{"id":35461,"date":"2022-09-19T16:47:40","date_gmt":"2022-09-19T11:17:40","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35461"},"modified":"2022-09-19T16:47:40","modified_gmt":"2022-09-19T11:17:40","slug":"inter-2nd-year-maths-2a-theory-of-equations-solutions-ex-4c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-theory-of-equations-solutions-ex-4c\/","title":{"rendered":"Inter 2nd Year Maths 2A Theory of Equations Solutions Ex 4(c)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Inter 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(c) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(c)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nForm the polynomial equation whose roots are
\n(i) 2 + 3i, 2 – 3i, 1 + i, 1 – i
\nSolution:
\nThe required equation is [x – (2 + 3i)] [x – (2 – 3i)] [x – (1 + i)][x – (1 – i)] = 0
\n\u21d2 [(x – 2) – 3i)] [(x – 2) + 3i] [(x – 1) – i] [(x – 1) + i] = 0
\n\u21d2 [(x – 2)2<\/sup> – 9i2<\/sup>] [(x – 1)2<\/sup> – i2<\/sup>] = 0
\n\u21d2 (x2<\/sup> – 4x + 4 + 9) (x2<\/sup> – 2x + 1 + 1) = 0
\n\u21d2 (x2<\/sup> – 4x + 13) (x2<\/sup> – 2x + 2) = 0
\n\u21d2 x4<\/sup> – 4x3<\/sup> + 13x2<\/sup> – 2x3<\/sup> + 8x2<\/sup> – 26x + 2x2<\/sup> – 8x + 26 = 0
\n\u21d2 x4<\/sup> – 6x3<\/sup> + 23x2<\/sup> – 34x + 26 = 0<\/p>\n

(ii) 3, 2, 1 + i, 1 – i
\nSolution:
\nRequired equation is (x – 3) (x – 2) [x – (1 + i)] [ x – (1 – i)] = 0
\n\u21d2 (x2<\/sup> – 5x + 6) [(x – 1) – i] [(x – 1) + i] = 0
\n\u21d2 (x2<\/sup> – 5x + 6) [(x – 1)2<\/sup> – i2<\/sup>] = 0
\n\u21d2 (x2<\/sup> – 5x + 6) (x2<\/sup> – 2x + 1 + 1) = 0
\n\u21d2 (x2<\/sup> – 5x + 6) (x2<\/sup> – 2x + 2) = 0
\n\u21d2 x4<\/sup> – 5x3<\/sup> + 6x2<\/sup> – 2x3<\/sup> + 10x2<\/sup> – 12x + 2x2<\/sup> – 10x + 12 = 0
\n\u21d2 x4<\/sup> – 7x3<\/sup> + 18x2<\/sup> – 22x + 12 = 0<\/p>\n

(iii) 1 + i, 1 – i, -1 + i, -1 – i
\nSolution:
\nRequired equation is [x – (1 + i)] [x – (1 – i)] [x – (-1 + i)] [x – (-1 – i)] = 0
\n\u21d2 [(x – 1) – i][(x – i) + i] [(x + 1) – i] [(x + 1) + i] = 0
\n\u21d2 [(x – 1)2<\/sup> – i2<\/sup>] [(x + 1)2<\/sup> – i2<\/sup>] = 0
\n\u21d2 (x2<\/sup> – 2x + 1 + 1) (x2<\/sup> + 2x + 1 + 1) = 0
\n\u21d2 (x2<\/sup> – 2x + 2) (x2<\/sup> + 2x + 2) = 0
\n\u21d2 x4<\/sup> – 2x3<\/sup> + 2x2<\/sup> + 2x3<\/sup> – 4x2<\/sup> + 4x + 2x2<\/sup> – 4x + 4 = 0
\n\u21d2 x4<\/sup> + 4 = 0<\/p>\n

(iv) 1 + i, 1 – i, 1 + i, 1 – i
\nSolution:
\nRequired equation is [x – (1 + i)] [x – (1 – i)] [x – (1 + i)] [x – (1 – i)] = 0
\n\u21d2 [(x – 1) – i]2<\/sup> [(x – 1) + i]2<\/sup> = 0
\n\u21d2 [(x – 1)2<\/sup> – i2<\/sup>] = 0
\n\u21d2 (x2<\/sup> – 2x + 1 + 1)2<\/sup> = 0
\n\u21d2 x4<\/sup> + 4x2<\/sup> + 4 – 4x3<\/sup> + 4x2<\/sup> – 8x = 0
\n\u21d2 x4<\/sup> – 4x3<\/sup> + 8x2<\/sup> – 8x + 4 = 0<\/p>\n

