{"id":22347,"date":"2024-01-16T14:59:46","date_gmt":"2024-01-16T09:29:46","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22347"},"modified":"2024-01-17T16:12:49","modified_gmt":"2024-01-17T10:42:49","slug":"inter-2nd-year-maths-2a-theory-of-equations-solutions-ex-4b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2a-theory-of-equations-solutions-ex-4b\/","title":{"rendered":"Inter 2nd Year Maths 2A Theory of Equations Solutions Ex 4(b)"},"content":{"rendered":"

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

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

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

Question 1.
\nSolve x3<\/sup> – 3x2<\/sup> – 16x + 48 = 0, given that the sum of two roots is zero.
\nSolution:
\nLet \u03b1, \u03b2, \u03b3 are the roots of x3<\/sup> – 3x2<\/sup> – 16x + 48 = 0
\n\u03b1 + \u03b2 + \u03b3 = 3
\nGiven \u03b1 + \u03b2 = 0 (\u2235 Sum of two roots is zero)
\n\u2234 \u03b3 = 3
\ni.e. x – 3 is a factor of x3<\/sup> – 3x2<\/sup> – 16x + 48 = 0
\n\"Inter
\nx2<\/sup> – 16 = 0
\n\u21d2 x2<\/sup> = 16
\n\u21d2 x = \u00b14
\n\u2234 The roots are -4, 4, 3<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the condition that x3<\/sup> – px2<\/sup> + qx – r = 0 may have the sum of its roots zero.
\nSolution:
\nLet \u03b1, \u03b2, \u03b3 be the roots of x3<\/sup> – px2<\/sup> + qx – r = 0
\n\u03b1 + \u03b2 + \u03b3 = p ………(1)
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = q ………..(2)
\n\u03b1\u03b2\u03b3 = r ………..(3)
\nGiven \u03b1 + \u03b2 = 0 (\u2235 Sum of two roots is zero)
\nFrom (1), \u03b3 = p
\n\u2234 \u03b3 is a root of x3<\/sup> – px2<\/sup> + qx – r = 0
\n\u03b33<\/sup> – p\u03b32<\/sup> + q\u03b3 – r = 0
\nBut \u03b3 = p
\np3<\/sup> – p(p2<\/sup>) + q(p) – r = 0
\np3<\/sup> – p2<\/sup> + qp – r = 0
\n\u2234 qp = r is the required condition.<\/p>\n

Question 3.
\nGiven that the roots of x3<\/sup> + 3px2<\/sup> + 3qx + r = 0 are in
\n(i) A.P., show that 2p3<\/sup> – 3qp + r = 0
\n(ii) G.P., show that p3<\/sup>r = q3<\/sup>
\n(iii) H.P., show that 2q3<\/sup> = r(3pq – r)
\nSolution:
\nGiven equation is x3<\/sup> + 3px2<\/sup> + 3qx + r = 0
\n(i) The roots are in A.P.
\nSuppose a – d, a, a + d are the roots
\nSum = a – d + a + a + d = -3p
\n\u21d2 3a = -3p
\n\u21d2 a = -p ……….(1)
\n\u2235 ‘a’ is a root of x3<\/sup> + 3px2<\/sup> + 3qx + r = 0
\n\u21d2 a3<\/sup> + 3pa2<\/sup> + 3qa + r = 0
\nBut a = -p
\n\u21d2 p3<\/sup> + 3p(-p)2<\/sup> + 3q(-p) + r = 0
\n\u21d2 2p3<\/sup> – 3pq + r = 0 is the required condition<\/p>\n

(ii) The roots are in G.P.
\nSuppose the roots be \\(\\frac{a}{R}\\), a, aR
\nGiven (\\(\\frac{a}{R}\\)) (a) (aR) = -r
\n\u21d2 a = -r
\n\u21d2 a = (-r)1\/3<\/sup>
\n\u2235 ‘a’ is a root of x3<\/sup> + 3px2<\/sup> + 3qx + r = 0
\n\u21d2 (-r1\/3<\/sup>)3<\/sup> + 3p(-r1\/3<\/sup>)2<\/sup> + 3q(-r1\/3<\/sup>) + r = 0
\n\u21d2 -r + 3pr2\/3<\/sup> – 3qr1\/3<\/sup> + r = 0
\npr2\/3<\/sup> = qr1\/3<\/sup>
\n\u21d2 pr1\/3<\/sup> = q
\n\u21d2 p1\/3<\/sup>r = q is the required condition<\/p>\n

