.. meta:: :author: Shiv Shankar Dayal :title: Quadratic Equations Solutions Part 4 :description: Quadratic Equations Solutions Part 4 :keywords: quadratic equations, algebra Quadratic Equations Solutions Part 4 ************************************ 151. Let :math:`y = \frac{x}{x^2 - 5x + 9}` :math:`\Rightarrow yx^2 - (5y + 1)x + 9y = 0` Since :math:`x` is real, discriminant of above equation has to be greater or equal to zero. :math:`\Rightarrow (5y + 1)^2 - 36y^2 \ge 0` :math:`\Rightarrow -11y^2 - 10y + 1 \ge 0` :math:`\Rightarrow -11y^2 - 11y + y + 1 \ge 0` Above is satisfied when :math:`y` lie between :math:`1` and :math:`-\frac{1}{11}`. 152. Let :math:`y = \frac{x^2 - 2x + p^2}{x^2 + 2x + p^2}` :math:`\Rightarrow (y - 1)x^2 + 2(y + 1)x + (y - 1)p^2 = 0` Since :math:`x` is real, discriminant of above equation has to be greater or equal to zero. :math:`\Rightarrow 4(y + 1)^2 - 4p^2(y - 1)^2 \ge 0` :math:`\Rightarrow (1 - p^2)y^2 + 2(1 + p^2)y + 1 - p^2 \ge 0` Since :math:`p > 1` coefficient of :math:`y^2` is negative and thus :math:`y` must lie between its roots for the above to be true. The roots are :math:`y = \frac{-2(1 + p^2) \pm \sqrt{4(1 + p^2)^2 - 4(1 - p^2)^2}}{2(1 - p^2)}` :math:`y = \frac{p - 1}{p + 1}, \frac{p + 1}{p - 1}` 153. Let :math:`y = \frac{x^2 + 2x + 1}{x^2 + 2x + 7}` :math:`\Rightarrow (y - 1)x^2 + 2(y - 1)x + 7y - 1 = 0` Since :math:`x` is real, following has to be true :math:`4(y - 1)^2 - 4(y - 1)(7y - 1) \ge 0` :math:`y^2 - 2y + 1 - 7y^2 + 8y - 1 \ge 0` :math:`-6y^2 + 6y \ge 0` Therefore, :math:`y` must lie between :math:`[0, 1]`. 154. Let :math:`y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}` :math:`\Rightarrow (y - 1)x^2 + 2(y - 7)x + 3(y - 3) = 0` Since :math:`x` is real, following must hold true :math:`4(y - 7)^2 - 12(y - 1)(y - 3) \ge 0` :math:`y^2 - 14y + 49 - 3y^2 + 12y - 9 \ge 0` :math:`-2y^2 - 2y + 40 \ge 0` :math:`y^2 + y - 20 \le 0` Therefore, least value is :math:`-5` and greatest value is :math:`4`. 155. Let :math:`y = \frac{x^2 -x + 1}{x^2 + x + 1}` :math:`\Rightarrow (y - 1)x^2 + (y + 1)x + y - 1 = 0` Since :math:`x` is real, following must hold true :math:`(y + 1)^2 - 4(y - 1)^2 \ge 0` :math:`y^2 + 2y + 1 - 4y^2 + 8y - 4 \ge 0` :math:`-3y^2 + 10y - 3 \ge 0` :math:`3y^2 - 10y + 3 \le 0` Therefore, least value if :math:`\frac{1}{3}` and greatest value is :math:`3`. 156. Let :math:`y = \frac{x^2 - 3x - 3}{2x^2 + 2x + 1}` :math:`\Rightarrow (2y - 1)x^2 + (2y + 3)x + y + 3 = 0` Since :math:`x` is real, following must hold true :math:`(2y + 3)^2 - 4(2y - 1)(y + 3) \ge 0` :math:`4y^2 + 12y + 9 - 8y^2 - 20y + 12 \ge 0` :math:`-4y^2 - 8y + 21 \ge 0` :math:`4y^2 + 8y - 21 \le 0` Therefore, least value is :math:`-\frac{7}{2}` and greatest value is :math:`\frac{3}{2}`. 157. Let :math:`y = \frac{2x^2 + x - 1}{x^2 + 4x + 2}` :math:`\Rightarrow (y - 2)x^2 + (4y - 1)x + 2y + 1 = 0` Since :math:`x` is real, following must hold true :math:`(4y - 1)^2 - 4(y - 2)(2y + 1) \ge 0` :math:`16y^2 - 8y + 1 - 8y^2 + 12y + 8 \ge 0` :math:`8y^2 + 4y + 9 \ge 0` Since discriminant of corresponding equation is negative :math:`y` is capable of having any real value. 