46. Quadratic Equations Solutions Part 4ΒΆ

  1. Let \(y = \frac{x}{x^2 - 5x + 9}\)

    \(\Rightarrow yx^2 - (5y + 1)x + 9y = 0\)

    Since \(x\) is real, discriminant of above equation has to be greater or equal to zero.

    \(\Rightarrow (5y + 1)^2 - 36y^2 \ge 0\)

    \(\Rightarrow -11y^2 - 10y + 1 \ge 0\)

    \(\Rightarrow -11y^2 - 11y + y + 1 \ge 0\)

    Above is satisfied when \(y\) lie between \(1\) and \(-\frac{1}{11}\).

  2. Let \(y = \frac{x^2 - 2x + p^2}{x^2 + 2x + p^2}\)

    \(\Rightarrow (y - 1)x^2 + 2(y + 1)x + (y - 1)p^2 = 0\)

    Since \(x\) is real, discriminant of above equation has to be greater or equal to zero.

    \(\Rightarrow 4(y + 1)^2 - 4p^2(y - 1)^2 \ge 0\)

    \(\Rightarrow (1 - p^2)y^2 + 2(1 + p^2)y + 1 - p^2 \ge 0\)

    Since \(p > 1\) coefficient of \(y^2\) is negative and thus \(y\) must lie between its roots for the above to be true.

    The roots are \(y = \frac{-2(1 + p^2) \pm \sqrt{4(1 + p^2)^2 - 4(1 - p^2)^2}}{2(1 - p^2)}\)

    \(y = \frac{p - 1}{p + 1}, \frac{p + 1}{p - 1}\)

  3. Let \(y = \frac{x^2 + 2x + 1}{x^2 + 2x + 7}\)

    \(\Rightarrow (y - 1)x^2 + 2(y - 1)x + 7y - 1 = 0\)

    Since \(x\) is real, following has to be true

    \(4(y - 1)^2 - 4(y - 1)(7y - 1) \ge 0\)

    \(y^2 - 2y + 1 - 7y^2 + 8y - 1 \ge 0\)

    \(-6y^2 + 6y \ge 0\)

    Therefore, \(y\) must lie between \([0, 1]\).

  4. Let \(y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}\)

    \(\Rightarrow (y - 1)x^2 + 2(y - 7)x + 3(y - 3) = 0\)

    Since \(x\) is real, following must hold true

    \(4(y - 7)^2 - 12(y - 1)(y - 3) \ge 0\)

    \(y^2 - 14y + 49 - 3y^2 + 12y - 9 \ge 0\)

    \(-2y^2 - 2y + 40 \ge 0\)

    \(y^2 + y - 20 \le 0\)

    Therefore, least value is \(-5\) and greatest value is \(4\).

  5. Let \(y = \frac{x^2 -x + 1}{x^2 + x + 1}\)

    \(\Rightarrow (y - 1)x^2 + (y + 1)x + y - 1 = 0\)

    Since \(x\) is real, following must hold true

    \((y + 1)^2 - 4(y - 1)^2 \ge 0\)

    \(y^2 + 2y + 1 - 4y^2 + 8y - 4 \ge 0\)

    \(-3y^2 + 10y - 3 \ge 0\)

    \(3y^2 - 10y + 3 \le 0\)

    Therefore, least value if \(\frac{1}{3}\) and greatest value is \(3\).

  6. Let \(y = \frac{x^2 - 3x - 3}{2x^2 + 2x + 1}\)

    \(\Rightarrow (2y - 1)x^2 + (2y + 3)x + y + 3 = 0\)

    Since \(x\) is real, following must hold true

    \((2y + 3)^2 - 4(2y - 1)(y + 3) \ge 0\)

    \(4y^2 + 12y + 9 - 8y^2 - 20y + 12 \ge 0\)

    \(-4y^2 - 8y + 21 \ge 0\)

    \(4y^2 + 8y - 21 \le 0\)

    Therefore, least value is \(-\frac{7}{2}\) and greatest value is \(\frac{3}{2}\).