(iii) The roots of x3<\/sup> + 3px2<\/sup> + 3qx + r = 0 are in H.P.
\nx3<\/sup> + 3px2<\/sup> + 3qx + r = 0 ……..(1)
\nLet y = \\(\\frac{1}{x}\\) so that \\(\\frac{1}{y^{3}}+\\frac{3 p}{y^{2}}+\\frac{3 q}{y}+r\\) = 0
\nRoots of ry3<\/sup> + 3qy2<\/sup> + 3py + 1 = 0 are in A.P.
\nry3<\/sup> + 3qy2<\/sup> + 3py + 1 = 0 ……..(2)
\nSuppose a – d, a, a + d be the roots of (2)
\nSum = a – d + a + a + d = \\(-\\frac{3 q}{r}\\)
\n3a = \\(-\\frac{3 q}{r}\\)
\na = \\(-\\frac{q}{r}\\) ………(1)
\n\u2235 ‘a’ is a root of ry3<\/sup> + 3qy2<\/sup> + 3py + 1 = 0
\n\u21d2 ra3<\/sup> + 3qa2<\/sup> + 3pa + 1 = 0
\nBut a = \\(-\\frac{q}{r}\\)
\n\u21d2 \\(r\\left(-\\frac{q}{r}\\right)^{3}+3 q\\left(-\\frac{q}{r}\\right)^{2}+3 p\\left(-\\frac{q}{r}\\right)+1=0\\)
\n\u21d2 \\(\\frac{-q^{3}}{r^{2}}+\\frac{3 q^{3}}{r^{2}}-\\frac{3 p q}{r}+1=0\\)
\n\u21d2 -q3<\/sup> + 3q3<\/sup> – 3pqr + r2<\/sup> = 0
\n\u21d2 2q3<\/sup> = r(3pq – r) is the required condition.<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the condition that x3<\/sup> – px2<\/sup> + qx – r = 0 may have the roots in G.P.
\nSolution:
\nLet \\(\\frac{a}{R}\\), a, aR be the roots of x3<\/sup> – px2<\/sup> + qx – r = 0
\nThe product of the roots = \\(\\frac{a}{R}\\) . a . aR = a3<\/sup>
\nproduct of the roots = r
\n\u21d2 a = r1\/3<\/sup>
\n\u2235 a is a root of x3<\/sup> – px2<\/sup> + qx – r = 0
\n\u21d2 a3<\/sup> – pa2<\/sup> + qa – r = 0
\nBut a = r1\/3<\/sup>
\n\u21d2 (r1\/3<\/sup>)3<\/sup> – p(r1\/3<\/sup>)2<\/sup> + q(r1\/3<\/sup>) – r = 0
\n\u21d2 r – p . r2\/3<\/sup> + q . r1\/3<\/sup> – r = 0
\n\u21d2 p . r2\/3<\/sup> = qr1\/3<\/sup>
\nBy cubing on both sides
\n\u21d2 p3<\/sup>r2<\/sup> = q3<\/sup>r
\n\u21d2 p3<\/sup>r = q3<\/sup> is the required condition<\/p>\n

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

Question 1.
\nSolve 9x3<\/sup> – 15x2<\/sup> + 7x – 1 = 0, given that two of its roots are equal.
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3 are the roots of 9x3<\/sup> – 15x3<\/sup> + 7x – 1 = 0
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{15}{9}=\\frac{5}{3}\\)
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{7}{9}\\)
\n\u03b1\u03b2\u03b3 = \\(\\frac{1}{9}\\)
\nGiven \u03b1 = \u03b2 (\u2235 two of its roots are equal)
\n2\u03b1 + \u03b3 = \\(\\frac{5}{3}\\)
\n\u21d2 \u03b3 = \\(\\frac{5}{3}\\) – 2\u03b1
\n\u03b12<\/sup> + 2\u03b1\u03b3 = \\(\\frac{7}{9}\\)
\n\u21d2 \u03b12<\/sup> + 2\u03b1 (\\(\\frac{5}{3}\\) – 2\u03b1) = \\(\\frac{7}{9}\\)
\n\u21d2 \u03b12<\/sup> + \\(\\frac{2 \\alpha(5-6 \\alpha)}{3}=\\frac{7}{9}\\)
\n\u21d2 9\u03b12<\/sup> + 6\u03b1(5 – 6\u03b1) = 7
\n\u21d2 9\u03b12<\/sup> + 30\u03b1 – 36\u03b12<\/sup> = 7
\n\u21d2 27\u03b12<\/sup> – 30\u03b1 + 7 = 0
\n\u21d2 (3\u03b1 – 1)(9\u03b1 – 7) = 0
\n\u21d2 \u03b1 = \\(\\frac{1}{3}\\) or \\(\\frac{7}{9}\\)
\n\"Inter
\nThe roots are \\(\\frac{1}{3}\\), \\(\\frac{1}{3}\\), 1<\/p>\n

