49. Quadratic Equations Solutions Part 2

51. Given \(r = k + \frac{s}{k}\)

    \(\Rightarrow r^2 = k^2 + \frac{s^2}{k^2} + 2s\)

    \(r^2 - 4s = k^2 + \frac{s^2}{k^2} + 2s - 4s\)

    \(r^ - 4s = k^2 + \frac{s^2}{k^2} - 2s = \left(k - \frac{s}{k}\right)^2\)

    Clearly, \(r^2 - 4s \geq 0\) if \(r, s, k\) are rationals which is discriminant of the given equation. Thus, roots will be rational provided given condition is met.

52. The given equation is \((x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0\)

    \(3x^2 -(a + b + b + c + c + a)x + ab + bc + ca = 0\)

    \(D = 4(a + b + c)^2 - 12(ab + bc + ca)\)

    \(= 4a^2 + 4b^2 + 4c^2 - 4ab - 4bc - 4ac\)

    \(= 2[(a - b)^2 + (b - c)^2 + (c - a)^2]\)

    This cannot be zero unless \(a = b = c\) which is the required condition for the roots to be equal.

53. Given equation is \(a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)x + c^2(a^2 - b^2) = 0\)

    \(D = b^4(c^2 - a^2)^2 - 4a^2c^2(b^2 - c^2)(a^2 - b^2)\)

    \(= b^4c^4 + b^4a^4 - 2b^4a^2c^2 - 4a^4b^2c^2 + 4a^2b^4c^2 - 4a^4c^4 + 4a^2b^2c^4\)

    \(= b^4c^4 + b^4a^4 + 2b^4a^2c^2 - 4a^4b^2c^2 - 4a^4c^4 + 4a^2b^2c^4\)

    \(= (b^2c^2 + b^2a^2 - 2a^2c^2)^2 \geq 0\)

    Thus, roots will be rational.

54. \(D = 16a^2b^2c^2d^2 - 4(a^4 + b^4)(c^4 + d^4)\)

    \(= 4[4a^2b^2c^22d^2 - a^4c^4 - a^4d^4 - b^4c^4 - b^4d^4]\)

    \(= -4[\{(a^2c^2 + b^2d^2)^2\}\{(a^2c^2 + b^2d^2)^2\}]\)

    Thus, if the roots are real then discriminant has to be zero because else it can be only negative and then roots wont remain real.

55. \(D = 4q^2 - 4pr = 4(q^2 - pr)\)

    Since \(p, q, r\) are in H. P. \(\Rightarrow q = \frac{2pr}{p + r}\)

    Substituting for \(q\) we get

    \(D = 4\left[\frac{4p^2r^2}{(p + r)^2} - pr\right]\)

    \(= 4\left[\frac{4p^2r^2 - p^3r - pr^3 - 2p^2r^2}{(p + r)^2}\right]\)

    \(= 4\left[\frac{2p^2r^2 - p^3 - r^3}{(p + r)^2}\right]\)

    \(= 4\left[\frac{pr(2pr - p^2 - r^2)}{(p + r)^2}\right]\)

    \(= 4\left[\frac{-pr(p - r)^2}{(p + r)^2}\right]\)

    Since \(p\) and \(r\) have the same sign discriminant is bound to be negative and roots will be complex numbers.

56. Discriminant of \(bx^2 + (b - c)x + (b - c - a) = 0, D_1 = (b - c)^2 -4b(b - c - a)\)

    \(= b^2 + c^2 -2bc -4b^2 + 4bc + 4ab\)

    Discriminant of \(ax^2 + 2bc + b = 0, D_2 = 4b^2 - 4ab\)

    Now, if \(D_2 < 0\)

    \(D_1 = (b + c)^2 - (4b^2 - 4ab) > 0\) and thus roots will be real.

    However, if \(D_1 < 0\) i.e. roots are imaginary then we have

    \(D_1 = (b + c)^2 - (4b^2 - 4ab) < 0 \Rightarrow 4b^2 - 4ab > 0 \because [(b + c)^2 > 0]\)

    Then roots of equation \(ax^2 + 2bx + b = 0\) will be real.