158. Let :math:`y = \frac{x^2 - 4x + 9}{x^2 + 4x + 9}` :math:`\Rightarrow (y - 1)x^2 + 4(y + 1)x + 9(y - 1) = 0` Since :math:`x` is real, following must hold true :math:`16(y + 1)^2 - 36(y - 1)^2 \ge 0` :math:`4y^2 + 8y + 4 - 9y^2 + 18y - 9 \ge 0` :math:`-5y^2 + 26y - 5 \ge 0` Therefore, least value is :math:`\frac{1}{5}` and greatest value is :math:`5`. 159. Let :math:`y = \frac{x^2 + 7x + 16}{x^2 - 5x + 16}` :math:`\Rightarrow (y - 1)x^2 - (5y + 7)x + 16(y - 1) = 0` Since :math:`x` is real, following must hold true :math:`(5y + 7)^2 - 64(y - 1)^2 \ge 0` :math:`25y^2 + 70y + 49 - 64y^2 + 128y - 64 \ge 0` :math:`-39y^2 + 198y - 15 \ge 0` :math:`13y^2 - 66y + 5 \le 0` Therefore, least value is :math:`\frac{1}{13}` and greatest value is :math:`5`. 160. Let :math:`y = \frac{6x^2 - 2x + 3}{2x^2 - 2x + 1}` :math:`\Rightarrow 2(y - 3)x^2 - 2(y - 1)x + y - 3 = 0` Since :math:`x` is real, following must hold true :math:`4(y - 1)^2 - 8(y - 3)^2 \ge 0` :math:`y^2 - 2y + 1 - 2y^2 + 12y - 18 \ge 0` :math:`y^2 - 10y + 17 \le 0` Therefore, least value is :math:`5 - 2\sqrt{2}` and greatest value is :math:`5 + 2\sqrt{2}`. 161. Let :math:`y = \frac{(x - 1)(x + 3)}{(x - 2)(x + 4)}` :math:`y = \frac{x^2 + 2x - 3}{x^2 + 2x - 8}` :math:`\Rightarrow (y - 1)x^2 + 2(y - 1)x^2 + 3 - 8y = 0` Since :math:`x` is real, following must hold true :math:`4(y - 1)^2 + 4(y - 1)(8y - 3) \ge 0` :math:`y^2 - 2y + 1 + 8y^2 - 11y + 3 \ge 0` :math:`9y^2 - 13y + 4 \ge 0` For above to be true :math:`y` must not lie between :math:`1` and :math:`\frac{4}{9}`. 162. Let :math:`y = \frac{2x^2 - 2x + 4}{x^2 - 4x + 3}` :math:`(y - 2)x^2 - 2(2y - 1)x + 3y - 4 = 0` Since :math:`x` is real, following must hold true :math:`4(2y - 1)^2 - 4(y - 2)(3y -4) \ge 0` :math:`\Rightarrow 4y^2 - 4y + 1 - 3y^2 + 10y - 8 \ge 0` :math:`\Rightarrow y^2 + 6y - 7 \ge 0` For above to be true :math:`y` cannot lie between :math:`1` and :math:`-7`. 163. Let :math:`y = \frac{x^2 + 2x - 11}{-x - 3}` :math:`\Rightarrow x^2 + (y + 2)x + 3y - 11 = 0` Since :math:`x` is real, following must hold true :math:`(y + 2)^2 - 12y + 44 \ge 0` :math:`y^2 - 8y + 48 \ge 0` For the above to be true :math:`y` cannot lie between :math:`4` and :math:`12`. 164. Let :math:`y = \frac{x}{x^2 + 1}` :math:`\Rightarrow yx^2 - x + y = 0` Since :math:`x` is real, following must hold true :math:`1 - 4y^2 \ge 0` :math:`y \le \frac{1}{2}` 165. Let :math:`y = \frac{x + a}{x^2 + bx + c^2}` :math:`\Rightarrow yx^2 + (by - 1)x - a + c^2y = 0` Since :math:`x` is real, following must hold true :math:`(by - 1)^2 - 4y(c^2y - a) \ge 0` :math:`b^2y^2 - 2by + 1 + 4ay - 4c^2y^2 \ge 0` :math:`(b^2 - 4c^2)y^2 + 2(2a - b)y + 1 \ge 0` Discriminant of corresponding equation is :math:`D = 4(2a - b)^2 - 4(b^2 - 4c^2)` :math:`= 4[4a^2 + b^2 - 4ab - b^2 + 4c^2] = 16(a^2 + c^2 - ab)` Given :math:`b^2 > 4c^2` and :math:`a^2 + c^2 > ab` therefore :math:`D < 0` and coefficient of :math:`y^2` is negative. Therefore, :math:`y` is capable of assuming any value. 166. Let :math:`y = \frac{x^2 - bc}{2x - b - c}` :math:`\Rightarrow x^2 - 2yx + (b + c)y - bc = 0` Since :math:`x` is real, following must hold true :math:`4y^2 - 4(b + c)y + 4bc \ge 0` :math:`y^2 - (b + c)y + bc \ge 0` For above to be true :math:`y` must not lie between :math:`b` and :math:`c`. 