{"id":4322,"date":"2023-12-18T08:44:05","date_gmt":"2023-12-18T03:14:05","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=4322"},"modified":"2023-12-19T17:50:09","modified_gmt":"2023-12-19T12:20:09","slug":"samacheer-kalvi-12th-maths-solutions-chapter-3-ex-3-5","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-maths-solutions-chapter-3-ex-3-5\/","title":{"rendered":"Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5"},"content":{"rendered":"

You can Download Samacheer Kalvi 12th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5<\/h2>\n

Question 1.
\nSolve the following equations:
\n(i) sin2<\/sup>x – 5sinx + 4 = 0
\n(ii) 12x3<\/sup> + 8x = 29x2<\/sup> – 4
\nSolution:
\n(i) sin2<\/sup>x – 5sinx + 4 = 0
\nLet y = sin x
\n(y2<\/sup> – 5y + 4 = 0
\n(y – 1) (y – 4) = 0
\n(y – 1) = o or (y – 4) = 0
\ny = 1 or y = 4
\nsin x = 1 or sin x = 4 [not possible since sin x \u2264 1]
\nsin x = sin \\(\\frac{\\pi}{2}\\)
\nx = n\u03c0 + (-1)n<\/sup> \u03b1, n \u2208 z.
\nx = n\u03c0 + (-1)n<\/sup> \\(\\frac{\\pi}{2}\\)
\n(ii) 12x3<\/sup> + 8x = 29x2<\/sup> – 4
\n12x3<\/sup> – 29x2<\/sup> + 8x + 4 = 0 ….. (1)
\nBy Trail and error method, (x – 2) is a factor of (1)
\nThe other factor is 12x2<\/sup> – 5x – 2
\nThe roots is 12x2<\/sup> – 5x – 2 = 0
\n(3x – 2) (4x + 1) = 0
\nx = \\(\\frac{2}{3}\\), x = \\(-\\frac{1}{4}\\)
\nThe roots are 2, \\(\\frac{2}{3}\\), \\(-\\frac{1}{4}\\)
\n\"Samacheer
\n[Here an<\/sub> = 12, a0<\/sub> = 4; Let \\(\\frac{p}{q}\\) be the root of the equation (1)
\nThe factors of a0 <\/sub>: \u00b11, \u00b12, \u00b14 (P must divisible by 4)
\nThe factor of an<\/sub> : \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b112
\nq must divide as (12)
\nUsing these p and q we can form \\(\\pm 1, \\pm 2, \\pm \\frac{1}{2}, \\pm \\frac{2}{3}, \\pm \\frac{1}{6}, \\pm \\frac{1}{4}, \\pm \\frac{3}{4}, \\pm 3\\) are the possible roots of equation. (1)]<\/p>\n

Question 2.
\nExamine for the rational roots of:
\n(i) 2x3<\/sup> – x2<\/sup> – 1 = 0
\n(ii) x8<\/sup> – 3x + 1 = 0
\nSolution:
\n(i) 2x3<\/sup> – x2<\/sup> – 1 = 0
\nSum of the co-efficients = 0
\n\u2234 (x – 1) is a factor
\nThe other factor is 2x2<\/sup> + x + 1.
\nThe root is (2x2<\/sup> + x + 1) = 0
\nHere \u2206 = b2<\/sup> – 4ac = (1)2<\/sup> – 4(2) (1) = 1 – 8 = -7 < 0
\nThe remaining roots are imaginary.
\nThe only rational root is x = 1
\n\"Samacheer
\n(ii) x8<\/sup> – 3x + 1 = 0 …. (1)
\nHere an<\/sub> = 1, a0<\/sub> = 1
\nIf \\(\\frac{p}{q}\\) is a rational root of (1)
\nThen q is a factor an<\/sub>, p is a factor of a0<\/sub>
\nThe possible values of p and q are \u00b1 1.
\nAmong the possible values 1, -1, [(p, q) = 1]
\nNone of them satisfies the equation (1)
\nThe above equation has no rational roots.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nSolve: \\(8 x^{\\frac{3}{2 n}}-8 x^{\\frac{-3}{2 n}}=63\\)
\nSolution:
\n\"Samacheer
\nSquaring on both sides
\n\"Samacheer
\nonly possible solution is x = 4n<\/sup><\/p>\n