57. From first equation \(x = \sqrt{\frac{1 - by^2}{a}}\)

    and from second equation \(x = \frac{1 - by}{a}\)

    Equating the values obtained

    \(\left(\frac{1 - by}{a}\right)^2 = \frac{1 - by^2}{a}\)

    \(1 + b^2y^2 - 2by = a - aby^2 \Rightarrow (b^2 + ab)y^2 - 2by + 1 - a = 0\)

    Values of \(x\) will be equal if values of \(y\) are equal i.e. discriminant of above equation is zero.

    \(\Rightarrow 42b^2 - 4(b^2 + ab)(1 - a) = 0\)

    \(\Rightarrow 4b^2 - 4b^2 + 4b^2a - 4ab + 4a^2b = 0\)

    \((a^2b + ab^2 - ab) = 0\)

    \(ab(a + b) = ab\)

    \(a + b = 1\)

58. Substituting \(y = mx + c\) in \(x^2 + y^2 = a^2,\) we get]

    \(x^2 + m^2x^2 + 2cmx + c^2 - a^2 = 0\)

    For roots to be equal, discriminant must be zero.

    \(D = 4c^2m^2 - 4(1 + m^2)(c^2 - a^2) = 0\)

    \(\Rightarrow c^2m^2 - c^2 + a^2 - c^2m^2 + a^2m^2 = 0\)

    \(c^2 = a^2(1 + m^2)\)

59. Clearly, roots are \(\alpha, \alpha + 1.\)

    Sum of roots \(= \alpha + \alpha + 1 = \frac{5a + 1}{4}\)

    \(\alpha = \frac{5a - 3}{8}\)

    Product of roots \(= \alpha(\alpha + 1) = \frac{5a}{4}\)

    Substituting value of \(\alpha\) from above

    \(\left(\frac{5a - 3}{8}\right)^2 + \frac{5a - 3}{8} = \frac{5a}{4}\)

    \(\frac{25a^2 - 30a + 9 + 40a - 24 - 80a}{64} = 0\)

    \(25a^2 - 70a - 15 = 0 \Rightarrow 5a^2 - 14a - 3 = 0\)

    \(\Rightarrow a = 3, -\frac{1}{5}\)

    If \(a = 3 \Rightarrow \alpha = \frac{3}{2}\)

    else if \(a = -\frac{1}{5} \Rightarrow \alpha = -\frac{1}{2}\)

    Now it is trivial to calculate the value of \(\beta.\)

60. Let one of the roots is \(\alpha\) then second root is \(\frac{1}{\alpha}\).

    Product of roots \(= \alpha * \frac{1}{\alpha} = \frac{k}{5} \Rightarrow k = 5\)

61. 1. The equation is \((5 + 4m)x^2 - (4 + 2m)x + 2 - m = 0\)

    For roots to be equal discriminant has to be zero.

    \(4(2 + m)^2 - 4(5 + 4m)(2 - m) = 0\)

    \(4 + 4m + m^2 - 10 - 3m + 4m^2 = 0\)

    \(5m^2 - m - 6 = 0 \Rightarrow m = 1, -\frac{6}{5}\)

    2. Product of roots \(= \frac{2 - m}{5 + 4m} = 2 \Rightarrow 2 - m = 10 + 8m \Rightarrow -\frac{8}{9}\)

    3. Sum of roots \(= \frac{4 + 2m}{5 + 4m} = 6 \Rightarrow m = -\frac{13}{11}\)

62. Let one root be \(\alpha\) then the second root is \(n\alpha.\)

    Sum of roots \((n + 1)\alpha = -\frac{b}{a} \Rightarrow \alpha = -\frac{b}{(n + 1)a}\)

    Product of roots \(n\alpha^2 = \frac{c}{a}\)

    Substituting value of \(\alpha\) from the earlier equation

    \(\frac{nb^2}{(n + 1)^2a^2} = \frac{c}{a} \Rightarrow (n + 1)^2 ca = nb^2\)

63. Following from previous problem \(n = \frac{3}{4}\) and substituting in final solution

    \(\left(\frac{3}{4} + 1\right)^2ca = \frac{3}{4}b^2 \Rightarrow 12b^2 = 49ac\)

64. Comparing this with problem 62’s first equation, we have \(a = 4, b = a, c = 3\) and \(n = \frac{1}{2}\)

    Substituting in the final relation we have, \(\frac{9}{4}*3*4 = \frac{1}{2}a^2\)

    \(a^2 = 54\)

    Discriminant of the second equation, \(D = 9 - 4(a^2 - 2a) < 0,\) and thus roots are imaginary.