\"Inter<\/p>\n

Question 2.
\nForm the polynomial equation with rational coefficients whose roots are
\n(i) 4\u221a3, 5 + 2i
\nSolution:
\nFor the polynomial equation with rational coeffs. the roots are conjugate surds and conjugate complex numbers.
\n4\u221a3, 5 + 2i
\nLet \u03b1 = 4\u221a3 then \u03b2 = -4\u221a3, and \u03b3 = 5 + 2i then \u03b4 = 5 – 2i
\n\u03b1, \u03b2, \u03b3, \u03b4 are the roots
\n\u03b1 + \u03b2 = 0, \u03b1\u03b2 = -48
\n\u03b3 + \u03b4 = 10, \u03b3\u03b4 = 25 + 4 = 29
\nThe required equation is [x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2] [x2<\/sup> – (\u03b3 + \u03b4)x + \u03b3\u03b4] = 0
\n\u21d2 (x2<\/sup> – 48) (x2<\/sup> – 10x + 29) = 0
\n\u21d2 x4<\/sup> – 10x3<\/sup> + 29x2<\/sup> – 48x2<\/sup> + 480x – 1932 = 0
\n\u21d2 x4<\/sup> – 10x3<\/sup> – 19x2<\/sup> + 480x – 1932 = 0<\/p>\n

(ii) 1 + 5i, 5 – i
\nSolution:
\nFor the polynomial equation with rational coeffs. the roots are conjugate surds and conjugate complex numbers.
\nLet \u03b1 = 1 + 5i then \u03b2 = 1 – 5i,
\nand \u03b3 = 5 + i then \u03b4 = 5 – i
\n\u03b1 + \u03b2 = 2, \u03b1\u03b2 = 26
\n\u03b3 + \u03b4 = 10, \u03b3\u03b4 = 26
\nThe required equation is [x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2] [x2<\/sup> – (\u03b3 + \u03b4)x + \u03b3\u03b4] = 0
\n\u21d2 (x2<\/sup> – 2x + 26) (x2<\/sup> – 10x + 26) = 0
\n\u21d2 x4<\/sup> – 12x3<\/sup> + 72x2<\/sup> – 312x + 676 = 0<\/p>\n

(iii) i – \u221a5
\nSolution:
\nFor the polynomial equation with rational coeffs. the roots are conjugate surds and conjugate complex numbers.
\nLet \u03b1 = i – \u221a5, \u03b2 = i + \u221a5, \u03b3 = -i – \u221a5, \u03b4 = -i + \u221a5 are the roots
\n\u03b1 + \u03b2 = 2i, \u03b1\u03b2 = -6
\n\u03b3 + \u03b4 = -2i, \u03b3\u03b4 = -6
\nThe required equation is [x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2][x2<\/sup> – (\u03b3 + \u03b4)x + \u03b3\u03b4] = 0
\n\u21d2 (x2<\/sup> – 2ix – 6) (x2<\/sup> + 2ix – 6) = 0
\n\u21d2 [(x2<\/sup> – 6) – 2ix] [(x2<\/sup> – 6) + 2ix] = 0
\n\u21d2 (x2<\/sup> – 6)2<\/sup> + 4x2<\/sup> = 0
\n\u21d2 x4<\/sup> + 36 – 12x2<\/sup> + 4x2<\/sup> = 0
\n\u21d2 x4<\/sup> – 8x2<\/sup> + 36 = 0<\/p>\n

(iv) -\u221a3 + i\u221a2
\nSolution:
\nLet \u03b1 = -\u221a3 + i\u221a2, \u03b2 = -\u221a3 – i\u221a2, \u03b3 = \u221a3 – i\u221a2, \u03b3 = \u221a3 + i\u221a2
\n\u03b1 + \u03b2 = -2\u221a3
\n\u03b1\u03b2 = (-\u221a3)2<\/sup> – (i\u221a2)2<\/sup>
\n= 3 – i2<\/sup> (2)
\n= 5
\n\u03b3 + \u03b4 = 2\u221a3
\n\u03b3\u03b4 = 5
\nThe required equation is [(x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2)] [(x2<\/sup> – (\u03b3 + \u03b4)x + \u03b3\u03b4)] = 0
\n\u21d2 (x2<\/sup> + 2\u221a3x + 5) (x2<\/sup> – 2\u221a3x + 5) = 0
\n\u21d2 (x2<\/sup> + 5)2<\/sup> – (2\u221a3x)2<\/sup> = 0
\n\u21d2 x4<\/sup> + 25 + 10x2<\/sup> – 12x2<\/sup> = 0
\n\u21d2 x4<\/sup> – 2x2<\/sup> + 25 = 0<\/p>\n