Question 4.
\nSolve: \\(2 \\sqrt{\\frac{x}{a}}+3 \\sqrt{\\frac{a}{x}}=\\frac{b}{a}+\\frac{6 a}{b}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 5.
\nSolve the equations:
\n(i) 6x4<\/sup> – 35x3<\/sup> + 62x2<\/sup> – 35x + 6 = 0
\n(ii) x4<\/sup> + 3x3<\/sup> – 3x – 1 = 0
\nSolution:
\n(i) 6x4<\/sup> – 35x3<\/sup> + 62x2<\/sup> – 35x + 6 = 0 ….. (1)
\nIt is a even degree reciprocal equation as p(x) = \\(x^{n} p\\left(\\frac{1}{x}\\right)\\)
\nDividing equation (1) by x2<\/sup>,
\n\"Samacheer
\n\"Samacheer
\n(ii) x4<\/sup> + 3x3<\/sup> – 3x – 1 = 0 …… (1)
\nIt is an even degree reciprocal function of type II.
\n1, -1 are the solution of equation (1)
\n(x – 1), (x + 1) are the factor of (1)
\n(x2<\/sup> – 1) is a factor of (1)
\nDividing (1) by (x2<\/sup> – 1)
\nwe get, x2<\/sup> + 3x + 1 = 0 is the other factor.
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 6.
\nFind all real numbers satisfying 4x<\/sup> – 3(2x+2<\/sup>) + 25<\/sup> = 0
\nSolution:
\n4x<\/sup> – 3 (2x+2<\/sup>) + 25<\/sup> = 0
\n\u21d2 (22<\/sup>)x<\/sup> – 3(2x<\/sup> . 22<\/sup>) + 25<\/sup> = 0
\n(22<\/sup>)x<\/sup> – 12 . 2x<\/sup>+ 32 = 0
\nLet y = 2x<\/sup>
\ny2<\/sup> – 12y + 32 = 0
\n\u21d2 (y – 4) (y – 8) = 0
\ny – 4 = 0 or y – 8 = 0
\nCase (i): 2x<\/sup> = 4
\n\u21d2 2x<\/sup> = (2)2<\/sup>
\n\u21d2 x = 2
\nCase (ii): 2x = 8
\n\u21d2 2x<\/sup> = (2)3<\/sup>
\n\u21d2 x = 3
\n\u2234 The roots are 2, 3<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nSolve the equation 6x4<\/sup> – 5x3<\/sup> – 38x2<\/sup> – 5x + 6 = 0 if it is known that \\(\\frac{1}{3}\\) is a solution.
\nSolution:
\n6x4<\/sup> – 5x3<\/sup> – 38x2<\/sup> – 5x + 6 = 0 …… (1)
\nx = \\(\\frac{1}{3}\\) is a Solution
\n\u2234 (3x – 1) is a factor of (1)
\n(1) is a Reciprocal equation even degree divide (1) by x2<\/sup>.
\n\"Samacheer
\n\"Samacheer<\/p>\n

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5 Additional Problems<\/h3>\n

Question 1.
\nSolve: 32x + 4<\/sup> + 1 = 2.3x + 2<\/sup>
\nSolution:
\n32x + 4<\/sup> = 3x + 2<\/sup> + 3x + 2<\/sup> – 1
\n32x + 4<\/sup> – 3x + 2<\/sup> = 3x + 2<\/sup> – 1 \u21d2 3x + 2<\/sup>[3x + 2<\/sup> – 1] = [3x + 2<\/sup> – 1]
\n3x + 2<\/sup>= 1 \u21d2 3x + 2<\/sup> = 30<\/sup>
\nx + 2 = 0 \u21d2 x = -2<\/p>\n

Question 2.
\nSolve: 2x<\/sup> – 2x + 3<\/sup> + 24<\/sup> = 0.
\nSolution:
\n22x<\/sup> – (2x<\/sup>.23<\/sup>) + 24<\/sup> = 0 since 23<\/sup> = 8 = 4 + 4 = 22<\/sup> + 22<\/sup>
\n22x<\/sup> – (4 + 4)2x<\/sup> + 24<\/sup> = 0 \u21d2 22x<\/sup> – (22<\/sup> + 22<\/sup>)2x<\/sup> + 24<\/sup> = 0
\n(2x<\/sup> – 22<\/sup>)(2x<\/sup> – 22<\/sup>) = 0 \u21d2 (2x<\/sup> – 22<\/sup>)2<\/sup> = 0
\n2x<\/sup> = 22<\/sup> \u21d2 x = 2<\/p>\n

Question 3.
\nSolve: (x – 4) (x + 2) (x + 3) (x – 3) + 8 = 0.
\nSolution:
\n(x – 4) (x + 3) (x + 2) (x – 3) + 8 = 0
\n(x2<\/sup> – x – 12) (x2<\/sup> – x – 6) + 8 = 0
\nLet y = x2<\/sup> – x
\n(y – 12) (y – 6) + 8 = 0 \u21d2 y2<\/sup> – 18y + 72 + 8 = 0
\ny2<\/sup> – 18y + 80 = 0 \u21d2 (y – 10)( y – 8) = 0
\nCase (i) y – 10 = 0
\nx2<\/sup> – x – 8 = 0
\n\"Samacheer<\/p>\n

Question 4.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 5.
\n5x – 1<\/sup> + 51 – x<\/sup> = 26
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nSolve: 12x4<\/sup> – 56x3<\/sup> + 89x2<\/sup> – 56x + 12 = 0
\nSolution:
\nSince the co-efficients of the equations are equal from both ends.
\nDivide the equation by x2<\/sup>
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 7.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 8.
\nSolve: x4<\/sup> + 4x3<\/sup> + 5x2<\/sup> + 4x + 1 = 0
\nSolution:
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5 Question 1. Solve the following equations: (i) sin2x – 5sinx + 4 = 0 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4322"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=4322"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4322\/revisions"}],"predecessor-version":[{"id":60171,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4322\/revisions\/60171"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=4322"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=4322"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=4322"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}