65. Let \(\alpha, \beta\) be the roots of the given equation.

    Sum of roots, \(\alpha + \beta = p\) and product of the roots \(\alpha\beta = q\)

    Given, \(\alpha + \beta = m(\alpha - \beta)\)

    Squaring, \((\alpha + \beta)^2 = m^2(\alpha - \beta)^2\)

    \(p^2 = m^2(\alpha + \beta)^2 - 4m^2\alpha\beta = m^2p^2 - 4m^2q \Rightarrow p^2(m^2 - 1) = 4m^2q\)

66. Let \(\alpha, \beta\) be the roots of the given equation.

    Sum of roots, \(\alpha + \beta = p\) and product of the roots \(\alpha\beta = q\)

    Given, \(\alpha - \beta = 1\)

    Squaring we have,

    \((\alpha - \beta)^2 = 1 \Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = 1\)

    \(\Rightarrow p^2 - 4q = 1\)

    Also, \([(\alpha - \beta)^2 + 2\alpha\beta]^2 = (1 + 2q)^2\)

    \((\alpha^2 + \beta^2)^2 = \alpha^4 + \beta^4 + 2\alpha^2\beta^2\)

    \(= \alpha^4 + \beta^4 - 2\alpha^2\beta^2 + 4\alpha^2\beta^2 = (\alpha^2 - \beta^2)^2 + 4q^2\)

    \([(\alpha + \beta)^2(\alpha - \beta)^2] + 4q^2 = p^2 + 4q^2\)

67. The given equation is \(a(x - b) + b(x - a) = m(x - a)(x - b)\)

    \(\Rightarrow mx^2 - xm(a + b) - mab - ax + ab - bx + ab = 0\)

    \(\Rightarrow mx^2 - x(m + 1)(a + b) - ab(m - 2) = 0\)

    If roots are equal in magnitude but opposite in sign then sum would be zero.

    \(\Rightarrow (m + 1)(a + b) = 0 \Rightarrow m = -1~\text{or}~a + b = 0\)

68. Let \(\alpha, \beta\) be the roots of the equation.

    Sum of roots, \(\alpha + \beta = -\frac{b}{a}\) and product of roots, \(\alpha\beta = \frac{c}{a}\)

    Difference of roots, \(\alpha - \beta = k\) as given.

    Squaring we get, \((\alpha - \beta)^2 = k^2 \Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = k^2\)

    \(\frac{b^2}{a^2} - 4\frac{c}{a} = k^2 \Rightarrow b^2 - 4ac = k^2a^2\)

69. Let \(\alpha\) be one of the roots of the equation \(ax^2 + bx + c = 0\)

    Clearly, \(\alpha^2\) will be the other root.

    Sum of roots, \(\alpha + \alpha^2 = -\frac{b}{a}\) and product of the roots \(\alpha^3 = \frac{c}{a}\)

    Cubing sum of roots,

    \(\frac{b^3}{a^3} = -\alpha^3(\alpha + 1)^3 = -\frac{c}{a}(\alpha^3 + 3\alpha(\alpha + 1) + 1)\)

    \(\frac{b^3}{a^3} = -\frac{c}{a}\left(\frac{c}{a} - \frac{3b}{a} + 1\right)\)

    Simplifying we get the desired relationship.

70. Let \(\alpha\) be one of the roots of the equation \(ax^2 + bx + c = 0\)

    Clearly, \(\alpha^2\) will be the other root.

