\)) (x – 2 + \\(\\sqrt{3}\\))
\n= [(x – 2) – \\(\\sqrt{3}\\)][(x – 2) + \\(\\sqrt{3}\\))]
\n= (x – 2)2<\/sup> – (\\(\\sqrt{3}\\))2<\/sup>
\n= x2<\/sup> – 4x + 4 – 3 = x2<\/sup> – 4x + 1
\n\u2234 It is factor of Polynomial apply division algorithm.
\n\"KSEEB
\nOther two zeroes are given by the Quotient
\nx2<\/sup> – 2x – 35 = 0
\nx2<\/sup> – 7x + 5x – 35 = 0
\nx(x – 7) + 5(x – 7) = 0
\n(x – 7)(x + 5) = 0
\nx – 7 = 0 (or) x + 5 = 0
\nx = 7 (or) x = – 5
\n\u2234 x = 7, – 5<\/p>\n

Question 5.
\nIf the polynomial x4<\/sup> – 6x3<\/sup> + 16x2<\/sup> – 25x + 10 is divided by another polynomial x2<\/sup> – 2x + k, the remainder comes out to be x + a, find k and a.
\nAnswer:
\nLet us apply the division algorithm to the given Polynomial
\n\"KSEEB
\nBut remainder is given to be x + a comparing coefficients of x in the remainder obtained and remainder given:
\n(2K – 9)x – K(8 – K)+ 10 = x + a
\n\u2234 2K – 9 = 1
\n\u2234 2K = 1 + 9
\nK = \\(\\frac{10}{2}\\)
\nK = 5 and – K(8 – K) + 10 = a
\nPut K = 5
\n– 5(8 – 5) + 10 = a
\n– 5(3) + 10 = a
\n– 15 + 10 = a
\na = – 5<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 9 Polynomials Ex 9.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.4 Question 1. Verify that the numbers …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25261"}],"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=25261"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25261\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25261"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25261"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25261"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}