\"Inter<\/p>\n

Question 2.
\nGiven that one root of 2x3<\/sup> + 3x2<\/sup> – 8x + 3 = 0 is double of another root, find the roots of the equation.
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3 are the roots of 2x3<\/sup> + 3x2<\/sup> – 8x + 3 = 0
\n\u03b1 + \u03b2 + \u03b3 = \\(-\\frac{3}{2}\\) ……..(1)
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = -4 ……..(2)
\n\u03b1\u03b2\u03b3 = \\(-\\frac{3}{2}\\)
\nGiven \u03b1 = 2\u03b2 (\u2235 one root is double the other)
\nSubstituting in (1)
\n3\u03b2 + \u03b3 = \\(-\\frac{3}{2}\\)
\n\u21d2 \u03b3 = \\(-\\frac{3}{2}\\) – 3\u03b2 …….(4)
\nSubstituting in (2)
\n\u03b1\u03b2 + \u03b3(\u03b1 + \u03b2) = -4
\n\u21d2 2\u03b22<\/sup> + 3\u03b2\u03b3 = -4
\n\u21d2 2\u03b22<\/sup> + 3\u03b2(\\(-\\frac{3}{2}\\) – 3\u03b2) = -4
\n\u21d2 2\u03b22<\/sup> – \\(\\frac{3 \\beta(3+6 \\beta)}{2}\\) = -4
\n\u21d2 4\u03b22<\/sup> – 9\u03b2 – 18\u03b22<\/sup> = -8
\n\u21d2 14\u03b22<\/sup> + 9\u03b2 – 8 = 0
\n\u21d2 (2\u03b2 – 1)(7\u03b2 + 8) = 0
\n\u21d2 2\u03b2 – 1 = 0 or 7\u03b2 + 8 = 0
\n\u21d2 \u03b2 = \\(\\frac{1}{2}\\) or \u03b2 = \\(-\\frac{8}{7}\\)
\n\"Inter
\n\u2234 The roots are \\(\\frac{1}{2}\\), 1 and -3<\/p>\n

Question 3.
\nSolve x3<\/sup> – 9x2<\/sup> + 14x + 24 = 0, given that two of the roots are in the ratio 3 : 2.
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3 are the roots of x3<\/sup> – 9x2<\/sup> + 14x + 24 = 0
\n\u03b1 + \u03b2 + \u03b3 = 9 ………(1)
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = 14 ……….(2)
\n\u03b1\u03b2\u03b3 = -24 ……….(3)
\n\u2235 two roots are in the ratio 3 : 2
\nLet \u03b1 : \u03b2 = 3 : 2
\n\u21d2 \u03b2 = \\(\\frac{2 \\alpha}{3}\\)
\nSubstituting in (1)
\n\\(\\frac{5 \\alpha}{3}\\) + \u03b3 = 9
\n\u21d2 \u03b3 = 9 – \\(\\frac{5 \\alpha}{3}\\) ………(4)
\nSubstituting in (2)
\n\u21d2 \\(\\frac{2}{3}\\) \u03b12<\/sup> + \u03b3(\u03b1 + \u03b2) = 14
\n\u21d2 \\(\\frac{2}{3} \\alpha^{2}+\\left(9-\\frac{5 \\alpha}{3}\\right) \\cdot \\frac{5 \\alpha}{3}\\) = 14
\n\u21d2 2\u03b12<\/sup> + 5\u03b1(9 – \\(\\frac{5 \\alpha}{3}\\)) = 42
\n\u21d2 2\u03b12<\/sup> + 5\u03b1 \\(\\frac{(27-5 \\alpha)}{3}\\) = 42
\n\u21d2 6\u03b12<\/sup> + 135\u03b1 – 25\u03b12<\/sup> = 126
\n\u21d2 19\u03b12<\/sup> – 135\u03b1 + 126 = 0
\n\u21d2 19\u03b12<\/sup> – 114\u03b1 – 21\u03b1 + 126 = 0
\n\u21d2 19\u03b1(\u03b1 – 6) – 21(\u03b1 – 6) = 0
\n\u21d2 (19\u03b1 – 21)(\u03b1 – 6) = 0
\n\u21d2 19\u03b1 – 21 = 0 or \u03b1 – 6 = 0
\n\u21d2 \u03b1 = \\(\\frac{21}{19}\\) or \u03b1 = 6
\nCase (i): \u03b1 = 6
\n\u03b2 = \\(\\frac{2 \\alpha}{3}\\)
\n= \\(\\frac{2}{3}\\) \u00d7 6
\n= 4
\n\u03b3 = 9 – \\(\\frac{5 \\alpha}{3}\\)
\n= 9 – \\(\\frac{5}{3}\\) \u00d7 6
\n= 9 – 10
\n= -1
\n\u03b1 = 6, \u03b2 = 4, \u03b3 = -1 satisfy \u03b1\u03b2\u03b3 = -24
\n\u2234 The roots are 6, 4, -1<\/p>\n