    Sum of roots, \(\alpha + \alpha^2 = -p\) and product of roots \(\alpha^3 = 1\)

    Thus, \(\alpha\) is cube root of unity.

    If \(\alpha = -1\) then \(p = -2\)

    else if it is one of the complex numbers then we know that \(1 + \omega + \omega^2 = 0\) which makes \(p = 1\)

71. Let \(\alpha\) be one of the roots of the equation \(ax^2 + bx + c = 0\)

    Clearly, \(\alpha^2\) will be the other root.

    Sum of roots, \(\alpha + \alpha^2 = -p\) and product of roots \(\alpha^3 = q\)

    \(p^3 = -\alpha^3(\alpha + 1)^3 = -q(\alpha^3 + 3\alpha(\alpha + 1) + 1) = -q(q - 3p + 1)\)

    \(p^3 - q(3p - 1) + q^2 = 0\)

72. 1. \(\alpha + \beta = -\frac{3}{2}\) and \(\alpha\beta = \frac{4}{2} = 2\)

    \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{9}{4} - 4 = -\frac{7}{4}\)

    2. \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}\)

    Substituting for numerator from previous part,

    \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{7}{8}\)

73. Sum of roots, \(\alpha + \beta = -\frac{b}{a}\) and product of roots, \(\alpha\beta = \frac{c}{a}\)

    \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}\)

    \(= \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta}\)

    \(= \frac{-\frac{b^3}{c^3} + \frac{3c}{a}\frac{b}{a}}{\frac{c}{a}}\)

    \(= \frac{3abc - b^3}{a^2c}\)

74. Sum of roots, \(\alpha + \beta = -\frac{b}{a}\) and product of roots, \(\alpha\beta = \frac{b}{a}\)

    Given expression is, \(\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\frac{b}{a}}\)

    \(= \frac{\alpha + \beta}{\sqrt{\alpha\beta}} + \sqrt{\frac{b}{a}}\)

    \(= \frac{-\frac{b}{a}}{\sqrt{\frac{b}{a}}} + \sqrt{\frac{b}{a}}\)

    \(= 0\)

75. Product of the roots of the first equation, \(= b^2\) and sum of roots of the second equation, \(= 2b\)

    Geometric mean of the roots of the first equation \(=~\text{square of product of roots}~= \sqrt{b^2} = b\)

    Arithmetic mean of the roots of the second equation \(= \text{half of sum of roots}~= \frac{2b}{2} = b\) and thus both are equal.

76. Let \(\alpha, \beta\) be the roots of the equation.

    Sum of roots, \(\alpha + \beta = -\frac{q}{p}\) and product of roots, \(\alpha\beta = \frac{r}{p}\)

    Given, sum of roots is equal to sum of square of roots.

    \(\therefore \alpha + \beta = \alpha^2 + \beta^2\)

    \(-\frac{q}{p} = (\alpha + \beta)^2 - 2\alpha\beta = \frac{q^2}{p^2} - \frac{2r}{p}\)

    \(2pr = pq + q^2\)

77. Let \(\alpha, \beta\) be the roots of the equation.

    Sum of roots, \(\alpha + \beta = p\) and product of roots, \(\alpha\beta = q\)

    \(\frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{\alpha^4 + \beta^4}{(\alpha\beta)^2}\)

    \(= \frac{(\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2}{\alpha^2\beta^2}\)

    \(= \frac{[(\alpha + \beta)^2 - 2\alpha\beta]^2}{\alpha^2\beta^2} - 2\)

    \(= \frac{(p^2 - 2q)^2}{q^2} - 2\)

    \(= \frac{p^4}{q^2} - \frac{4p^2}{q} + 2\)

78. Let \(\alpha, \beta\) be the roots of the equation.

    Sum of roots, \(\alpha + \beta = -\frac{b}{a}\) and product of roots, \(\alpha\beta = \frac{c}{a}\)

    \(\frac{1}{(a\alpha + b)^2} + \frac{1}{(a\beta + b)^2} = \frac{(a\alpha + b)^2 + (a\beta + b)^2}{[(a\alpha + b)(a\beta + b)]^2}\)

