Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Theory of Equations Important Questions Very Short Answer Type to help strengthen their preparations for exams.<\/p>\nTS Inter Second Year Maths 2A Theory of Equations Important Questions Very Short Answer Type<\/h2>\n
Question 1.
\nForm the polynomial equation of the lowest degree whose roots are 1, – 1, 3. [May ’06]
\nSolution:
\nLet \u03b1 = 1; \u03b2 = – 1, \u03b3 = 3
\nThe equation having roots \u03b1, \u03b2, \u03b3 is
\n(x – \u03b1) (x – \u03b2) (x – \u03b3) = 0
\n\u21d2 (x – 1) (x + 1) ( x – 3) = 0
\n\u21d2 (x2<\/sup> – 1) (x – 3) = 0
\n\u21d2 x3<\/sup> – 3x2<\/sup> – x + 3 = 0.<\/p>\nQuestion 2.
\nForm the polynomial equation of the lowest degree whose roots are 2 \u00b1 \u221a3, 1 \u00b1 2i. [May \u201902, ’01]
\nSolution:
\nLet
\n\u03b1 = 2 + \u221a3, \u03b2 = 2 – \u221a3, \u03b3 = 1 + 2i, \u03b4 = 1 – 2i
\nThe equation having \u03b1, \u03b2, \u03b3, \u03b4 is
\n(x – \u03b1) (x – \u03b2) (x – \u03b3) (x – \u03b4) = 0
\n\u21d2 (x – 2 – \u221a3) (x – 2 + \u221a3) ((x – 1) – 2i) ((x – 1) + 2i) = 0
\n\u21d2 ((x – 2) – \u221a3) ((x – 2) + \u221a3) ((x – 1) – 2i) ((x – 1) + 2i) = 0
\n\u21d2 (x2<\/sup> + 4 – 4x – 3) (x2<\/sup> + 1 – 2x + 4) = 0
\n\u21d2 (x2<\/sup> – 4x + 1) (x2<\/sup> – 2x + 5) = 0
\n\u21d2 x4<\/sup> + 5x2<\/sup> – 2x3<\/sup> – 4x3<\/sup> + 8x2<\/sup> – 20x + x2<\/sup> – 2x + 5 = 0
\n\u21d2 x4<\/sup> – 6x3<\/sup> + 14x2<\/sup> – 22x + 5 = 0<\/p>\nQuestion 3.
\nIf 1, 1, \u03b1 are the roots of x3<\/sup> – 6x2<\/sup> + 9x – 4 = 0. then find \u03b1. [AP – Mar. ’18; TS – May 2016; May ’11]
\nSolution:
\nGiven equation is x3<\/sup> – 6x2<\/sup> + 9x – 4 = 0
\nComparing this equation with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nwe get, a = 1; b = – 6; c = 9; d = – 4
\nSince 1, 1, \u03b1 are the roots of x3<\/sup> – 6x2<\/sup> + 9x – 4 = 0
\nthen sum of the roots = s1<\/sub> = \\(\\frac{-b}{a}\\)
\n\u21d2 1 + 1 + \u03b1 = \\(\\frac{-(-6)}{1}\\) = 6
\n\u21d2 2 + \u03b1 = 6
\n\u21d2 \u03b1 = 4.<\/p>\n<\/p>\n
Question 4.
\nIf – 1, 2 and \u03b1 are the roots of 2x3<\/sup> + x2<\/sup> – 7x – 6 = 0, then find \u03b1.
\n[AP-Mar. 18; Mar. ’14, ’13, ’10, ’06, May’ 12, ’10]
\nSolution:
\nGiven equation is 2x3<\/sup> + x2<\/sup> – 7x – 6 = 0
\nComparing this with ax3<\/sup> + bx2<\/sup> + cx + d = 0 we get,
\na = 2; b = 1; c = – 7; d = – 6
\nSince – 1, 2, \u03b1 are the roots of 2x3<\/sup> + x2<\/sup> – 7x – 6 = 0
\nThe sum of the roots = s1<\/sub> = \\(\\frac{-b}{a}\\)
\n– 1 + 2 + \u03b1 = – \\(\\frac{1}{2}\\)
\n1 + \u03b1 = – \\(\\frac{1}{2}\\)
\n\u21d2 \u03b1 = – 1 – \\(\\frac{1}{2}\\)
\n\u03b1 = – \\(\\frac{3}{2}\\)<\/p>\nQuestion 5.