Case (ii): \u03b1 = \\(\\frac{21}{19}\\)
\n\u03b2 = \\(\\frac{2}{3} \\times \\frac{21}{19}=\\frac{14}{19}\\)
\n\u03b3 = 9 – \\(\\frac{5 \\alpha}{3}\\)
\n= 9 – \\(\\frac{5}{3} \\cdot \\frac{21}{19}\\)
\n= \\(\\frac{136}{19}\\)
\nThese values do not satisfy \u03b1\u03b2\u03b3= -24
\n\u2234 The roots are 6, 4, -1.<\/p>\n

\"Inter<\/p>\n

Question 4.
\nSolve the following equations, given that the roots of each are in A.P.
\n(i) 8x3<\/sup> – 36x2<\/sup> – 18x + 81 = 0
\nSolution:
\nGiven the roots of 8x3<\/sup> – 36x2<\/sup> – 18x + 81 = 0 are in A.P.
\nLet the roots be a – d, a, a + d
\nSum of the roots = \\(\\frac{36}{8}\\)
\n\u21d2 a – d + a + a + d = \\(\\frac{9}{2}\\)
\n\u21d2 3a = \\(\\frac{9}{2}\\)
\n\u21d2 a = \\(\\frac{3}{2}\\)
\n\u2234 (x – \\(\\frac{3}{2}\\)) is a factor of 8x3<\/sup> – 36x2<\/sup> – 18x + 81 = 0
\n\"Inter
\n\u21d2 8x2<\/sup> – 24x – 54 = 0
\n\u21d2 4x2<\/sup> – 12x – 27 = 0
\n\u21d2 4x2<\/sup> – 18x + 6x – 27 = 0
\n\u21d2 2x(2x – 9) + 3(2x – 9) = 0
\n\u21d2 (2x + 3) (2x – 9) = 0
\n\u21d2 x = \\(-\\frac{3}{2}, \\frac{9}{2}\\)
\n\u2234 The roots are \\(-\\frac{3}{2}, \\frac{3}{2}, \\frac{9}{2}\\)<\/p>\n

(ii) x3<\/sup> – 3x2<\/sup> – 6x + 8 = 0
\nSolution:
\nThe roots of x3<\/sup> – 3x2<\/sup> – 6x + 8 = 0 are in A.P
\nSuppose a – d, a, a + d be the roots
\nSum = a – d + a + a + d = 3
\n\u21d2 3a = 3
\n\u21d2 a = 1
\n\u2234 (x – 1) is a factor of x3<\/sup> – 3x2<\/sup> – 6x + 8 = 0
\n\"Inter
\n\u21d2 x2<\/sup> – 2x – 8 = 0
\n\u21d2 x2<\/sup> – 4x + 2x – 8 = 0
\n\u21d2 x(x – 4)+ 2(x – 4) = 0
\n\u21d2 (x – 4) (x + 2) = 0
\n\u21d2 x = 4, -2
\n\u2234 The roots are -2, 1, 4<\/p>\n

Question 5.
\nSolve the following equations, given that the roots of each are in G.P.
\n(i) 3x3<\/sup> – 26x2<\/sup> + 52x – 24 = 0
\nSolution:
\nGiven equation is 3x3<\/sup> – 26x2<\/sup> + 52x – 24 = 0
\nThe roots are in G.P.
\nSuppose \\(\\frac{a}{r}\\), a, ar are the roots.
\nProduct = \\(\\frac{a}{r}\\) . a . ar = \\(-\\left(-\\frac{24}{3}\\right)\\)
\n\u21d2 a3<\/sup> = 8
\n\u21d2 a = 2
\n\u2234 (x – 2) is a factor of 3x3<\/sup> – 26x2<\/sup> + 52x – 24
\n\"Inter
\nHint: 3 \u00d7 12 = 3 \u00d76 \u00d7 2 = (-18)(-2)
\n\u21d2 3x2<\/sup> – 20x + 12 = 0
\n\u21d2 3x2<\/sup> – 18x – 2x + 12 = 0
\n\u21d2 3x(x – 6) – 2(x – 6) = 0
\n\u21d2 (3x – 2) (x – 6) = 0
\n\u21d2 x = \\(\\frac{2}{3}\\), 6
\n\u2234 The roots are \\(\\frac{2}{3}\\), 2, 6.<\/p>\n