  7. Let \(y = \frac{2x^2 + x - 1}{x^2 + 4x + 2}\)

    \(\Rightarrow (y - 2)x^2 + (4y - 1)x + 2y + 1 = 0\)

    Since \(x\) is real, following must hold true

    \((4y - 1)^2 - 4(y - 2)(2y + 1) \ge 0\)

    \(16y^2 - 8y + 1 - 8y^2 + 12y + 8 \ge 0\)

    \(8y^2 + 4y + 9 \ge 0\)

    Since discriminant of corresponding equation is negative \(y\) is capable of having any real value.

  8. Let \(y = \frac{x^2 - 4x + 9}{x^2 + 4x + 9}\)

    \(\Rightarrow (y - 1)x^2 + 4(y + 1)x + 9(y - 1) = 0\)

    Since \(x\) is real, following must hold true

    \(16(y + 1)^2 - 36(y - 1)^2 \ge 0\)

    \(4y^2 + 8y + 4 - 9y^2 + 18y - 9 \ge 0\)

    \(-5y^2 + 26y - 5 \ge 0\)

    Therefore, least value is \(\frac{1}{5}\) and greatest value is \(5\).

  9. Let \(y = \frac{x^2 + 7x + 16}{x^2 - 5x + 16}\)

    \(\Rightarrow (y - 1)x^2 - (5y + 7)x + 16(y - 1) = 0\)

    Since \(x\) is real, following must hold true

    \((5y + 7)^2 - 64(y - 1)^2 \ge 0\)

    \(25y^2 + 70y + 49 - 64y^2 + 128y - 64 \ge 0\)

    \(-39y^2 + 198y - 15 \ge 0\)

    \(13y^2 - 66y + 5 \le 0\)

    Therefore, least value is \(\frac{1}{13}\) and greatest value is \(5\).

  10. Let \(y = \frac{6x^2 - 2x + 3}{2x^2 - 2x + 1}\)

    \(\Rightarrow 2(y - 3)x^2 - 2(y - 1)x + y - 3 = 0\)

    Since \(x\) is real, following must hold true

    \(4(y - 1)^2 - 8(y - 3)^2 \ge 0\)

    \(y^2 - 2y + 1 - 2y^2 + 12y - 18 \ge 0\)

    \(y^2 - 10y + 17 \le 0\)

    Therefore, least value is \(5 - 2\sqrt{2}\) and greatest value is \(5 + 2\sqrt{2}\).

  11. Let \(y = \frac{(x - 1)(x + 3)}{(x - 2)(x + 4)}\)

    \(y = \frac{x^2 + 2x - 3}{x^2 + 2x - 8}\)

    \(\Rightarrow (y - 1)x^2 + 2(y - 1)x^2 + 3 - 8y = 0\)

    Since \(x\) is real, following must hold true

    \(4(y - 1)^2 + 4(y - 1)(8y - 3) \ge 0\)

    \(y^2 - 2y + 1 + 8y^2 - 11y + 3 \ge 0\)

    \(9y^2 - 13y + 4 \ge 0\)

    For above to be true \(y\) must not lie between \(1\) and \(\frac{4}{9}\).

  12. Let \(y = \frac{2x^2 - 2x + 4}{x^2 - 4x + 3}\)

    \((y - 2)x^2 - 2(2y - 1)x + 3y - 4 = 0\)

    Since \(x\) is real, following must hold true

    \(4(2y - 1)^2 - 4(y - 2)(3y -4) \ge 0\)

    \(\Rightarrow 4y^2 - 4y + 1 - 3y^2 + 10y - 8 \ge 0\)

    \(\Rightarrow y^2 + 6y - 7 \ge 0\)

    For above to be true \(y\) cannot lie between \(1\) and \(-7\).