\nIf 1, – 2 and 3 are the roots of x3<\/sup> – 2x2<\/sup> + ax + 6 = 0, then find \u2018a\u2019. [TS – May 2015; March \u201804]
\nSolution:
\nGiven equation is x3<\/sup> – 2x2<\/sup> + ax + 6 = 0
\nSince 1, – 2, 3 are the roots of x3<\/sup> – 2x2<\/sup> + ax + 6
\nNow 1 is a root of given equation then
\n13<\/sup> – 2(1)2<\/sup> + a(1) + 6 = 0
\n\u21d2 1 – 2 + a + 6 = 0
\na + 5 = 0
\na = – 5<\/p>\nQuestion 6.
\nIf the product of the roots of 4x3<\/sup> + 16x2<\/sup> – 9x – a = 0, is 9, then find ‘a’. [AP – Mar. 19, 17; TS – Mar. 16; May 13, 12, 08]
\nSolution:
\nGiven equation is 4x3<\/sup> + 16x2<\/sup> – 9x – a = 0
\nComparing this equation with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nwe get, a = 4, b = 16, c = – 9, d = – a
\nGiven that, the product of the roots = 9
\ns3<\/sub> = 9
\n\u21d2 \\(\\frac{-\\mathrm{d}}{\\mathrm{a}}\\) = 9
\n\u21d2 \\(\\frac{-(-a)}{4}\\) = 9
\n\u21d2 a = 36.<\/p>\n<\/p>\n
Question 7.
\nIf \u03b1, \u03b2 and 1 are the roots of x3<\/sup> – 2x2<\/sup> – 5x + 6 = 0, then find \u03b1 and \u03b2. [AP – May, Mar. 2016; May 09, March 08]
\nSolution:
\nGiven equation is x3<\/sup> – 2x2<\/sup> – 5x + 6 = 0
\nComparing this equation with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nwe get a = 1, b = – 2, c = – 5, d = 6
\nSince \u03b1, \u03b2 and 1 are the roots of x3<\/sup> – 2x2<\/sup> – 5x + 6 = 0
\nthen s1<\/sub> = \u03b1 + \u03b2 + 1
\n= \\(\\frac{-b}{a}=\\frac{-(-2)}{1}\\) = 2
\n\u21d2 \u03b1 + \u03b2 = 1 …………(i)
\ns3<\/sub> = \u03b1\u03b2 . 1 = \\(\\frac{-d}{a}\\)
\n\u21d2 \u03b1\u03b2 = \\(\\frac{-6}{1}\\) =
\n\u21d2 \u03b1\u03b2 = – 6
\n(\u03b1 – \u03b2)2<\/sup> = (\u03b1 + \u03b2)2<\/sup> – 4\u03b1\u03b2
\n= 1 – 4 ( -6)
\n= 1 + 24 = 25
\n\u21d2 \u03b1 – \u03b2 = 5 ……………(2)
\nSolve (1) and (2);<\/p>\n<\/p>\n
Question 8.
\nIf \u03b1, \u03b2 and \u03b3 are the roots of x3<\/sup> + 2x2<\/sup> + 3x – 4 = 0, then find \u03b12<\/sup>\u03b22<\/sup>. [May ’07]
\nSolution:
\nGiven equation is x3<\/sup> + 2x2<\/sup> + 3x – 4 = 0
\nSince \u03b1, \u03b2 and \u03b3 are the roots of x3<\/sup> + 2x2<\/sup> + 3x – 4 = 0 then
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{-(-2)}{1}\\) = 2
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{3}{1}\\) = 3
\n\u03b1\u03b2\u03b3 = \\(\\frac{-(-4)}{1}\\) = 4
\n\u03a3\u03b12<\/sup>\u03b22<\/sup> = \u03b12<\/sup>\u03b22<\/sup> + \u03b22<\/sup>\u03b32<\/sup> + \u03b32<\/sup>\u03b12<\/sup>
\n= (\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1)2<\/sup> – 2\u03b1\u03b2\u03b3 (\u03b1 + \u03b2 + \u03b3)
\n= 32<\/sup> – 2 . 4(2)
\n= 9 – 16 = – 7.<\/p>\nQuestion 9.