(ii) 54x3<\/sup> – 39x2<\/sup> – 26x + 16 = 0
\nSolution:
\nGiven equation is 54x3<\/sup> – 39x2<\/sup> – 26x + 16 = 0
\nThe roots are in G.P.
\nSuppose \\(\\frac{a}{r}\\), a, ar be the roots.
\nProduct = \\(\\frac{a}{r}\\) . a . ar = \\(-\\frac{16}{54}\\)
\n\"Inter
\nHint: 18 \u00d7 8 = 9 \u00d7 2 \u00d7 8 = (-9) (-16)
\n\u21d2 54x2<\/sup> – 75x + 24 = 0
\n\u21d2 18x2<\/sup> – 25x + 8 = 0
\n\u21d2 18x2<\/sup> – 9x – 16x + 8 = 0
\n\u21d2 9x(2x – 1) – 8(2x- 1) = 0
\n\u21d2 (9x – 8) (2x – 1) = 0
\n\u21d2 x = \\(\\frac{8}{9}, \\frac{1}{2}\\)
\n\u2234 The roots are \\(\\frac{8}{9},-\\frac{2}{3}, \\frac{1}{2}\\)<\/p>\n

\"Inter<\/p>\n

Question 6.
\nSolve the following equations, given that the roots of each are in H.P.
\n(i) 6x3<\/sup> – 11x2<\/sup> + 6x – 1 = 0
\nSolution:
\nGiven equation is 6x3<\/sup> – 11x2<\/sup> + 6x – 1 = 0 …….(1)
\nPut y = \\(\\frac{1}{x}\\) so that \\(\\frac{6}{\\mathrm{y}^{3}}-\\frac{11}{\\mathrm{y}^{2}}+\\frac{6}{\\mathrm{y}}-1\\) = 0
\n\u21d2 6 – 11y + 6y2<\/sup> – y3<\/sup> = 0
\n\u21d2 y3<\/sup> – 6y2<\/sup> + 11y – 6 = 0 ………(2)
\nRoots of (1) are in H.P.
\n\u21d2 Roots of (2) are in A.P.
\nLet a – d, a, a + d be the roots of (2)
\nSum = a – d + a + a + d = 6
\n\u21d2 3a = 6
\n\u21d2 a = 2
\nProduct = a(a2<\/sup> – d2<\/sup>) = 6
\n\u21d2 2(4 – d2<\/sup>) = 6
\n\u21d2 4 – d2<\/sup> = 3
\n\u21d2 d2<\/sup> = 1
\n\u21d2 d = 1
\na – d = 2 – 1 = 1,
\na = 2
\na + d = 2 + 1 = 3
\nThe roots of (2) are 1, 2, 3
\nThe roots of (1) are 1, \\(\\frac{1}{2}\\), \\(\\frac{1}{3}\\)<\/p>\n

(ii) 15x3<\/sup> – 23x2<\/sup> + 9x – 1 = 0
\nSolution:
\nGiven equation is 15x3<\/sup> – 23x2<\/sup> + 9x – 1 = 0 …….(1)
\nPut y = \\(\\frac{1}{x}\\) so that \\(\\frac{15}{y^{3}}-\\frac{23}{y^{2}}+\\frac{9}{y}-1\\) = 0
\n\u21d2 15 – 23y + 9y2<\/sup> – y3<\/sup> = 0
\n\u21d2 y3<\/sup> – 9y2<\/sup> + 23y – 15 = 0 ………(2)
\nRoots of (1) are in H.P. So that roots of (2) are in A.P.
\nLet a – d, a, a + d be the roots of (2)
\nSum = a – d + a + a + d = 9
\n\u21d2 3a = 9
\n\u21d2 a = 3
\nProduct = a(a2<\/sup> – d2<\/sup>) = 15
\n\u21d2 3(9 – d2<\/sup>) = 15
\n\u21d2 9 – d2<\/sup> = 5
\n\u21d2 d2<\/sup> = 4
\n\u21d2 d = 2
\na – d = 3 – 2 = 1
\na = 3
\na + d = 3 + 2 = 5
\nRoots of (2) are 1, 3, 5
\nHence roots of (1) are 1, \\(\\frac{1}{3}\\), \\(\\frac{1}{5}\\)<\/p>\n

