{"id":1283,"date":"2020-09-15T09:07:49","date_gmt":"2020-09-15T03:37:49","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1283"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-9-ex-9-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-9-ex-9-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.2<\/h2>\n

Question 1.
\nFind the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients
\n(i) x2<\/sup> – 2x – 8
\n(ii) 4s2<\/sup> – 4s + 1
\n(iii) 6x2<\/sup> – 3 – 7x
\n(iv) 4u2<\/sup> – 8u
\n(v) t2<\/sup> – 15
\n(vi) 3x2<\/sup> – x – 4
\nSolution:
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB<\/p>\n

(iv) 4u2<\/sup> – 8u
\n= 4u2<\/sup> – 8u + 0
\n= 4u (u – 2)
\nIf 4u = 0, then u = 0
\nIf u – 2 = 0, then u = 2
\n\u2234 Zeroes are 0 and 2
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

(v) t2<\/sup> – 15
\n= t2<\/sup> + 0 – 15
\n\"KSEEB<\/p>\n

(vi) 3x2<\/sup> – x – 4
\n= 3x2<\/sup> – 4x + 3x – 4
\n= x(3x – 4) + 1 (3x – 4)
\n= (3x – 4) (x + 1)
\nIf 3x – 4 = 0, then x = \\(\\frac{4}{3}\\)
\nIf x + 1 = 0, then x = -1
\n\u2234 Zeroes are \\(\\frac{4}{3}\\) and -1.
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
\n(i) \\(\\frac{1}{4}\\), 1
\n(ii) \\(\\sqrt{2}, \\frac{1}{3}\\)
\n(iii) 0, \\(\\sqrt{5}\\)
\n(iv) 1, 1
\n(v) \\(-\\frac{1}{4}, \\frac{1}{4}\\)
\n(vi) 4, 1
\nSolution:
\n(i) \\(\\frac{1}{4}\\), 1
\nHere m + n = \\(\\frac{1}{4}\\), mn = -1
\n\u2234 Quadratic Equation
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

(ii) \\(\\sqrt{2}, \\frac{1}{3}\\)
\nHere m + n = \\(\\sqrt{2}\\), mn = \\(\\frac{1}{3}\\)
\n\u2234 Quadratic Equation
\n\"KSEEB<\/p>\n

(iii) 0, \\(\\sqrt{5}\\)
\nHere m + n = 0, mn = \\(\\sqrt{5}\\)
\n\u2234 Quadratic Equation
\n\"KSEEB<\/p>\n

(iv) Standard form of quadratic polynomial
\nsum and product of its zeroes is
\nK(x2<\/sup> – sum of the zeroes) x + product of zeroes.
\n= K(x2<\/sup> – 1x + 1)
\nTaking K = 1
\n= x2<\/sup> – x + 1<\/p>\n

(v) \\(-\\frac{1}{4}, \\frac{1}{4}\\)
\nHere m + n = \\(-\\frac{1}{4}\\), mn = \\(\\frac{1}{4}\\)
\n\u2234 Quadratic Equation
\n\"KSEEB<\/p>\n

(vi) Standard form of quadratic polynomial sum and product of its zeroes is
\n= K[x2<\/sup> – (sum of the zeroes) x + Product of zeroes]
\n= K(x2<\/sup> – 4x + 1)
\nTaking K = 1
\n= 1(x2<\/sup> – 4x + 1)
\n= x2<\/sup> – 4x + 1<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1. Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.2 Question 1. Find the zeroes of the following quadratic …<\/p>\n","protected":false},"author":2,"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\/1283"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1283"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1283\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1283"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1283"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}