    \(\frac{a(\alpha^2 + \beta^2) + 2ab(\alpha + \beta) + 2b^2}{(a^2\alpha\beta + 2ab(\alpha + \beta) + b^2)^2}\)

    Substituting for sum of roots, product of roots and \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) and simplifying

    \(= \frac{b^2 - 2ac}{c^2a^2}\)

79. Rewriting the equation we have \(\lambda x^2 + x(1 - \lambda) + 5 = 0\)

    Since \(\alpha\) and \(\beta\) are the roots therefore, we have \(\alpha + \beta = \frac{\lambda - 1}{\lambda}\) and \(\alpha\beta = \frac{5}{\lambda}\)

    Given, \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{5}\)

    \(\frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}\)

    \(\frac{(\lambda - 1)^2 - 10\lambda}{5\lambda} = \frac{4}{5}\)

    \((\lambda - 1)^2 - 10\lambda = 4\lambda \Rightarrow \lambda^2 - 16\lambda + 1 = 0\)

    \(\therefore \lambda_1 + \lambda_2 = 16\) and \(\lambda_1\lambda_2 = 1\)

    1. \(\frac{\lambda_1}{\lambda_2} + \frac{\lambda_2}{\lambda_1} = \frac{(\lambda_1 + \lambda_2)^2 - 2\lambda_1\lambda_2}{\lambda_1\lambda_2}\)

       Substituting the values for sum and product we have, result as \(254\)

    2. This is similar to previous part and has been left as an exercise.

80. For the first equation \(\alpha + \beta = -p\) and \(\alpha\beta = q\) and similarly for the second \(\gamma + \delta = -r\) and \(\gamma\delta = s\)

    1. \((\alpha + \gamma)(\alpha + \delta)(\beta + \gamma)(\beta + \delta)\)

       \(= [\alpha^2 + \alpha(\gamma + \delta) + \gamma\delta][\beta^2 + \beta(\gamma + \delta) + \gamma\delta]\)

       \(= (\alpha^2 - r\alpha + s)(\beta^2 - r\beta + s)\)

       \(= (\alpha^2\beta^2 - r\alpha\beta^2 + s\beta^2 - r\alpha^2\beta - r^2\alpha\beta - rs\beta + s\alpha^2 - rs\alpha + s^2)\)

       \(= q^2 - r\alpha\beta(\alpha + \beta) + s(\alpha^2 + \beta^2) + r^2p - rs(\alpha + \beta) + s^2\)

       \(= q^2 + prs + s(p^2 - 2q) + r^2p - rsq + s^2\)

    2. and iii. are similar in nature and has been left as an exercise.

81. \(\alpha + \beta = p\) and \(\alpha\beta = q\)

    Now, \(R. H. S. = (\alpha + \beta)(\alpha^n + \beta^n) - \alpha\beta(\alpha^{n - 1} + \beta^{n - 1})\)

    \(= \alpha^{n + 1} + \beta^{n + 1} = L. H. S.\)

82. This is similar to 80 and has been left as an exercise.

83. Clearly, \(\alpha + \beta = 2p, \alpha\beta = q\) and \(\gamma + \delta = 2r, \gamma\delta = s\) i. \(\frac{\alpha}{\beta} = \frac{\gamma}{\delta}\)

    By componendo and dividendo

    \(\frac{\alpha + \beta}{\alpha - \beta} = \frac{\gamma + \delta}{\gamma - \delta}\)

    Squaring,

    \(\left(\frac{\alpha + \beta}{\alpha - \beta}\right)^2 = \left(\frac{\gamma + \delta}{\gamma - \delta}\right)^2\)

    \(1 - \frac{4\alpha\beta}{(\alpha + \beta)^2} = 1 - \frac{4\gamma\delta}{(\gamma + \delta)^2}\)

    \(\frac{q}{p^2} = \frac{s}{r^2}\)

    2. Since \(\alpha, \beta, \gamma, \delta\) are in G. P. Hence, \(\frac{\alpha}{\beta} = \frac{\gamma}{\delta}\) and then we can proceed like previous part.