  13. Let \(y = \frac{x^2 + 2x - 11}{-x - 3}\)

    \(\Rightarrow x^2 + (y + 2)x + 3y - 11 = 0\)

    Since \(x\) is real, following must hold true

    \((y + 2)^2 - 12y + 44 \ge 0\)

    \(y^2 - 8y + 48 \ge 0\)

    For the above to be true \(y\) cannot lie between \(4\) and \(12\).

  14. Let \(y = \frac{x}{x^2 + 1}\)

    \(\Rightarrow yx^2 - x + y = 0\)

    Since \(x\) is real, following must hold true

    \(1 - 4y^2 \ge 0\)

    \(y \le \frac{1}{2}\)

  15. Let \(y = \frac{x + a}{x^2 + bx + c^2}\)

    \(\Rightarrow yx^2 + (by - 1)x - a + c^2y = 0\)

    Since \(x\) is real, following must hold true

    \((by - 1)^2 - 4y(c^2y - a) \ge 0\)

    \(b^2y^2 - 2by + 1 + 4ay - 4c^2y^2 \ge 0\)

    \((b^2 - 4c^2)y^2 + 2(2a - b)y + 1 \ge 0\)

    Discriminant of corresponding equation is \(D = 4(2a - b)^2 - 4(b^2 - 4c^2)\)

    \(= 4[4a^2 + b^2 - 4ab - b^2 + 4c^2] = 16(a^2 + c^2 - ab)\)

    Given \(b^2 > 4c^2\) and \(a^2 + c^2 > ab\) therefore \(D < 0\) and coefficient of \(y^2\) is negative. Therefore, \(y\) is capable of assuming any value.

  16. Let \(y = \frac{x^2 - bc}{2x - b - c}\)

    \(\Rightarrow x^2 - 2yx + (b + c)y - bc = 0\)

    Since \(x\) is real, following must hold true

    \(4y^2 - 4(b + c)y + 4bc \ge 0\)

    \(y^2 - (b + c)y + bc \ge 0\)

    For above to be true \(y\) must not lie between \(b\) and \(c\).

  17. Given expression is \(\frac{1}{x + 1} + \frac{1}{3x + 1} - \frac{1}{(x + 1)(3x + 1)}\)

    \(= \frac{4x + 1}{3x^2 + 4x + 1}\)

    Let \(y = \frac{4x + 1}{3x^2 + 4x + 1}\)

    \(\Rightarrow 3yx^2 + 4(y - 1)x + y - 1 = 0\)

    Since \(x\) is real, following must hold true

    \(16(y - 1)^2 - 12y(y - 1) \ge 0\)

    \(\Rightarrow 4y^2 - 8y + 4 - 3y^2 + 3y \ge 0\)

    \(\Rightarrow y^2 - 5y + 4 \ge 0\)

    Above is true provided \(y\) does not lie between \(1\) and \(4\).

  18. Let \(y = \frac{2x^2 + x - 3}{3x + 1}\)

    \(\Rightarrow 2x^2 + (1 - 3y)x - (y + 3) = 0\)

    Since \(x\) is real, following must hold true

    \((1 - 3y)^2 - 8(y + 3) \ge 0\)

    \(1 - 6y + 9y^2 - 8y - 24 \ge 0\)

    \(9y^2 - 14y - 23 \ge 0\)

    Discriminant of corresponding equation is negative and coefficient of \(y^2\) is positive therefore \(y\) is capable of assuming any real value.

  19. Let \(y = \frac{2x^2 + 4x + 1}{x^2 + 4x + 2}\)

    \((y - 2)x^2 + 4(y - 1)x + 2y - 1 = 0\)

    Since \(x\) is real, following must hold true

    \(16(y - 1)^2 - 4(y - 2)(2y - 1) \ge 0\)

    \(4y^2 - 8y + 4 - 2y^2 + 5y - 2 \ge 0\)

    \(2y^2 - 3y + 2 \ge 0\)

    Discriminant of corresponding equation is negative and coefficient of \(y^2\) is positive therefore \(y\) is capable of assuming any real value.

