{"id":35463,"date":"2022-09-19T16:48:24","date_gmt":"2022-09-19T11:18:24","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35463"},"modified":"2022-09-19T16:48:24","modified_gmt":"2022-09-19T11:18:24","slug":"inter-2nd-year-maths-2a-theory-of-equations-solutions-ex-4d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-theory-of-equations-solutions-ex-4d\/","title":{"rendered":"Inter 2nd Year Maths 2A Theory of Equations Solutions Ex 4(d)"},"content":{"rendered":"

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

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

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

Question 1.
\nFind the algebraic equation whose roots are 3 times the roots of x3<\/sup> + 2x2<\/sup> – 4x + 1 = 0
\nSolution:
\nGiven equation is f(x) = x3<\/sup> + 2x2<\/sup> – 4x + 1 = 0
\nWe require an equation whose roots are 3 times the roots of f(x) = 0
\ni,e., Required equation is f(\\(\\frac{x}{3}\\)) = 0
\n\u21d2 \\(\\left(\\frac{x}{3}\\right)^{3}+2\\left(\\frac{x}{3}\\right)^{2}-\\frac{4 x}{3}+1=0\\)
\n\u21d2 \\(\\frac{x^{3}}{27}+\\frac{2}{9} x^{2}-\\frac{4}{3} x+1=0\\)
\nMultiplying with 27, required equation is x3<\/sup> + 6x2<\/sup> – 36x + 27 = 0<\/p>\n

Question 2.
\nFind the algebraic equation whose roots are 2 times the roots of x5<\/sup> – 2x4<\/sup> + 3x3<\/sup> – 2x2<\/sup> + 4x + 3 = 0
\nSolution:
\nGiven equation is f(x) = x5<\/sup> – 2x4<\/sup> + 3x3<\/sup> – 2x2<\/sup> + 4x + 3 = 0
\nWe require an equation whose roots are 2 times the roots of f(x) = 0
\nRequired equation is f(\\(\\frac{x}{2}\\)) = 0
\n\u21d2 \\(\\left(\\frac{x}{2}\\right)^{5}-2\\left(\\frac{x}{2}\\right)^{4}+3\\left(\\frac{x}{2}\\right)^{3}-2\\left(\\frac{x}{2}\\right)^{2}+4\\left(\\frac{x}{2}\\right)\\) + 3 = 0
\n\u21d2 \\(\\frac{x^{5}}{32}-2 \\cdot \\frac{x^{4}}{16}+3 \\cdot \\frac{x^{3}}{8}-2 \\cdot \\frac{x^{2}}{4}+4 \\cdot \\frac{x}{2}+3=0\\)
\nMultiplying with 32, the required equation is
\n\u21d2 x5<\/sup> – 4x4<\/sup> + 12x3<\/sup> – 16x2<\/sup> + 64x + 96 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the transformed equation whose roots are the negative of the roots of x4<\/sup> + 5x3<\/sup> + 11x + 3 = 0
\nSolution:
\nGiven f(x) = x4<\/sup> + 5x3<\/sup> + 11x + 3 = 0
\nWe want an equation whose roots are -\u03b11<\/sub>, -\u03b12<\/sub>, -\u03b13<\/sub>, -\u03b14<\/sub>
\nRequired equation f(-x) = 0
\n\u21d2 (-x)4<\/sup> + 5(-x)3<\/sup> + 11(-x) + 3 = 0
\n\u21d2 x4<\/sup> – 5x3<\/sup> – 11x + 3 = 0<\/p>\n

Question 4.
\nFind the transformed equation whose roots are the negatives of the roots of x7<\/sup> + 3x5<\/sup> + x3<\/sup> – x2<\/sup> + 7x + 2 = 0
\nSolution:
\nGiven f(x) = x7<\/sup> + 3x5<\/sup> + x3<\/sup> – x2<\/sup> + 7x + 2 = 0
\nWe want an equation whose roots are -\u03b11<\/sub>, -\u03b12<\/sub>, ………., -\u03b1n<\/sub>
\nRequired equation is f(-x) = 0
\n\u21d2 (-x)7<\/sup> + 3(-x)5<\/sup> + (-x)3<\/sup> – (-x)2<\/sup> + 7(-x) + 2 = 0
\n\u21d2 -x7<\/sup> – 3x5<\/sup> – x3<\/sup> – x2<\/sup> – 7x + 2 = 0
\n\u21d2 x7<\/sup> + 3x5<\/sup> + x3<\/sup> + x2<\/sup> + 7x – 2 = 0<\/p>\n