    3. Since \(\alpha, \beta, \gamma, \delta\) are in A. P. Hence, \(\alpha - \beta = \gamma - \delta\)

       \((\alpha + \beta)^2 - 4\alpha\beta = (\gamma + \delta)^2 - 4\gamma\delta\)

       \(4p^2 - 4q = 4r^2 - 4s \Rightarrow s - q = r^2 - p^2\)

84. Clearly, \(\alpha + \beta = -\frac{2b}{a}\) and \(\alpha\beta = \frac{c}{a}\) for \(ax^2 + 2bx + c = 0\)

    and \(\alpha + \beta + 2k = -\frac{2B}{A}\) and \((\alpha + k)(\beta + k) = \frac{c}{a}\)

    Given expression can be rewritten as

    \(\frac{b^2}{a^2} - \frac{c}{a} = \frac{B^2}{A^2} - \frac{C}{A}\)

    \(\frac{(\alpha + \beta)^2}{4} - \alpha\beta = \frac{(\alpha + \beta + 2k)^2}{4} - (\alpha + k)(\beta + k)\)

    \((\alpha - \beta)^2 = (\alpha + k - \beta - k)^2\)

85. This problem is similar to 84 and has been left as an exercise to the reader.

86. Let \(\alpha, \beta\) be the roots of \(x^2 + 2px + q = 0\) and \(\gamma, \delta\) be the roots of \(x^2 + 2qx + p = 0\)

    \(\alpha + \beta = -2p\) and \(\gamma + \delta = -2q.\) Also, \(\alpha\beta = q\) and \(\gamma\delta = p\)

    Given that roots differ by a constant term say \(k\). \(\therefore \alpha + k = \gamma\) and \(\beta + k = \delta\)

    Thus, \(\alpha + \beta + 2k = -2q \Rightarrow -2p + 2k = -2q \Rightarrow k = p - q\)

    \(\gamma\delta = \alpha\beta + (\alpha + \beta)k + k^2 = p\)

    \(q - 2pk + k^2 = p \Rightarrow -2p + k = 1 \Rightarrow p + q + 1 = 0\)

87. Clearly, \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\)

    1. Sum of these roots \(= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}\)

       \(= \frac{b^2 - 2ac}{ac}\)

       Product of these roots \(= 1\)

       Therefore, such an equation is \(x^2 -\frac{b^2 - 2ac}{ac}x + 1 = 0\)

    2. Sum of these roots \(= \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta} = \frac{3abc - b^3}{a^2c}\)

       Product of these roots \(= \alpha\beta = \frac{c}{a}\)

       Therefore, an equation whose roots were these is \(x^2 - \frac{3abc - b^3}{a^2c}x + \frac{c}{a} = 0\)

    Rest of the parts are left as an exercise to the reader. 88, 89, 90 are also similar to 87 and have been left as an exercise.

91. Since complex roots appear in pair, if one of the roots is \(5 + 3i\) the other root would be \(5 - 3i\)

    \(\therefore p = -10\) and \(q = 34\)

92. Conjugate complex root is \(3 - 4i\). Sum of roots \(= 6\) and product of roots \(= 25\)

    Therefore, equation whose roots are \(3 \pm 4i\) is \(x^2 - 6x + 25 = 0\)

93. Discriminant \(D = 20\) and thus roots will be irrational pair. One of the roots \(\alpha = \frac{-2 + 2\sqrt{5}}{8}\)

    \(\alpha = \frac{-1 + \sqrt{5}}{4}\)

    Now it is trivial to prove that the other root is \(4\alpha^3 - 3\alpha\).

94. Equation whose roots are \(\alpha\) and \(\beta\) is \(x^2 - 5x + 3 = 0\) as it is satisfied by both of them.

    \(\therefore \alpha + \beta = 5\) and \(\alpha\beta = 3\)

    \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{19}{3}\)

    \(\frac{\alpha}{\beta} . \frac{\beta}{\alpha} = 1\)

    Therefore, the desired equation is \(x^2 - \frac{19}{3}x + 1 = 0\)

    Rest of the problems are left as exercises.