\"Inter<\/p>\n

Question 7.
\nSolve the following equations, given that they have multiple roots.
\n(i) x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 24x + 36 = 0
\nSolution:
\n(i) Let f(x) = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 24x + 36
\n\u21d2 f'(x) = 4x3<\/sup> – 18x2<\/sup> + 26x – 24
\n\u21d2 f'(3) = 4(27) – 18(9) + 26(3) – 24
\n\u21d2 f'(3) = 108 – 162 + 78 – 24
\n\u21d2 f'(3) = 0
\nf(3) = 81 – 162 + 117 – 72 + 36 = 0
\nHint: Choose the value of x from the factors of the G.C.D of constant terms in f(x) and f'(x).
\n\u2234 x – 3 is a factor of f'(x) and f(x)
\n\u2234 3 is the repeated foot of f(x)
\n\"Inter
\nx2<\/sup> + 4 = 0
\n\u21d2 x = \u00b12i
\n\u2234 The roots of the given equation are 3, 3, \u00b12i<\/p>\n

(ii) 3x4<\/sup> + 16x3<\/sup> + 24x2<\/sup> – 16 = 0
\nSolution:
\nLet f(x) = 3x4<\/sup> + 16x3<\/sup> + 24x2<\/sup> – 16
\nf(x) = 12x3<\/sup> + 48x2<\/sup> + 48x
\n= 12x(x2<\/sup> + 4x + 4)
\n= 12x (x + 2)2<\/sup>
\nf'(-2) = 0
\nf(-2) = 3(16) + 16(-8) + 24(4) – 16 = 0
\n\u2234 x + 2 is a factor of f'(x) and f(x)
\n\u2234 -2 is a multiple root of f(x) = 0
\n\"Inter
\n3x2<\/sup> + 4x – 4 = 0
\n\u21d2 3x2<\/sup> + 6x – 2x – 4 = 0
\n\u21d2 3x(x + 2) – 2(x + 2) = 0
\n\u21d2 (x + 2) (3x – 2) = 0
\n\u21d2 x = -2, \\(\\frac{2}{3}\\)
\n\u2234 The roots of the given equation are -2, -2, -2, \\(\\frac{2}{3}\\)<\/p>\n

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

Question 1.
\nSolve x4<\/sup> + x3<\/sup> – 16x2<\/sup> – 4x + 48 = 0, given that the product of two of the roots is 6.
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3, \u03b4 are the roots of
\nx4<\/sup> + x3<\/sup> – 16x2<\/sup> – 4x + 48 = 0 ………..(1)
\nSum of the roots = -1
\n\u21d2 \u03b1 + \u03b2 + \u03b3 + \u03b4 = -1
\nand Product of the roots = \u03b1\u03b2\u03b3\u03b4 = 48
\n\u2235 Product of two roots is 6
\nLet \u03b1\u03b2 = 6
\nFrom (1), \u03b3\u03b4 = \\(\\frac{48}{\\alpha \\beta}=\\frac{48}{6}\\) = 8
\nLet \u03b1 + \u03b2 = p and \u03b3 + \u03b4 = q
\nThe equation having roots \u03b1, \u03b2 is x2<\/sup> – (\u03b1 + \u03b2) x + \u03b1\u03b2 = 0
\n\u21d2 x2<\/sup> – px + 6 = 0 ………..(2)
\nThe equation having the roots \u03b3, \u03b4 is x2<\/sup> – (\u03b3 + \u03b4) x + \u03b3\u03b4 = 0
\n\u21d2 x2<\/sup> – qx + 8 = 0 ……….(3)
\n\u2234 From (1), (2) and (3)
\nx4<\/sup> + x3<\/sup> – 16x2<\/sup> – 4x + 48 = (x2<\/sup> – px + 6) (x2<\/sup> – qx + 8)
\n= x4<\/sup> – (p + q) x3<\/sup> + (pq + 14) x2<\/sup> – (8p + 86q) x + 48
\nComparing the like terms,
\np + q = -1
\n8p + 6q = 4 \u21d2 4p + 3q = 2
\nSolving, q = -6
\n\u2234 p = -1 + 6 = 5
\nSubstitute the value of p in eq. (2),
\nx2<\/sup> – 5x + 6 = 0 \u21d2 x = 2, 3
\nSubstitute\/the value of q in eq. (3),
\nx2<\/sup> + 6x + 8 = 0 \u21d2 x = -2, – 4
\n\u2234 The roots of the given equation are -4, -2, 2, 3<\/p>\n