  20. Let \(y = \frac{ax^2 + 3x - 4}{3x - 4x^2 + a}\)

    \(\Rightarrow (a + 4y)x^2 + 3(1 - y)x - (4 + ay) = 0\)

    Since \(x\) is real, discriminant is greater than or equal to zero

    \(\Rightarrow 9(1 - y)^2 + 4(a + 4y)(4 + ay) \ge 0\)

    \(\Rightarrow 9y^2 - 18y + 9 + 16a + 64y + 4a^2y + 16ay^2 \ge 0\)

    \(\Rightarrow (9 + 16a)y^2 + (46 + 4a^2)y + 9 + 16a \ge 0\)

    So for \(y\) to assume any real value \(9 + 16a > 0\) and \(D < 0\)

    \(\Rightarrow 4(23 + 2a^2)^2 - (9 + 16a)^2 < 0\)

    \((a + 4)(a - 1)(a - 7) < 0\)

    But since \(a > -\frac{9}{16}\) therefore \(a\) must lie between \(1\) and \(7\).

  21. Let \(y = \frac{m^2}{1 + x} - \frac{n^2}{1 - x}\)

    \(y = \frac{m^2 - n^2 - (m^2 + n^2)x }{1 - x^2}\)

    \(yx^2 - (m^2 + n^2)x + m^2 - n^2 - y = 0\)

    Since \(x\) is real, discriminant is greater than or equal to zero

    \((m^2 + n^2)^2 - 4y(m^2 - n^2 - y) \ge 0\)

    \(4y^2 - 4(m^2 - n^2)y + (m^2 + n^2)^2 \ge 0\)

    For \(y\) to be able to assume to any value discriminant of corresponding equation has to be negative since coefficient of \(y^2\) is positive.

    \(16(m^2 - n^2)^2 - 16(m^2 + n^2)^2 < 0\)

    which is true.

  22. Let \(y = \frac{4x}{x^2 + 16}\)

    \(\Rightarrow yx^2 - 4x + 16y = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \(16 - 64y^2 \ge 0\)

    \(y^2 le \frac{1}{4}\)

    \(-\frac{1}{2} \le y \le \frac{1}{2}\)

    \(\left|\frac{4x}{x^2 + 16}\right| < \frac{1}{2}\)

  23. Given \(x^2 - xy + y^2 - 4x - 4y + 16 = 0\)

    \(x^2 - (y + 4)x + y^2 - 4y + 16 = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \((y + 4)^2 - 4(y^2 - 4y + 16) \ge 0\)

    \(y^2 + 8y + 16 - 4y^2 + 16y - 64 \ge 0\)

    \(-3y^2 + 24y - 48 \ge 0\)

    \(y^2 - 8y + 16 \le 0\)

    \((y - 4)^2 \le 0\)

    The above inequality is only satisfied by \(y = 4\)

    However, if \(y = 4\) the given equation becomes

    \(x^2 - 8x + 16 = 0\) which is again only satisfied by \(x = 4\)

  24. Given \(x^2 + 12xy + 4y^2 + 4x + 8y + 20 = 0\)

    \(x^2 + 4(1 + 3y)x + 4(y^2 + 2y + 5) = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \(16(1 + 3y)^2 - 16(y^2 + 2y + 5) \ge 0\)

    \(1 + 6y + 9y^2 - y^2 - 2y - 5 \ge 0\)

    \(8y^2 + 4y - 4 \ge 0\)

    \(2y^2 + y - 1 \ge 0 \Rightarrow (2y - 1)(y + 1) \ge 0\)

    Therefore, \(y\) cannot lie between \(-1\) and \(\frac{1}{2}\).

