{"id":509,"date":"2023-10-23T11:30:43","date_gmt":"2023-10-23T06:00:43","guid":{"rendered":"https:\/\/samacheer-kalvi.com\/?p=509"},"modified":"2023-11-10T10:47:03","modified_gmt":"2023-11-10T05:17:03","slug":"samacheer-kalvi-10th-maths-guide-chapter-3-ex-3-13","status":"publish","type":"post","link":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-10th-maths-guide-chapter-3-ex-3-13\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13"},"content":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.13 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.13<\/h2>\n

Question 1.
\nDetermine the nature of the roots for the following quadratic equations
\n(i) 15x2<\/sup> + 11x + 2 = 0
\nAnswer:
\nHere a = 15, b = 11, c = 2
\n\u2206 = b2<\/sup> – 4ac
\n\u2206 = 112<\/sup> – 4(15) \u00d7 2
\n= 121 – 120
\n\u2206 = 1 > 0
\nSo the equation will have real and unequal roots<\/p>\n

(ii) x2<\/sup> – x – 1 = 0
\nAnswer:
\nHere a = 1, b = -1, c = -1
\n\u2206 = b2<\/sup> – 4ac
\n= (-1)2<\/sup> – 4(1)(-1)
\n= 1 + 4 = 5
\n\u2206 = 1 > 0
\nSo the equation will have real and unequal roots.<\/p>\n

\"Samacheer<\/p>\n

(iii) \\(\\sqrt { 2 }\\) t2<\/sup> – 3t + 3\\(\\sqrt { 2 }\\) = 0
\nAnswer:
\nHere a = \\(\\sqrt { 2 }\\) , b = -3, c = 3\\(\\sqrt { 2 }\\)
\n\u2206 = b2<\/sup> – 4ac
\n= (-3)2<\/sup> – 4(\\(\\sqrt { 2 }\\)) (3\\(\\sqrt { 2 }\\))
\n= 9 – 24 = -15
\n\u2206 = -15 < 0
\nSo the equation will have no real roots.<\/p>\n

(iv) 9y2<\/sup> – 6\\(\\sqrt { 2y }\\) + 2 = 0
\nAnswer:
\nHere a = 9, b = -6\\(\\sqrt { 2 }\\), c = 2
\n\u2206 = b2<\/sup> – 4ac
\n= (-6\\(\\sqrt { 2 }\\))2<\/sup> – 4(9) (2)
\n= 72 – 72
\n= 0
\nSo the equation will have real and equal roots.<\/p>\n

(v) 9a2<\/sup>b2<\/sup>x2<\/sup> – 24abcdx + 16c2<\/sup>d2<\/sup> = 0, a \u2260 0, b \u2260 0
\nAnswer:
\nHere a = 9a2<\/sup>b2<\/sup>; b = -24 abed, c = 16c2<\/sup>d2<\/sup>
\n\u2206 = b2<\/sup> – 4ac
\n= (-24abcd)2<\/sup> – 4(9a2<\/sup>b2<\/sup>) (16c2<\/sup>d2<\/sup>)
\n= 576a2<\/sup>b2<\/sup>c2<\/sup>d2<\/sup> – 576a2<\/sup>b2<\/sup>c2<\/sup>d2<\/sup>
\n\u2206 = 0
\nSo the equation will have real and equal roots.<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nFind the value(s) of ‘k\u2019 for which the roots of the following equations are real and equal.
\n(i) (5k – 6)2<\/sup> + 2kx + 1 = 0
\n(ii) kx2<\/sup> + (6k + 2)x + 16 = 0
\nSolution:
\n(5k – 6)x2<\/sup> + 2kx + 1 :
\na = (5k – 6), b = 2, c = 1
\n\u0394 = b2<\/sup> – 4ac
\n\u21d2 (2k)2<\/sup> – 4 (5k – 6)(1)
\n\u21d2 4k2<\/sup> – 20k + 24 = 0 [\u2235 Since the roots are real and equal)
\n\u21d2 k2<\/sup> – 5k + 6 = 0
\n\u21d2 (k – 3)(k – 2) = 0
\nk = 3, 2<\/p>\n