Question 5.
\nFind the polynomial equation whose roots are the reciprocals of the roots of x4<\/sup> – 3x3<\/sup> + 7x2<\/sup> + 5x – 2 = 0
\nSolution:
\nGiven equation is f(x) = x4<\/sup> – 3x3<\/sup> + 7x2<\/sup> + 5x – 2 = 0
\nRequired equation is f(\\(\\frac{1}{x}\\)) = 0
\ni.e., \\(\\frac{1}{x^{4}}-\\frac{3}{x^{3}}+\\frac{7}{x^{2}}+\\frac{5}{x}-2=0\\)
\nMultiplying with x4<\/sup>
\n\u21d2 1 – 3x + 7x2<\/sup> + 5x3<\/sup> – 2x4<\/sup> = 0
\n\u21d2 2x4<\/sup> – 5x3<\/sup> – 7x2<\/sup> + 3x – 1 = 0<\/p>\n

Question 6.
\nFind the polynomial equation whose roots are the reciprocals of the roots of x5<\/sup> + 11x4<\/sup> + x3<\/sup> + 4x2<\/sup> – 13x + 6 = 0
\nSolution:
\nGiven equation is f(x) = x5<\/sup> + 11x4<\/sup> + x3<\/sup> + 4x2<\/sup> – 13x + 6 = 0
\nRequired equation is f(\\(\\frac{1}{x}\\)) = 0
\n\\(\\frac{1}{x^5}+\\frac{11}{x^4}+\\frac{1}{x^3}+\\frac{4}{x^2}-\\frac{13}{x}+6=0\\)
\nMultiplying by x5<\/sup>
\n\u21d2 1 + 11x + x2<\/sup> + 4x3<\/sup> – 13x4<\/sup> + 6x5<\/sup> = 0
\n\u21d2 6x5<\/sup> – 13x4<\/sup> + 4x3<\/sup> + x2<\/sup> + 11x + 1 = 0<\/p>\n

\"Inter<\/p>\n

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

Question 1.
\nFind the polynomial equation whose roots are the squares of the roots of x4<\/sup> + x3<\/sup> + 2x2<\/sup> + x + 1 = 0
\nSolution:
\nGiven equation is f(x) = x4<\/sup> + x3<\/sup> + 2x2<\/sup> + x + 1 = 0
\nRequired equation f(\u221ax) = 0
\n\u21d2 x2<\/sup> + x\u221ax + 2x + \u221ax + 1 = 0
\n\u21d2 \u221ax(x + 1) = -(x2<\/sup> + 2x + 1)
\nSquaring both sides,
\n\u21d2 x(x + 1)2<\/sup> = (x2<\/sup> + 2x + 1)2<\/sup>
\n\u21d2 x(x2<\/sup> + 2x + 1) = x4<\/sup> + 4x2<\/sup> + 1 + 4x3<\/sup> + 4x + 2x2<\/sup>
\n\u21d2 x3<\/sup> + 2x2<\/sup> + x = x4<\/sup> + 4x3<\/sup> + 6x2<\/sup> + 4x + 1
\n\u21d2 x4<\/sup> + 3x3<\/sup> + 4x2<\/sup> + 3x + 1 = 0<\/p>\n

Question 2.
\nForm the polynomial equation whose roots are the squares of the roots of x3<\/sup> + 3x2<\/sup> – 7x + 6 = 0
\nSolution:
\nGiven equation is f(x) = x3<\/sup> + 3x2<\/sup> – 7x + 6 = 0
\nRequired equation is f(\u221ax) = 0
\n\u21d2 x\u221ax + 3x – 7\u221ax + 6 = 0
\n\u21d2 \u221ax(x – 7) = -(3x + 6)
\nSquaring on both sides,
\n\u21d2 x(x – 7)2<\/sup> = (3x + 6)2<\/sup>
\n\u21d2 x(x2<\/sup> – 14x + 49) = 9x2<\/sup> + 36 + 36x
\n\u21d2 x3<\/sup> – 14x2<\/sup> + 49x – 9x2<\/sup> – 36x – 36 = 0
\n\u21d2 x3<\/sup> – 23x2<\/sup> + 13x – 36 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nForm the polynomial equation whose roots are cubes of the roots of x3<\/sup> + 3x2<\/sup> + 2 = 0
\nSolution:
\nGiven equation is x3<\/sup> + 3x2<\/sup> + 2 = 0
\nPut y = x3<\/sup> so that x = y1\/3<\/sup>
\n\u2234 y + 3y2\/3<\/sup> + 2 = 0
\n\u2234 3y2\/3<\/sup> = -(y + 2)
\nCubing on both sides,
\n27y2<\/sup> = -(y + 2)3<\/sup> = -(y3<\/sup> + 6y2<\/sup> + 12y + 8)
\n\u2234 y3<\/sup> + 6y2<\/sup> + 12y + 8 = 0
\n\u21d2 y3<\/sup> + 33y2<\/sup> + 12y + 8 = 0
\nRequired equation is x3<\/sup> + 33x2<\/sup> + 12x + 8 = 0<\/p>\n