\nIf \u03b1, \u03b2 and \u03b3 are the roots of 4x3<\/sup> – 6x2<\/sup> + 7x + 3 = 0, then find the value of \u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1. [TS – Mar. 2019]
\nSolution:
\nGiven equation is 4x3<\/sup> – 6x2<\/sup> + 7x + 3 = 0
\nComparing this equation with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nwhere a = 4; b = – 6; c = 7; d = 3
\nSince \u03b1, \u03b2, \u03b3 are the roots of 4x3<\/sup> – 6x2<\/sup> + 7x + 3 = 0 then
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = S2<\/sub> = \\(\\frac{c}{a}=\\frac{7}{4}\\).<\/p>\nQuestion 10.
\nFind the relations between the roots and the coefficients of the cubic equation 3x3<\/sup> – 10x2<\/sup> + 7x + 10 = 0.
\nSolution:
\nGiven cubic equation is 3x3<\/sup> – 10x2<\/sup> + 7x + 10 = 0
\nComparing this equation with
\nax3<\/sup> + bx2<\/sup> + cx + d = 0, we get
\na = 3, b = – 10, c = 7, d = 10
\nLet \u03b1, \u03b2, \u03b3 be the roots of given equation
\ns1<\/sub> = \\(\\frac{-b}{a}\\)
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{-(-10)}{3}=\\frac{10}{3}\\)
\ns2<\/sub> = \u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1
\n= \\(\\frac{c}{a}=\\frac{7}{3}\\)
\ns3<\/sub> = \u03b1\u03b2\u03b3
\n= \\(\\frac{-\\mathrm{d}}{\\mathrm{a}}=\\frac{-10}{3}\\).<\/p>\n<\/p>\n
Question 11.
\nWrite down the relations between the roots and the coefficients of the biquadratic equation x4<\/sup> – 2x3<\/sup> + 4x2<\/sup> + 6x – 21 = 0.
\nSolution:
\nGiven biquadratic equation is
\nx4<\/sup> – 2x3<\/sup> + 4x2<\/sup> + 6x – 21 = 0 ……………… (1)
\nComparing this equation with
\nax4<\/sup> + bx3<\/sup> + cx2<\/sup> + dx + e = 0,
\nwe get a = 1, b = – 2; c = 4; d = 6; e = – 21
\nLet \u03b1, \u03b2, \u03b3, \u03b4 are the roots of equation (1) then
\ni) s1<\/sub> = \u03b1 + \u03b2 + \u03b3 + \u03b4
\n= \u03a3\u03b1 = \\(\\frac{-b}{a}=\\frac{-(-2)}{1}\\) = 2
\nii) s2<\/sub> = \u03a3\u03b1\u03b2
\n= \\(\\frac{c}{a}=\\frac{4}{1}\\) = 4
\niii) s3 = \u03a3\u03b1\u03b2\u03b3
\n= \\(\\frac{-d}{a}=\\frac{-6}{1}\\) = – 6
\niv) s = \u03a3\u03b1\u03b2\u03b3\u03b4
\n= \\(\\frac{\\mathrm{e}}{\\mathrm{a}}=\\frac{-21}{1}\\) = – 21<\/p>\nQuestion 12.
\nIf 1, 2, 3 and 4 are the roots of x4<\/sup> + ax3<\/sup>+ bx2<\/sup> + cx + d = 0, then find the values of a, b, c and d. [AP – May 2015]
\nSolution:
\nGiven that the roots of the polynomial equation are 1, 2, 3 and 4.