(ii) kx2<\/sup> + (6k + 2)x + 16 = 0
\na = k, b = (6k + 2), c = 16
\n\u0394 = b2<\/sup> – 4ac [\u2235 the roots are real and equal)
\n\u21d2 (6k + 2)2<\/sup> – 4 \u00d7 k \u00d7 16 = 0
\n\u21d2 36k2<\/sup> + 24k + 4 – 64k = 0
\n\u21d2 36k2<\/sup> – 40k + 4 =0
\n\u21d2 36k2<\/sup> – 36k – 4k + 4 =0
\n\u21d2 36k (k – 1) – 4 (k – 1) = 0
\n\u21d2 4 (k – 1) (9k – 1) =0
\n\u21d2 k = 1 or k = \\(\\frac{1}{9}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nIf the roots of (a – b)x2<\/sup> + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.
\nAnswer:
\n(a – b) x2 + (b – c) x + (c – a) = 0
\nHere a = (a – b);b = b – c ; c = c – a
\n\"Samacheer
\nSince the equation has real and equal roots \u2206 = 0
\n\u2234 b2<\/sup> – 4ac = 0
\n(b – c)2<\/sup> – 4(a – b)(c – a) = 0
\nb2<\/sup> + c2<\/sup> – 2bc -4 (ac – a2<\/sup> – bc + ab) = 0
\nb2<\/sup> + c2<\/sup> – 2bc – 4ac + 4a2<\/sup> + 4bc – 4ab = 0
\nb2<\/sup> + c2<\/sup> + 2bc -4a (b + c) + 4a2<\/sup> = 0
\n(b + c)2<\/sup> – 4a (b + c) + 4a2<\/sup> = 0
\n[(b+c) – 2a]2<\/sup> = 0 [using a2<\/sup> – 2ab + b2<\/sup> = (a – b)2<\/sup>]
\nb + c – 2a = 0
\nb + c = 2a
\nb + c = a + a
\nc – a = a – b (t2<\/sub> – t1<\/sub> = t3<\/sub> – t2<\/sub>)
\nb,a,c are in A.P.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nIf a, b are real then show that the roots of the equation (a – b)2<\/sup> – 6(a + b)x – 9(a – b) = 0 are real and unequal.
\nSolution:
\n(a – b)x2<\/sup> – 6(a + b)x – 9(a – b) = 0
\n\u0394 = b2<\/sup> – 4ac
\n= (-6(a + b)2<\/sup> – 4(a – b)(-9(a – b))
\n= 36(a + b)2<\/sup> + 36(a – b)2<\/sup>
\n= 36 (a2<\/sup> + 2ab + b2<\/sup>) + 36(a2<\/sup> – 2ab + b2<\/sup>)
\n= 72a2<\/sup> + 12b2<\/sup>
\n= 72(a2<\/sup> + b2<\/sup>) > 0
\n\u2234 The roots are real and unequal.<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nIf the roots of the equation (c2<\/sup> – ab)x2<\/sup> – 2(a2<\/sup> – bc)x + b2<\/sup> – ac = 0 are real and equal prove that either a = 0 (or) a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc
\nAnswer:
\n(c2<\/sup> – ab)x2<\/sup> – 2(a2<\/sup> – bc)x + b2<\/sup> – ac = 0
\nHere a = c2<\/sup> – ab ; b = – 2 (a2<\/sup> – bc); c = b2<\/sup> – ac
\nSince the roots are real and equal
\n\u2206 = b2<\/sup> – 4ac
\n[-2 (a2<\/sup> – bc)]2<\/sup> – 4(c2<\/sup> – ab) (b2<\/sup> – ac) = 0
\n4(a2<\/sup> – bc)2<\/sup> – 4[c2<\/sup> b2<\/sup> – ac3<\/sup> – ab3<\/sup> + a2<\/sup>bc] = 0
\nDivided by 4 we get
\n(a2<\/sup> – bc)2<\/sup> – [c2<\/sup> b2<\/sup> – ac3<\/sup> – ab3<\/sup> + a2<\/sup>bc] = 0
\na4<\/sup> + b2<\/sup> c2<\/sup> – 2a2<\/sup> bc – c2<\/sup>b2<\/sup> + ac3<\/sup> + ab3<\/sup> – a2<\/sup>bc = 0
\na4<\/sup> + ab3<\/sup> + ac3<\/sup> – 3a2<\/sup>bc = 0
\n= a(a3<\/sup> + b3<\/sup> + c3<\/sup>) = 3a2<\/sup>bc
\na3<\/sup> + b3<\/sup> + c3<\/sup> = \\(\\frac{3 a^{2} b c}{a}\\)
\na3<\/sup> + b3<\/sup> + c3<\/sup> = 3 abc
\nHence it is proved<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.13 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.13 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/509"}],"collection":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/comments?post=509"}],"version-history":[{"count":1,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/509\/revisions"}],"predecessor-version":[{"id":49466,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/509\/revisions\/49466"}],"wp:attachment":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/media?parent=509"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/categories?post=509"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/tags?post=509"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}