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

Question 1.
\nFind the polynomial equation whose roots are the translates of those of the equation x4<\/sup> – 5x3<\/sup> + 7x2<\/sup> – 17x + 11 = 0 by -2.
\nSolution:
\nGiven equation is f(x) = x4<\/sup> – 5x3<\/sup> + 7x2<\/sup> – 17x + 11 = 0
\nThe required equation is f(x + 2) = 0
\n(x + 2)4<\/sup> – 5(x + 2)3<\/sup> + 7(x + 2)2<\/sup> – 17(x + 2) + 11 = 0
\n\"Inter
\nRequired equation is x4<\/sup> + 3x3<\/sup> + x2<\/sup> – 17x – 19 = 0<\/p>\n

Question 2.
\nFind the polynomial equation whose roots are the translates of those of x5<\/sup> – 4x4<\/sup> + 3x2<\/sup> – 4x + 6 = 0 by -3.
\nSolution:
\nGiven equation is f(x) = x5<\/sup> – 4x4<\/sup> + 3x2<\/sup> – 4x + 6 = 0
\nRequired equation is f(x + 3) = 0
\n(x + 3)5<\/sup> – 4(x + 3)4<\/sup> + 3(x + 3)2<\/sup>
\n\"Inter
\nRequired equation is x5<\/sup> + 11x4<\/sup> + 42x3<\/sup> + 57x2<\/sup> – 13x – 60 = 0<\/p>\n

Question 3.
\nFind the polynomial equation whose roots are the translates of the roots of the equation x4<\/sup> – x3<\/sup> – 10x2<\/sup> + 4x + 24 = 0 by 2.
\nSolution:
\nGiven f(x) = x4<\/sup> – x3<\/sup> – 10x2<\/sup> + 4x + 24 = 0
\nRequired equation is f(x – 2) = 0
\n(x – 2)4<\/sup> – (x – 2)3<\/sup> – 10(x – 2)2<\/sup> + 4(x – 2) + 24 = 0
\n\"Inter
\nRequired equation is x4<\/sup> – 9x3<\/sup> + 20x2<\/sup> = 0<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the polynomial equation whose roots are the translates of the equation 3x5<\/sup> – 5x3<\/sup> + 7 = 0 by 4.
\nSolution:
\nGiven f(x) = 3x5<\/sup> – 5x3<\/sup> + 7 = 0
\nRequired equation is f(x – 4) = 0
\n3(x – 4)5<\/sup> – 5(x – 4)3<\/sup> + 7 = 0
\n\"Inter
\nRequired equation is x5<\/sup> – 60x4<\/sup> + 475x3<\/sup> – 1860x2<\/sup> + 3600x – 2745 = 0<\/p>\n

Question 5.
\nTransform each of the following equations into ones in which of the coefficients of the second highest power of x is zero and also find their transformed equations.
\n(i) x3<\/sup> – 6x2<\/sup> + 10x – 3 = 0
\nSolution:
\nGiven equation is x3<\/sup> – 6x2<\/sup> + 10x – 3 = 0
\nTo remove the second term diminish the roots by \\(-\\frac{a_1}{n a_0}=\\frac{6}{3}=2\\)
\n\"Inter
\nRequired equation is x3<\/sup> – 2x + 1 = 0<\/p>\n

(ii) x4<\/sup> + 4x3<\/sup> + 2x2<\/sup> – 4x – 2 = 0
\nSolution:
\nGiven equation is x4<\/sup> + 4x3<\/sup> + 2x2<\/sup> – 4x – 2 = 0
\nDiminishing the roots by \\(-\\frac{a_1}{n a_0}=\\frac{-4}{4}=-1\\)
\n\"Inter
\nRequired equation is x4<\/sup> – 4x2<\/sup> + 1 = 0<\/p>\n