    Rewriting the equation in terms of \(y\)

    \(4y^2 + 4(3x + 2)y + x^2 + 4x + 20 = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \((3x + 2)^2 - x^2 - 4x - 20 \ge 0\)

    \(8x^2 + 8x - 16 \ge 0 \Rightarrow x^2 + x - 2 \ge 0\)

    Therefore, \(x\) cannot lie between \(-2\) and \(1\).

  25. Corresponding equation is \(x^2 - 5mx + 4m^2 + 1 = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \(\Rightarrow 25m^2 - 16m^2 - 4 \ge 0\)

    \(9m^2 - 4 \ge 0\)

    \(-\frac{2}{3} \le m \le \frac{2}{3}\)

  26. Let \(y = -3x^2 + x + 2\)

    \(3x^2 - x - 2 + y = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \(1 - 12(-2 + y) \ge 0\)

    \(25 - 12y \ge 0\)

    \(y \le \frac{25}{12}\)

  27. Let the positive number be \(x\) and \(y = x + \frac{1}{x}\)

    \(x^2 - yx + 1 = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \(y^2 - 4 \ge 0\)

    Therefore, least value of \(y\) is \(2\) as \(y\) cannot be negative since \(x\) is positive.

  28. Let \(x\) be the length and \(y\) be the breadth then \(x + 2y = 600\) and we have to maximize \(xy\)

    \(xy = x\frac{600 - x}{2} = z\) (say)

    \(x^2 - 600x + 2z = 0\)

    Since \(x\) is real, discriminant has to be greater than or equal to zero.

    \(360000 - 8z \ge 0\)

    \(z \le 45000\)

    Thus, maximum area is \(45000\) mt. sq.

    Substituting

    \(x^2 - 600x + 90000 = 0\)

    \((x - 300)^2 = 0 \Rightarrow x = 300 \Rightarrow y = 150\)

  29. If \(y - mx\) is a factor then equation reduces to \(bm^2 + 2hm + a = 0\)

    and if \(my + x\) is a factor then it reduces to \(am^2 - 2hm + b = 0\)

    By cross-multiplication we have

    \(\frac{m^2}{-2h(a + b)} = \frac{m}{a^2 - b^2} = \frac{1}{2h(a + b)}\)

    Thus, condition becomes \(a + b = 0\) or \(4h^2 + (a^2 - b^2) = 0\)

  30. Expression \((4 - k)x^2 + 2(k + 2)x + 8k + 1\) will be perfect square if discriminant of corresponding equation will be equal to zero.

    \(\Rightarrow 4(k + 2)^2 - 4(4 - k)(8k + 1) = 0\)

    \(4k^2 + 16k + 16 - 128k + 4k - 16 + 32k^2 = 0\)

    \(36k^2 - 108k = 0\)

    \(\Rightarrow k = 0, 3\)

  31. \(3x^2 - xy - 2y^2 + mx + y + 1\) is resolvable into two linear factors if its discriminant is a perfect square.

    Rewriting the expression \(3x^2 + (m - y)x - 2y^2 + y + 1\)

    The discriminant, \((m - y)^2 - 12(- 2y^2 + y + 1)\) has to be a perfect square.

    \(y^2 - 2my + m^2 + 24y^2 - 12y - 12\) has to be a perfect square which means corresponding discriminant has to be equal to be zero.

    Solving for that we arrive at the values for \(m = 4, -\frac{7}{2}\)

  32. Proceeding as previous problem we arrive at the solution as \(7 , \frac{98}{3}\)

  33. If \(x - \alpha\) is a factor then it has to satisfy both the equations i.e.

    \(a_1\alpha^2 + b_1\alpha + c_1 = 0\) and \(a_2\alpha^2 + b_2\alpha + c_1 = 0\)

    By cross-multiplication we have

    \(\frac{\alpha^2}{b_1c_1 - b_2c_1} = \frac{\alpha}{a_2c_1 - a_1c_1} = \frac{1}{a_1b_2 - a_2b_1}\)

    From first two we have the required condition \(\alpha(a_1 - a_2) = b_2 - b_1\)