\"Inter<\/p>\n

Question 2.
\nSolve 8x4<\/sup> – 2x3<\/sup> – 27x2<\/sup> + 6x + 9 = 0 given that two roots have the same absolute value, but are opposite in signs.
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3, \u03b4 are the roots of the equation
\n8x4<\/sup> – 2x3<\/sup> – 27x2<\/sup> + 6x + 9 = 0
\n\u21d2 \\(x^{4}-\\frac{1}{4} x^{3}-\\frac{27}{8} x^{2}+\\frac{3}{4} x+\\frac{9}{8}=0\\) …………(1)
\nSum of the roots = \u03b1 + \u03b2 + \u03b3 + \u03b4 = \\(\\frac{1}{4}\\) and
\nProduct of the roots = \u03b1\u03b2\u03b3\u03b4 = \\(\\frac{9}{8}\\)
\nGiven \u03b2 = -\u03b1
\n\u21d2 \u03b1 + \u03b2 = 0
\n\u2234 0 + \u03b3 + \u03b4 = \\(\\frac{1}{4}\\)
\n\u21d2 \u03b3 + \u03b4 = \\(\\frac{1}{4}\\)
\nLet \u03b1\u03b2 = p, \u03b3\u03b4 = q, so that pq = \\(\\frac{9}{8}\\)
\nThe equation having the roots \u03b1, \u03b2 is x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0
\n\u21d2 x2<\/sup> + p = 0 ……….(2)
\nThe equation having the roots \u03b3, \u03b4 is x2<\/sup> – (\u03b3 + \u03b4)x + \u03b3\u03b4 = 0
\n\u21d2 x2<\/sup> – \\(\\frac{1}{4}\\) x + q = 0 ……..(3)
\nFrom (1), (2) and (3)
\n\\(x^{4}-\\frac{1}{4} x^{3}-\\frac{27}{8} x^{2}+\\frac{3}{4} x+\\frac{9}{8}\\) = (x2<\/sup> + p) \\(\\left(x^{2}-\\frac{1}{4} x+q\\right)\\)
\n= \\(x^{4}-\\frac{1}{4} x^{3}+(p+q) x^{2}-\\frac{p}{4} x+p q\\)
\nComparing the coefficients of x and constants
\n\\(\\frac{-p}{4}=\\frac{3}{4}\\) \u21d2 p = -3
\npq = \\(\\frac{9}{8}\\)
\n\u21d2 q = \\(\\frac{9}{8} \\times \\frac{-1}{3}=\\frac{-3}{8}\\)
\nSubstitute the value of p in eq. (2),
\nx2<\/sup> – 3 = 0 \u21d2 x = \u00b1\u221a3
\nSubstitute the value of q in eq. (3),
\n\\(x^{2}-\\frac{1}{4} x-\\frac{3}{8}=0\\)
\n\u21d2 8x2<\/sup> – 2x – 3 = 0
\n\u21d2 (2x + 1) (4x – 3) = 6
\n\u21d2 x = \\(-\\frac{1}{2}, \\frac{3}{4}\\)
\n\u2234 The roots of the given equation are -\u221a3, \\(-\\frac{1}{2}, \\frac{3}{4}\\), \u221a3<\/p>\n

Question 3.
\nSolve 18x3<\/sup> + 81x2<\/sup> + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots.
\nSolution:
\nSuppose \u03b1, \u03b2, \u03b3 are the roots of 18x3<\/sup> + 81x2<\/sup> + 121x + 60 = 0
\nsum = \u03b1 + \u03b2 + \u03b3 = \\(\\frac{-81}{18}=\\frac{-9}{2}\\) ……….(1)
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{121}{18}\\) …………(2)
\n\u03b1\u03b2\u03b3 = \\(\\frac{-60}{18}=\\frac{-10}{3}\\) ………(3)
\n\u2235 One root is equal to half of the sum of the remaining two,
\nLet \u03b1 = \\(\\frac{1}{2}\\) (\u03b2 + \u03b3)
\nSubstitute in (1)
\n\"Inter<\/p>\n

Question 4.
\nFind the condition in order that the equation ax4<\/sup> + 4bx3<\/sup> + 6cx2<\/sup> + 4dx + e = 0 may have a pair of equal roots.
\nSolution:
\nLet \u03b1, \u03b1, \u03b2, \u03b2 are the roots of the equation.
\nax4<\/sup> + 4bx3<\/sup> + 6cx2<\/sup> + 4dx + e = 0
\n\u21d2 \\(x^{4}+\\frac{4 b}{a} x^{2}+\\frac{6 c}{a} x^{2}+\\frac{4 d}{a} x+\\frac{e}{a}=0\\)
\nSum of the roots, 2(\u03b1 + \u03b2) = \\(-\\frac{4 b}{a}\\)
\n\u21d2 \u03b1 + \u03b2 = \\(-\\frac{2 b}{a}\\)
\n\u21d2 \u03b1\u03b2 = k (say)
\nEquation having roots \u03b1, \u03b2 is x2<\/sup> – (\u03b1 + \u03b2) x + \u03b1\u03b2 = 0
\n\"Inter<\/p>\n