(iii) x3<\/sup> – 6x2<\/sup> + 4x – 7 = 0
\nSolution:
\nGiven equation is x3<\/sup> – 6x2<\/sup> + 4x – 7 = 0
\nDiminishing the roots by \\(-\\frac{a_1}{n a_0}=\\frac{6}{3}=2\\)
\n\"Inter
\nRequired equation is x3<\/sup> – 8x – 15 = 0<\/p>\n

(iv) x3<\/sup> + 6x2<\/sup> + 4x + 4 = 0
\nSolution:
\nGiven equation is x3<\/sup> + 6x2<\/sup> + 4x + 4 = 0
\nTo remove the second term diminish the roots by \\(\\frac{-a_1}{n a_0}=-\\frac{6}{3}=-2\\)
\n\"Inter
\nRequired equation is x3<\/sup> – 8x + 12 = 0<\/p>\n

\"Inter<\/p>\n

Question 6.
\nTransform each of the following equations into ones in which the coefficients of the third highest power of x are zero.
\nHint: To remove the rth<\/sup> term in an equation f(x) = 0 of degree n diminish the roots by ‘h’ such that \\(f^{(n-r+1)}(h)=0\\)
\n(i) x4<\/sup> + 2x3<\/sup> – 12x2<\/sup> + 2x – 1 = 0
\nSolution:
\nLet f(x) = x4<\/sup> + 2x3<\/sup> – 12x2<\/sup> + 2x – 1
\nTo remove the 3rd<\/sup> term, diminish the roots by h such that f”(h) = 0
\nf'(x) = 4x3<\/sup> + 6x2<\/sup> – 24x + 2
\nf”(x) = 12x2<\/sup> + 12x – 24
\nf”(h) = 0
\n\u21d2 12h2<\/sup> + 12h – 24 = 0
\n\u21d2 h2<\/sup> + h – 2 = 0
\n\u21d2 (h + 2) (h – 1) = 0
\n\u21d2 h = -2 or 1
\nCase (i):
\n\"Inter
\nTransformed equation is x4<\/sup> – 6x3<\/sup> + 42x – 53 = 0
\nCase (ii):
\n\"Inter
\nTransformed equation is x4<\/sup> – 6x3<\/sup> – 12x – 8 = 0
\n\u2234 The required equation is x4<\/sup> – 6x3<\/sup> + 42x – 53 = 0
\nor x4<\/sup> + 6x3<\/sup> – 12x – 8 = 0<\/p>\n

(ii) x3<\/sup> + 2x2<\/sup> + x + 1 = 0
\nSolution:
\nLet f(x) = x3<\/sup> + 2x2<\/sup> + x + 1
\nTo remove the 3rd<\/sup> term, diminish the roots by h such that f'(h) = 0, f'(x) = 3x2<\/sup> + 4x + 1
\nf'(h) = 0
\n\u21d2 3h2<\/sup> + 4h + 1 = 0
\n\u21d2 (3h + 1) (h + 1)
\n\u21d2 h = -1, \\(-\\frac{1}{3}\\)
\nCase (i):
\n\"Inter
\nTransformed equation is x3<\/sup> – x2<\/sup> + 1 = 0
\nCase (ii):
\n\"Inter
\nTransformed equation is x3<\/sup> + x2<\/sup> + \\(\\frac{23}{27}\\) = 0
\n\u21d2 27x3<\/sup> + 27x2<\/sup> + 23 = 0
\n\u2234 The required equation is x3<\/sup> – x2<\/sup> + 1 = 0 or 27x3<\/sup> + 27x2<\/sup> + 23 = 0<\/p>\n