  34. Corresponding equation is \(6x^2 + 7xy + 2y^2 + 11x + 7y + 3 = 0\)

    \(\Rightarrow 6x^2 + (7y + 11)x + 2y^2 + 7y + 3 = 0\)

    \(x = \frac{-7y - 11 \pm\sqrt{(7y + 11)^2 - 24(2y^2 + 7y + 3)}}{12}\)

    \(x = \frac{-7y - 11 \pm\sqrt{49y^2 + 154y + 121 - 48y^2 - 168y - 72}}{12}\)

    \(x = \frac{-7y - 11 \pm\sqrt{y^2 - 14y + 49}}{12}\)

    Therefore, factors are \(2x + y + 3\) and \(3x + 2y + 1\)

  35. Corresponding equation is \(x^2 - 5xy + 4y^2 + x + 2y -2 = 0\)

    \(\Rightarrow x^2 + (1 - 5y)x + 4y^2 + 2y - 2 = 0\)

    \(x = \frac{5y - 1 \pm \sqrt{(1 - 5y)^2 - 16y^2 - 8y + 8}}{2}\)

    \(x = \frac{5y - 1 \pm \sqrt{1 - 10y + 25y^2 - 16y^2 - 8y + 8}}{2}\)

    \(x = \frac{5y - 1 \pm \sqrt{9y^2 - 18y + 9}}{2}\)

    Therefore, factors are \(x - 4y + 2\) and \(x - y - 1\)

  36. Corresponding equation is \(2x^2 + 5xy - 3y^2 + x + 17y - 10 = 0\)

    \(\Rightarrow 2x^2 + (5y + 1)x - 3y^2 + 17y - 10 = 0\)

    \(x = \frac{-5y - 1 \pm \sqrt{25y^2 + 10y + 1 + 24y^2 - 136y + 80}}{4}\)

    \(x = \frac{-5y - 1 \pm \sqrt{(7y - 9)^2}}{4}\)

    Therefore, factors are \(2x - y + 5\) and \(x + 3y - 2\)

  37. Corresponding equation is \(3x^2 + 5xy - 2y^2 - 3x + 8y - 6 = 0\)

    Proceeding as previous problems factors can be found as \(3x - y + 3\) and \(x + 2y - 2\)

  38. We know that \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\) can be resolved into two linear factors if and only if

    \(abc + 2fgh - af^2 - bg^2 - ch^2 = 0\)

    Given expression is \(3x^2 + 2\alpha xy + 2y^2 + 2ax - 4y + 1\)

    Comparing we find coefficients as \(a = 3, h = \alpha, b = 2, g = a, f = -2, c = 1\)

    \(\therefore 6 - 4a\alpha - 12 - 2a^2 - \alpha^2 = 0\)

    Clearly, \(\alpha\) has to be a root of the equation \(x^2 + 4ax + 2a^2 + 6 = 0\)

  39. Let \(D_1\) be discriminant of the equation \(x^2 + px + q = 0\) and \(D_2\) be discriminant of the equation \(x^2 + rx + s = 0\)

    \(D_1 + D_2 = p^2 - 4q + r^2 - 4s > p^2 + r^2 - pr [\because 4(q + s) < pr]\)

    If both \(p\) and \(r\) are zero, then \(D_1 + D_2 > 0\)

    In case one of \(p\) or \(r\) is non-zero then

    \(D_1 + D_2 > r^2\left\{\left(\frac{p}{r}\right)^2 - \frac{p}{r} + 1\right\}\) if \(r \ne 0\)

    \(D_1 + D_2 > p^2\left\{\left(\frac{r}{p}\right)^2 - \frac{r}{p} + 1\right\}\) if \(p \ne 0\)

    Since for real \(x, x^2 - x + 1 > 0\) as corresponding equation has imaginary roots.

    Thus, in all cases \(D_1 + D_2 > 0\) therefore at least one of them is greater than zero i.e. roots of at least one of the given equations are real.