\nThen(x – 1) (x – 2) (x – 3) (x – 4) = 0
\n(x2<\/sup> – 3x + 2) (x2<\/sup> – 7x + 12) = 0
\nx4<\/sup> – 7x3<\/sup> + 12x2<\/sup> – 3x3<\/sup> + 21x2<\/sup> – 36x + 2x2<\/sup> – 14x + 24 = 0
\nx4<\/sup> – 10x3<\/sup> + 35x2<\/sup> – 50x + 24 = 0
\nNow, comparing this equation with
\nx4<\/sup> + ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nwe get a = – 10; b = 35; c = – 50; d = 24.<\/p>\nQuestion 13.
\nIf a, b, c are the roots of x3<\/sup> – px2<\/sup> + qx – r = 0 and r \u2260 0, then find \\(\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}\\) in terms of p, q, r.
\nSolution:
\nGiven equation is x3<\/sup> – px2<\/sup> + qx – r = 0
\nSince a, b and c are the roots of the equation x3<\/sup> – px2<\/sup> + qx – r = 0 then
\ns1<\/sub> = a + b + c
\n= \\(\\frac{-(-p)}{1}\\) = p;
\ns2<\/sub> = ab + bc + ca
\n= \\(\\frac{q}{1}\\) = q;
\ns3<\/sub> = abc
\n= \\(\\frac{-(-r)}{1}\\) = r
\n\\(\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}=\\frac{b^2 c^2+a^2 c^2+a^2 b^2}{a^2 b^2 c^2}\\)
\n= \\(\\frac{(a b+b c+c a)^2-2 a b c(a+b+c)}{(a b c)^2}\\)
\n= \\(\\frac{q^2-2 p \\cdot r}{r^2}=\\frac{q^2-2 p r}{r^2}\\).<\/p>\n<\/p>\n
Question 14.
\nFind the sum of the squares and the sum of the cubes of the roots of the equation x3<\/sup> – px2<\/sup> + qx – r = o in terms of p, q, r.
\nSolution:
\nGiven equation is x3<\/sup> – px2<\/sup> + qx – r = 0
\nLet \u03b1, \u03b2, \u03b3 are the roots of the equation x3<\/sup> – px2<\/sup> + qx – r = 0 then
\ns1<\/sub> = \u03b1 + \u03b2 + \u03b3
\n= \\(\\frac{-\\mathrm{b}}{\\mathrm{a}}=\\frac{-(-\\dot{\\mathrm{p}})}{1}\\) = p;
\ns2<\/sub> = \u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1
\n= \\(\\frac{q}{1}\\) = q;
\ns3<\/sub> = \u03b1\u03b2\u03b3
\n= \\(\\frac{-(-r)}{1}\\) = r
\ni) The sum of the squares of the roots of the equation = \u03b12<\/sup> + \u03b22<\/sup> + \u03b32<\/sup>
\n= (\u03b1 + \u03b2 + \u03b3)2<\/sup> – 2(\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1)
\n= p2<\/sup> – 2q
\nii) The sum of the cubes of the roots is \u03b13<\/sup> + \u03b23<\/sup> + \u03b33<\/sup> = (\u03b1 + \u03b2 + \u03b3)
\n(\u03b12<\/sup> + \u03b22<\/sup> + \u03b32<\/sup> – \u03b1\u03b2 – \u03b2\u03b3 – \u03b3\u03b1) + 3\u03b1\u03b2\u03b3
\n= p(p2<\/sup> – 2q – q) + 3r
\n= p3<\/sup> – 3pq + 3r.<\/p>\nQuestion 15.
\nLet \u03b1, \u03b2, \u03b3 be the roots of x3<\/sup> + px2<\/sup> + qx + r = 0. Then find \u03a3\u03b13<\/sup>. [March ’03]
\nSolution:
\nGiven equation is x3<\/sup> + px2<\/sup> + qx + r = 0
\nSince \u03b1, \u03b2, \u03b3 are the roots of the equation x3<\/sup> + px2<\/sup> + qx + r = 0 then
\ns1<\/sub> = \u03b1 + \u03b2 + \u03b3
\n= \\(\\frac{-\\mathrm{p}}{1}\\) = – p;
\ns2<\/sub> = \u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1
\n= \\(\\frac{q}{1}\\) = q;
\ns3<\/sub> = \u03b1\u03b2\u03b3
\n= \\(\\frac{-r}{1}\\) = – r
\n\u03a3\u03b13<\/sup> = \u03b13<\/sup> + \u03b23<\/sup> + \u03b33<\/sup>
\n= (\u03b1 + \u03b2 + \u03b3) (\u03b12<\/sup> + \u03b22<\/sup> + \u03b32<\/sup> – \u03b1\u03b2 – \u03b2\u03b3 – \u03b3\u03b1) + 3\u03b1\u03b2\u03b3
\n= (- p) ((p2<\/sup> – 2q) – (q)) + 3 (- r)
\n= – p (p2<\/sup> – 3q) – 3r
\n= – p3<\/sup> + 3pq – 3r<\/p>\nQuestion 16.