\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nSolve the equation x4<\/sup> + 2x3<\/sup> – 5x2<\/sup> + 6x + 2 = 0 given that 1 + i is one of its roots.
\nSolution:
\nLet 1 + i be one root \u21d2 1 – i be another root
\nThe equation having roots 1 \u00b1 i is x2<\/sup> – 2x + 2 = 0
\n\u2234 x2<\/sup> – 2x + 2 is a factor of x4<\/sup> + 2x3<\/sup> – 5x2<\/sup> + 6x + 2 = 0
\n\"Inter
\n\u2234 The roots of the given equation are 1 \u00b1 i, -2 + \u221a3<\/p>\n

Question 2.
\nSolve the equation 3x3<\/sup> – 4x2<\/sup> + x + 88 = 0 which has 2 – \u221a-7 as a root.
\nSolution:
\nLet 2 – \u221a-7 (i.e) 2 – \u221a7i is one root
\n\u21d2 2 + \u221a7i is another root.
\nThe equation having roots 2 \u00b1 \u221a7i is x2<\/sup> – 4x + 11 = 0
\n\u2234 x2<\/sup> – 4x + 11 is a factor of the given equation.
\n\"Inter
\n3x + 8 = 0
\n\u21d2 x = \\(-\\frac{8}{3}\\)
\n\u2234 The roots of the given equation are 2 \u00b1 \u221a7i, \\(-\\frac{8}{3}\\)<\/p>\n

\"Inter<\/p>\n

Question 3.
\nSolve x4<\/sup> – 4x2<\/sup> + 8x + 35 = 0, given that 2 + i\u221a3 is a root.
\nSolution:
\nLet 2 + i\u221a3 be one root
\n\u21d2 2 – i\u221a3 is another root.
\nThe equation having roots 2 \u00b1 i\u221a3 is x2<\/sup> – 4x + 7 = 0
\n\u2234 x2<\/sup> – 4x + 7 is a factor of x4<\/sup> – 4x2<\/sup> + 8x + 35
\n\"Inter
\n\u2234 The roots of the given equation are 2 \u00b1 i\u221a3, -2 \u00b1 i<\/p>\n

Question 4.
\nSolve the equation x4<\/sup> – 6x3<\/sup> + 11x2<\/sup> – 10x + 2 = 0, given that 2 + \u221a3 is a root of the equation.
\nSolution:
\n2 + \u221a3 is one root
\n\u21d2 2 – \u221a3 is another root.
\nThe equation having the roots of 2 \u00b1 \u221a3 is x2<\/sup> – 4x + 1 = 0
\n\u2234 x2<\/sup> – 4x + 1 is a factor of x4<\/sup> – 6x3<\/sup> + 11x2<\/sup> – 10x + 2 = 0
\n\"Inter
\n\u2234 The roots of the required equation are 2 \u00b1 \u221a3, 1 \u00b1 i<\/p>\n

Question 5.
\nGiven that -2 + \u221a-7 is a root of the equation x4<\/sup> + 2x2<\/sup> – 16x + 77 = 0, solve it completely.
\nSolution:
\n-2 – \u221a-7 (i.e) -2 + i\u221a7 is one root
\n\u21d2 -2 – i\u221a7 is another root.
\nThe equation having the roots of -2 \u00b1 i\u221a7 is x2<\/sup> + 4x + 11 = 0
\n\u2234 x2<\/sup> + 4x + 11 is a factor of x4<\/sup> + 2x2<\/sup> – 16x + 77
\n\"Inter
\n\u2234 The roots of the required equation are -2 \u00b1 i\u221a7, 2 \u00b1 \u221a3i<\/p>\n

\"Inter<\/p>\n

Question 6.
\nSolve the equations x4<\/sup> + 2x3<\/sup> – 16x2<\/sup> – 22x + 7 = 0, given that 2 – \u221a3 is a root of it.
\nSolution:
\n2 + \u221a3 is a root \u21d2 2 – \u221a3 is also a root.
\nThe equation having roots 2 \u00b1 \u221a3 is
\nx2<\/sup> – (2 + \u221a3 + 2 – \u221a3)x + (2 + \u221a3) (2 – \u221a3) = 0
\n\u21d2 x2<\/sup> – 4x + 1 = 0
\n\"Inter
\nx2<\/sup> + 6x + 7 = 0
\n\u21d2 x = \\(\\frac{-6 \\pm \\sqrt{36-28}}{2}\\)
\n\u21d2 x = -3 \u00b1 \u221a2
\n\u2234 Roots are 2 \u00b1 \u221a3, -3 \u00b1 \u221a2<\/p>\n