\"Inter<\/p>\n

Question 7.
\nSolve the following equations.
\n(i) x4<\/sup> – 10x3<\/sup> + 26x2<\/sup> – 10x + 1 = 0
\nSolution:
\nThis is a standard reciprocal equation.
\nDividing with x2<\/sup>
\n\\(x^2-10 x+26-\\frac{10}{x}+\\frac{1}{x^2}=0\\)
\n\\(\\left(x^2+\\frac{1}{x^2}\\right)-10\\left(x+\\frac{1}{x}\\right)+26=0\\) …….(1)
\nput a = x + \\(\\frac{1}{x}\\)
\n\\(x^2+\\frac{1}{x^2}=\\left(x+\\frac{1}{x}\\right)^2-2\\) = a2<\/sup> – 2
\nSubstituting in (1)
\na2<\/sup> – 2 – 10a + 26 = 0
\n\u21d2 a2<\/sup> – 10a + 24 = 0
\n\u21d2 (a – 4)(a – 6) = 0
\n\u21d2 a = 4 or 6
\nCase (i): a = 4
\nx + \\(\\frac{1}{x}\\) = 4
\n\u21d2 x2<\/sup> + 1 = 4x
\n\u21d2 x2<\/sup> – 4x + 1 = 0
\n\u21d2 x = \\(\\frac{4 \\pm \\sqrt{16-4}}{2}=\\frac{4 \\pm 2 \\sqrt{3}}{2}\\)
\n\u21d2 x = 2 \u00b1 \u221a3
\nCase (ii): a = 6
\nx + \\(\\frac{1}{x}\\) = 6
\n\u21d2 x2<\/sup> + 1 = 6x
\n\u21d2 x2<\/sup> – 6x + 1 = 0
\n\u21d2 x = \\(\\frac{6 \\pm \\sqrt{36-4}}{2}\\)
\n\u21d2 x = \\(\\frac{2(3 \\pm 2 \\sqrt{2)}}{2}\\)
\n\u21d2 x = 3 \u00b1 2\u221a2
\n\u2234 The roots are 3 \u00b1 2\u221a2, 2 \u00b1 \u221a3<\/p>\n

(ii) 2x5<\/sup> + x4<\/sup> – 12x3<\/sup> – 12x2<\/sup> + x + 2 = 0
\nSolution:
\nGiven f(x) = 2x5<\/sup> + x4<\/sup> – 12x3<\/sup> – 12x2<\/sup> + x + 2 = 0
\nThis is an odd-degree reciprocal equation of the first type.
\n\u2234 -1 is a root.
\nDividing f(x) with x + 1
\n\"Inter
\nDividing f(x) by (x + 1), we get
\n2x4<\/sup> – x3<\/sup> – 11x2<\/sup> – x + 2 = 0
\nDividing by x2<\/sup>
\n\\(2 x^2-x-11-\\frac{1}{x}+\\frac{2}{x^2}=0\\)
\n\\(2\\left(x^2+\\frac{1}{x^2}\\right)-\\left(x+\\frac{1}{x}\\right)-11=0\\) ……..(1)
\nPut a = x + \\(\\frac{1}{x}\\) so that
\n\\(x^2+\\frac{1}{x^2}=a^2-2\\)
\nSubstituting in (1), the required equation is
\n\u21d2 2(a2<\/sup> – 2) – a – 11 = 0
\n\u21d2 2a2<\/sup> – 4 – a – 11 = 0
\n\u21d2 2a2<\/sup> – a – 15 = 0
\n\u21d2 (a – 3) (2a + 5) = 0
\n\u21d2 a = 3 or \\(-\\frac{5}{2}\\)
\nCase (i): a = 3
\nx + \\(\\frac{1}{x}\\) = 3
\n\u21d2 x2<\/sup> + 1 = 3x
\n\u21d2 x2<\/sup> – 3x + 1 = 0
\n\u21d2 x = \\(\\frac{3 \\pm \\sqrt{9-4}}{2}=\\frac{3 \\pm \\sqrt{5}}{2}\\)
\nCase (ii): a = \\(-\\frac{5}{2}\\)
\n\u21d2 \\(x+\\frac{1}{x}=-\\frac{5}{2}\\)
\n\u21d2 \\(\\frac{x^2+1}{x}=-\\frac{5}{2}\\)
\n\u21d2 2x2<\/sup> + 2 = -5x
\n\u21d2 2x2<\/sup> + 5x + 2 = 0
\n\u21d2 (2x + 1) (x + 2) = 0
\n\u21d2 x = \\(-\\frac{1}{2}\\), -2
\n\u2234 The roots are -1, \\(-\\frac{1}{2}\\), -2, \\(\\frac{3 \\pm \\sqrt{5}}{2}\\)<\/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(d) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(d) I. Question 1. Find the algebraic equation whose roots are 3 times the roots of x3 …<\/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\/35463"}],"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=35463"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35463\/revisions"}],"predecessor-version":[{"id":35464,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35463\/revisions\/35464"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35463"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35463"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35463"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}