  40. Roots of equation \(P(x)Q(x) = 0\) will be the roots of equation \(P(x) = 0\) i.e. \(ax^2 + bx + c = 0\) and \(Q(x) = -ax^2 + bx + c = 0\)

    Let \(D_1\) and \(D_2\) be the discriminants of two equations, then

    \(D_1 + D_2 = b^2 - 4ax + b^2 + 4ac = 2b^2 > 0\)

    Hence, at least one of \(D_1\) and \(D_2\) will be zero. Hence, \(P(x)Q(x) = 0\) has at least two real roots.

  41. Let \(D_1\) be the discriminant of \(bx^2 + (b - c)x + b - c - a = 0\) and \(D_2\) be discriminant of \(ax^2 + 2bx + b = 0\), then

    \(D_1 + D_2 = (b - c)^2 - 4b(b - c - a) + 4b^2 - 4ab = (b + c)^2 \ge 0\)

    Hence, if \(D_2 < 0\), then \(D_1 > 0\).

    Therefore, roots of \(bx^2 + (b - c)x + b - c - a = 0\) will be real if roots of \(ax^2 + 2bx + b = 0\) are imaginary and vice versa.

  42. Let \(a = 2m + 1, b = 2n + 1, c = 2r + 1\)

    Now \(D = (2n + 1)^2 - 4(2m + 1)(2r + 1)\)

    \(= (\text{an odd number}) - (\text{an even number}) =\) an odd number

    If possible let \(D\) be a perfect square then it has to be square of an odd number.

    \((2k + 1)^2 = (2n + 1)^2 - 4(2m + 1)(2r + 1)\)

    \((2m + 1)(2r + 1) = (n + k + 1)(n - k)\)

    If \(n\) and \(k\) are both odd or even then \(n - k\) will be even or zero. However, if one is odd and one is even then \((n + k + 1)\) will be even. So, R. H. S. is an even while L. H. S. is an odd number. Thus, \(D\) cannot be a perfect square. Hence, roots cannot be a rational numbers.

  43. Let \(D_1\) be discriminant of \(ax^2 + 2bx + c = 0\) then \(D_1 = 4b^2 - 4ac = 4k,\) where \(k = b^2 - ac\)

    Let \(D_2\) is discriminant of \((a + c)(ax^2 + 2bx + c) = 2(ac - b^2)(x^2 + 1)\)

    \(D_2 = 4(a + c)^2b^2 - 4(a^2 + b^2 + k)(b^2 + c^2 + k)\)

    \(= -D1[4b^2 + (a - c)^2] \Rightarrow D_2 < 0 \because D_1 > 0\)

    Therefore, roots of second equation are non-real complex numbers.

  44. \(D = 4[(^nC_r)^2 - ^nC_{r - 1}^nC_{r + 1}]\)

    \(= 4(a - b),\) where \(a = ^(nC_r)^2, b = ^nC_{r - 1}^nC_{r + 1}\)

    \(\frac{a}{b} = \left(1 + \frac{1}{r}\right)\left(1 + \frac{1}{n - r}\right) > 1\)

    \(a > b \Rightarrow D > 0\)

    Thus, roots of given equation are real and distinct.

  45. Let \(D\) be the discriminant of the given equation.

    \(D = (2m - 1)^2 - 4m(m - 2)\)

    \(= 4m + 1 =\) an odd number.

    For roots to be rational discriminant must be a perfect square.

    \((2k + 1)^2 = 4m + 1,\) where \(k \in I\)

    \(m = k(k + 1)\)

  46. Let \(y = e^{\sin x}\) then given equation becomes

    \(y - \frac{1}{y} - 4 = 0\)

    \(y = 2\pm \sqrt{5} \therefore e^{\sin x} = 2 \pm \sqrt{5}\)

    \(\sin x = \log_e (2 - \sqrt{5})\) is not defined.

    \(\sin x = \log_e (2 + \sqrt{5}) > 1\) is not possible.