Question 7.
\nSolve the equation 3x5<\/sup> – 4x4<\/sup> – 42x3<\/sup> + 56x2<\/sup> + 27x – 36 = 0 given that \u221a2 + \u221a5 is one of its roots.
\nSolution:
\n\u221a2 + \u221a5 is a root
\n\u21d2 -\u221a2 – \u221a5, -\u221a2 + \u221a5, -\u221a2 – \u221a5 are also roots.
\nThe equation haying roots \u221a2 + \u221a5 is
\nx2<\/sup> – (\u221a2 + \u221a5 + \u221a2 – \u221a5) + (\u221a2 + \u221a5) (\u221a2 – \u221a5) = 0
\n\u21d2 x2<\/sup> – 2\u221a2x – 3 = 0
\nThe equation having roots -\u221a2 \u00b1 \u221a5 is
\nx2<\/sup> – (-\u221a2 + \u221a5 – \u221a2 – \u221a5) + (\u221a2 + \u221a5)(-\u221a2 – \u221a5) = 0
\n\u21d2 x2<\/sup> + 2\u221ax – 3 = 0
\nThe equation having roots \u00b1\u221a2 \u00b1 \u221a5 is (x2<\/sup> + 2\u221a2x – 3) (x2<\/sup> – 2\u221a2x – 3) = 0
\n\u21d2 (x2<\/sup> – 3)2<\/sup> – (2\u221a2x)2<\/sup> = 0
\n\u21d2 x4<\/sup> – 6x2<\/sup> + 9 – 8x2<\/sup> = 0
\n\u21d2 x4<\/sup> – 14x2<\/sup> + 9 = 0
\n\u2234 3x5<\/sup> – 4x4<\/sup> – 42x3<\/sup> + 56x2<\/sup> + 27x – 36 = 0
\n\u21d2 3x(x4<\/sup> – 14x2<\/sup> + 9) – 4(x4<\/sup> – 14x2<\/sup> + 9) = 0
\n\u21d2 (x4<\/sup> – 14x2<\/sup> + 9) (3x – 4) = 0
\n\u21d2 x = \u00b1\u221a2 \u00b1 \u221a5 or 4\/3
\n\u2234 The roots are \u00b1\u221a2 \u00b1 \u221a5, 4\/3<\/p>\n

\"Inter<\/p>\n

Question 8.
\nSolve the equation x4<\/sup> – 9x3<\/sup> + 27x2<\/sup> – 29x + 6 = 0, given that one root of it is 2 – \u221a3.
\nSolution:
\n2 – \u221a3 is one root \u21d2 2 + \u221a3 is another root.
\nThe equation having the roots of 2 \u00b1 \u221a3 is x2<\/sup> – 4x + 1 = 0
\n\u2234 x2<\/sup> – 4x + 1 is a factor of the given equation.
\n\"Inter
\nx2<\/sup> – 5x + 6 = 0
\n\u21d2 (x – 2) (x – 3) = 0
\n\u21d2 x = 2, 3
\n\u2234 The roots of the required equations are 2 \u00b1 \u221a3, 2, 3<\/p>\n

Question 9.
\nShow that the equation \\(\\frac{a^{2}}{x-a^{\\prime}}+\\frac{b^{2}}{x-b^{\\prime}}+\\frac{c^{2}}{x-c^{\\prime}}+\\ldots+\\frac{k^{2}}{x-k^{\\prime}}\\)= x – m Where a, b, c ….k, m, a’, b’, c’…. k’ are all real numbers, cannot have a non-real root.
\nSolution:
\nLet \u03b1 + i\u03b2 be the root of the given equation.
\nSuppose if \u03b2 \u2260 0, then \u03b1 – i\u03b2 is also a root of the given equation.
\nSubstitute \u03b1 + i\u03b2 in the given equation, and we get
\n\"Inter
\n\\(\\left[\\frac{a^{2}}{\\left(\\alpha-a^{\\prime}\\right)^{2}+\\beta^{2}}+\\frac{b^{2}}{\\left(\\alpha-b^{\\prime}\\right)^{2}+\\beta^{2}}+\\ldots \\frac{k^{2}}{\\left(\\alpha-k^{\\prime}\\right)^{2}+\\beta^{2}}+1\\right]\\)
\n= 0
\n\u21d2 \u03b2 = 0
\nThis is a contradiction.
\n\u2234 The given equation cannot have non-real roots.<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(c) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(c) I. Question 1. Form the polynomial equation whose roots are (i) 2 + 3i, 2 – …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35461"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35461"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35461\/revisions"}],"predecessor-version":[{"id":35462,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35461\/revisions\/35462"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35461"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35461"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}