\nFind s1<\/sub>, s2<\/sub>, s3<\/sub> and s4<\/sub> for the equation x4<\/sup> – 16x3<\/sup> + 86x2<\/sup> – 176x + 105 = 0.
\nSolution:
\nGiven equation is x4<\/sup> – 16x3<\/sup> + 86x2<\/sup> – 176x + 105 = 0
\nComparing this equation with ax4<\/sup> + bx3<\/sup> + cx2<\/sup> + dx + e = 0
\nwe get a = 1; b = – 16; c = 86; d = – 176; e = 105
\nNow, s1<\/sub> = \\(=\\frac{-b}{a}=\\frac{-(-16)}{1}\\) = 16;
\ns2<\/sub> = \\(\\frac{c}{\\mathrm{a}}=\\frac{86}{1}\\) = 86;
\ns3<\/sub> = \\(\\frac{-\\mathrm{d}}{\\mathrm{a}}=\\frac{-(-176)}{1}\\) = 176;
\ns4<\/sub> = \\(\\frac{\\mathrm{e}}{\\mathrm{a}}=\\frac{105}{1}\\) = 105<\/p>\n<\/p>\n
Question 17.
\nFind the algebraic equation whose roots are 2 times the roots of x5<\/sup> – 2x4<\/sup> + 3x3<\/sup> – 2x2<\/sup> + 4x + 3 = 0. [Board Paper]
\nSolution:
\nLet f(x) = x5<\/sup> – 2x4<\/sup> + 3x3<\/sup> – 2x2<\/sup> + 4x + 3 = 0
\n\u2234 Required equation is f(\\(\\frac{x}{2}\\)) = 0
\n\u21d2 \\(\\frac{x^5}{32}-\\frac{2 x^4}{16}+\\frac{3 x^3}{8}-\\frac{2 x^2}{4}+\\frac{4 x}{2}+3\\) = 0
\n\u21d2 x5<\/sup> – 2x4<\/sup> + 3x3<\/sup> – 2x2<\/sup> + 4x + 3 = 0<\/p>\nQuestion 18.
\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:
\nLet f(x) = x7<\/sup> + 3x5<\/sup> + x3<\/sup> – x2<\/sup> + 7x + 2 = 0
\nRequired equation is f(- x) = 0
\n(- x)7<\/sup> + 3(- x)5<\/sup> + (- x)3<\/sup> – (- x)2<\/sup> + 7(- x) + 2 = 0
\nx7<\/sup> + 3x5<\/sup> + x3<\/sup> + x2<\/sup> + 7x – 2 = 0<\/p>\nQuestion 19.
\nFind the polynomial equation whose roots are the reciprocals of the roots of x4<\/sup> – 3x3<\/sup> + 7x2<\/sup> + 5x – 2 = 0. [March ’11] [TS – Mar. 2015]
\nSolution:
\nLet f(x) = x4<\/sup> – 3x3<\/sup> + 7x2<\/sup> + 5x – 2 = 0
\nThe required equation is f(\\(\\frac{1}{x}\\)) = 0
\n\u21d2 \\(\\frac{1}{x^4}-\\frac{3}{x^3}+\\frac{7}{x^2}+\\frac{5}{x}-2\\) = 0
\n\u21d2 1 – 3x + 7x2<\/sup> + 5x3<\/sup> – 2x4<\/sup> = 0
\n\u21d2 2x4<\/sup> – 5x3<\/sup> – 7x2<\/sup> + 3x – 1 = 0<\/p>\nQuestion 20.