Question 5.
\n(i) Show that x5<\/sup> – 5x3<\/sup> + 5x2<\/sup> – 1 = 0 has three equal roots and find this root.
\nSolution:
\nLet f(x) = x5<\/sup> – 5x3<\/sup> + 5x2<\/sup> – 1
\nf'(x) = 5x4<\/sup> – 15x2<\/sup> + 10x = 5x(x3<\/sup> – 3x + 2)
\nf'(1) = 5(1) (1 – 3 + 2) = 0
\nf(1) = 1 – 5 + 5 – 1 = 0
\nx – 1 is a factor of f'(x) and f(x)
\n\u2234 1 is a repeated root of f(x).
\n\"Inter
\nx3<\/sup> + 2x2<\/sup> – 2x – 1 = 0
\n\u21d2 1 is a root of the above equation (\u2235 sum of the coefficients is zero)
\n\u2234 1 is the required root.<\/p>\n

(ii) Find the repeated roots of x5<\/sup> – 3x4<\/sup> – 5x3<\/sup> + 27x2<\/sup> – 32x + 12 = 0
\nSolution:
\nLet f(x) = x5<\/sup> – 3x4<\/sup> – 5x3<\/sup> + 27x2<\/sup> – 32x + 12
\nf'(x) = 5x4<\/sup> – 12x3<\/sup> – 15x2<\/sup> + 54x – 32
\nf'(2) = 5(2)4<\/sup> – 12(2)3<\/sup> – 15(2)2<\/sup> + 54(2) – 32
\n= 80 – 96 – 60 + 108 – 32
\n= 0
\nf(2) = (2)5<\/sup> – 3(2)4<\/sup> – 5(2)3<\/sup> + 27(2)2<\/sup> – 32(2) + 12
\n= 32 – 48 – 40 + 108 – 64 + 12
\n= 152 – 152
\n= 0
\n\u2234 x – 2 is a common factor of f'(x) and f(x)
\n2 is a multiple root of f(x) = 0
\n\"Inter
\nLet g(x) = x3<\/sup> + x2<\/sup> – 5x + 3
\ng'(x) = 3x2<\/sup> + 2x – 5 = (3x + 5) (x – 1)
\ng(1) = 1 + 1 – 5 + 3 = 0
\n\u2234 x – 1 is a common factor of g'(x) and g(x)
\n\u2234 1 is a multiple root of g(x) = 0
\n\"Inter
\nx + 3 = 0 \u21d2 x = -3
\n\u2234 The roots are 2, 2, 1, 1, -3<\/p>\n

\"Inter<\/p>\n

Question 6.
\nSolve the equation 8x3<\/sup> – 20x2<\/sup> + 6x + 9 = 0 given that the equation has multiple roots.
\nSolution:
\nLet f(x) = 8x3<\/sup> – 20x2<\/sup> + 6x + 9
\nf'(x) = 24x2<\/sup> – 40x + 6
\n= 2 (12x2<\/sup> – 20x + 3)
\n= 2[12x2<\/sup> – 18x – 2x + 3]
\n= 2[6x(2x – 3) – 1(2x – 3)]
\n= 2(2x – 3) (6x – 1)
\nf'(x) = 0
\n\u21d2 x = \\(\\frac{3}{2}\\), x = \\(\\frac{1}{6}\\)
\n\\(f\\left(\\frac{3}{2}\\right)=8\\left(\\frac{3}{2}\\right)^{3}-20\\left(\\frac{3}{2}\\right)^{2}+6\\left(\\frac{3}{2}\\right)+9\\)
\n= 27 – 45 + 9 + 9
\n= 0
\nHence x – \\(\\frac{3}{2}\\) is a factor of f(x) and f'(x)
\n\u2234 \\(\\frac{3}{2}\\) is a multiple root of f(x) = 0
\n\"Inter
\n8x – 4 = 0
\n\u21d2 x = \\(\\frac{4}{8}=\\frac{1}{2}\\)
\n\u2234 The roots of the equation f(x) = 0 are \\(\\frac{3}{2}, \\frac{3}{2}, \\frac{1}{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(b) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(b) I. Question 1. Solve x3 – 3×2 – 16x + 48 = 0, given that the … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22347"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22347"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22347\/revisions"}],"predecessor-version":[{"id":22348,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22347\/revisions\/22348"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22347"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22347"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22347"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}