    Hence, roots of given equation cannot be real.

  47. Given equation is \(az^2 + bz + c + i = 0\)

    \(z = \frac{-b \pm \sqrt{b^2 - 4a(c + i)}}{2a} = \frac{-b \pm(p + iq)}{2a}\)

    where \(\sqrt{b^2 - 4a(c + i)} = p + iq\)

    \(b^2 - 4ac = p^2 - q^2\) and \(-4a = 2qp\)

    Since \(z\) is purely imaginary \(\frac{-b \pm p}{2a} = 0 \Rightarrow \pm p = b\)

    \(-4a = 2(\pm)q \Rightarrow q = \pm \frac{2a}{b}\)

    Then, \(b^2 - 4ac = b^2 - \frac{4a^2}{b^2}\)

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

  48. \(D = a^2 - 4b\)

    Let \(a\) be an odd number then \(D\) is an odd number and a perfect square as roots are rational. Let \(D = (2n + 1)^2,\) and \(a = 2m + 1\) where \(m, n \in I\)

    Now roots \(= \frac{-(2m + 1)\pm (2n + 1)}{2} = \frac{\text{an even no.}}{2} = \text{an integer}\)

    Similarly, it can be proven when \(a\) is an even no. then roots are integers.

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

    \(\alpha + \beta = -7\) and \(\alpha\beta = 14(q^2 + 1)\)

    \(\frac{\alpha\beta}{7} = 2(q^2 + 1) = \text{an integer}\)

    \(\therefore \alpha\beta\) is divisible by \(7\) and \(7\) is a prime number.

    \(\therefore\) at least one of \(\alpha\) and \(\beta\) must be a multiple of \(7\).

    Let \(\alpha = 7k,\) where \(k \in I\)

    \(\beta = -7(k + 1)\)

    Thus, \(-\frac{2(q^2 + 1)}{7} = k(k + 1) = \text{an integer}\)

    Let \(f(q) = q^2 + 1\) then it can be shown that \(f(1), f(2), ..., f(7)\) are not divisible by \(7\).

    \(f(q + 7) = q^2 + 1 + 14q + 49\) which is not divisible by \(7\) as \(q^2 + 1\) is not divisible by \(7\).

    Hence, \(\alpha, \beta\) cannot be integers.

  50. Given equation is \([a^3(b - c) + b^3(c - a) + c^3(a - b)]x^2 - [a^3(b^2 - c^2) + b^3(c^2 - a^2) + c^3(a^2 - b^2)]x + abc[a^2(b - c) + b^2(c - a) + c^2(a - b)] = 0\)

    But \(a^3(b - c) + b^3(c - a) + c^3(a - b) = -(a - b)(b - c)(c - a)(a + b + c)\) and

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

    \(a^2(b - c) + b^2(c - a) + c^2(a - b) = -(a - b)(b - c)(c - a)\) the above equation becomes

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

    Roots are \(\frac{(ab + bc + ca \pm \sqrt{(ab + bc + ca)^2 - 4abc(a + b + c)})}{2(a + b + c)}\)

    Roots will be equal if \(D = 0\)

    If \(\frac{1}{\sqrt{a}}\pm \frac{1}{\sqrt{b}}\pm \frac{1}{c} = 0\)

    \(\frac{\sqrt{bc} \pm \sqrt{ca} \pm \sqrt{ab}}{\sqrt{abc}} = 0\)

    \(\Rightarrow \sqrt{bc} \pm \sqrt{ca} \pm \sqrt{ab} = 0\)

    Squaring

    \(bc + ca + ab \pm 2\sqrt{abc}(\sqrt{a}\pm \sqrt{b} \pm \sqrt{c}) = 0\)

    \((bc + ca + ab)^2 = 4abc(a + b + c + \sqrt{bc} \pm \sqrt{ca} \pm \sqrt{ab})\)

    \(\Rightarrow D = 0\) i.e. roots are equal.