167. Given expression is :math:`\frac{1}{x + 1} + \frac{1}{3x + 1} - \frac{1}{(x + 1)(3x + 1)}` :math:`= \frac{4x + 1}{3x^2 + 4x + 1}` Let :math:`y = \frac{4x + 1}{3x^2 + 4x + 1}` :math:`\Rightarrow 3yx^2 + 4(y - 1)x + y - 1 = 0` Since :math:`x` is real, following must hold true :math:`16(y - 1)^2 - 12y(y - 1) \ge 0` :math:`\Rightarrow 4y^2 - 8y + 4 - 3y^2 + 3y \ge 0` :math:`\Rightarrow y^2 - 5y + 4 \ge 0` Above is true provided :math:`y` does not lie between :math:`1` and :math:`4`. 168. Let :math:`y = \frac{2x^2 + x - 3}{3x + 1}` :math:`\Rightarrow 2x^2 + (1 - 3y)x - (y + 3) = 0` Since :math:`x` is real, following must hold true :math:`(1 - 3y)^2 - 8(y + 3) \ge 0` :math:`1 - 6y + 9y^2 - 8y - 24 \ge 0` :math:`9y^2 - 14y - 23 \ge 0` Discriminant of corresponding equation is negative and coefficient of :math:`y^2` is positive therefore :math:`y` is capable of assuming any real value. 169. Let :math:`y = \frac{2x^2 + 4x + 1}{x^2 + 4x + 2}` :math:`(y - 2)x^2 + 4(y - 1)x + 2y - 1 = 0` Since :math:`x` is real, following must hold true :math:`16(y - 1)^2 - 4(y - 2)(2y - 1) \ge 0` :math:`4y^2 - 8y + 4 - 2y^2 + 5y - 2 \ge 0` :math:`2y^2 - 3y + 2 \ge 0` Discriminant of corresponding equation is negative and coefficient of :math:`y^2` is positive therefore :math:`y` is capable of assuming any real value. 170. Let :math:`y = \frac{ax^2 + 3x - 4}{3x - 4x^2 + a}` :math:`\Rightarrow (a + 4y)x^2 + 3(1 - y)x - (4 + ay) = 0` Since :math:`x` is real, discriminant is greater than or equal to zero :math:`\Rightarrow 9(1 - y)^2 + 4(a + 4y)(4 + ay) \ge 0` :math:`\Rightarrow 9y^2 - 18y + 9 + 16a + 64y + 4a^2y + 16ay^2 \ge 0` :math:`\Rightarrow (9 + 16a)y^2 + (46 + 4a^2)y + 9 + 16a \ge 0` So for :math:`y` to assume any real value :math:`9 + 16a > 0` and :math:`D < 0` :math:`\Rightarrow 4(23 + 2a^2)^2 - (9 + 16a)^2 < 0` :math:`(a + 4)(a - 1)(a - 7) < 0` But since :math:`a > -\frac{9}{16}` therefore :math:`a` must lie between :math:`1` and :math:`7`. 171. Let :math:`y = \frac{m^2}{1 + x} - \frac{n^2}{1 - x}` :math:`y = \frac{m^2 - n^2 - (m^2 + n^2)x }{1 - x^2}` :math:`yx^2 - (m^2 + n^2)x + m^2 - n^2 - y = 0` Since :math:`x` is real, discriminant is greater than or equal to zero :math:`(m^2 + n^2)^2 - 4y(m^2 - n^2 - y) \ge 0` :math:`4y^2 - 4(m^2 - n^2)y + (m^2 + n^2)^2 \ge 0` For :math:`y` to be able to assume to any value discriminant of corresponding equation has to be negative since coefficient of :math:`y^2` is positive. :math:`16(m^2 - n^2)^2 - 16(m^2 + n^2)^2 < 0` which is true. 172. Let :math:`y = \frac{4x}{x^2 + 16}` :math:`\Rightarrow yx^2 - 4x + 16y = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`16 - 64y^2 \ge 0` :math:`y^2 le \frac{1}{4}` :math:`-\frac{1}{2} \le y \le \frac{1}{2}` :math:`\left|\frac{4x}{x^2 + 16}\right| < \frac{1}{2}` 173. Given :math:`x^2 - xy + y^2 - 4x - 4y + 16 = 0` :math:`x^2 - (y + 4)x + y^2 - 4y + 16 = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`(y + 4)^2 - 4(y^2 - 4y + 16) \ge 0` :math:`y^2 + 8y + 16 - 4y^2 + 16y - 64 \ge 0` :math:`-3y^2 + 24y - 48 \ge 0` :math:`y^2 - 8y + 16 \le 0` :math:`(y - 4)^2 \le 0` The above inequality is only satisfied by :math:`y = 4` However, if :math:`y = 4` the given equation becomes :math:`x^2 - 8x + 16 = 0` which is again only satisfied by :math:`x = 4` 174. Given :math:`x^2 + 12xy + 4y^2 + 4x + 8y + 20 = 0` :math:`x^2 + 4(1 + 3y)x + 4(y^2 + 2y + 5) = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`16(1 + 3y)^2 - 16(y^2 + 2y + 5) \ge 0` :math:`1 + 6y + 9y^2 - y^2 - 2y - 5 \ge 0` :math:`8y^2 + 4y - 4 \ge 0` :math:`2y^2 + y - 1 \ge 0 \Rightarrow (2y - 1)(y + 1) \ge 0` Therefore, :math:`y` cannot lie between :math:`-1` and :math:`\frac{1}{2}`. Rewriting the equation in terms of :math:`y` :math:`4y^2 + 4(3x + 2)y + x^2 + 4x + 20 = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`(3x + 2)^2 - x^2 - 4x - 20 \ge 0` :math:`8x^2 + 8x - 16 \ge 0 \Rightarrow x^2 + x - 2 \ge 0` Therefore, :math:`x` cannot lie between :math:`-2` and :math:`1`. 175. Corresponding equation is :math:`x^2 - 5mx + 4m^2 + 1 = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`\Rightarrow 25m^2 - 16m^2 - 4 \ge 0` :math:`9m^2 - 4 \ge 0` :math:`-\frac{2}{3} \le m \le \frac{2}{3}` 176. Let :math:`y = -3x^2 + x + 2` :math:`3x^2 - x - 2 + y = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`1 - 12(-2 + y) \ge 0` :math:`25 - 12y \ge 0` :math:`y \le \frac{25}{12}` 177. Let the positive number be :math:`x` and :math:`y = x + \frac{1}{x}` :math:`x^2 - yx + 1 = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`y^2 - 4 \ge 0` Therefore, least value of :math:`y` is :math:`2` as :math:`y` cannot be negative since :math:`x` is positive. 178. Let :math:`x` be the length and :math:`y` be the breadth then :math:`x + 2y = 600` and we have to maximize :math:`xy` :math:`xy = x\frac{600 - x}{2} = z` (say) :math:`x^2 - 600x + 2z = 0` Since :math:`x` is real, discriminant has to be greater than or equal to zero. :math:`360000 - 8z \ge 0` :math:`z \le 45000` Thus, maximum area is :math:`45000` mt. sq. Substituting :math:`x^2 - 600x + 90000 = 0` :math:`(x - 300)^2 = 0 \Rightarrow x = 300 \Rightarrow y = 150` 179. If :math:`y - mx` is a factor then equation reduces to :math:`bm^2 + 2hm + a = 0` and if :math:`my + x` is a factor then it reduces to :math:`am^2 - 2hm + b = 0` By cross-multiplication we have :math:`\frac{m^2}{-2h(a + b)} = \frac{m}{a^2 - b^2} = \frac{1}{2h(a + b)}` Thus, condition becomes :math:`a + b = 0` or :math:`4h^2 + (a^2 - b^2) = 0` 180. Expression :math:`(4 - k)x^2 + 2(k + 2)x + 8k + 1` will be perfect square if discriminant of corresponding equation will be equal to zero. :math:`\Rightarrow 4(k + 2)^2 - 4(4 - k)(8k + 1) = 0` :math:`4k^2 + 16k + 16 - 128k + 4k - 16 + 32k^2 = 0` :math:`36k^2 - 108k = 0` :math:`\Rightarrow k = 0, 3` 181. :math:`3x^2 - xy - 2y^2 + mx + y + 1` is resolvable into two linear factors if its discriminant is a perfect square. Rewriting the expression :math:`3x^2 + (m - y)x - 2y^2 + y + 1` The discriminant, :math:`(m - y)^2 - 12(- 2y^2 + y + 1)` has to be a perfect square. :math:`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 :math:`m = 4, -\frac{7}{2}` 182. Proceeding as previous problem we arrive at the solution as :math:`7 , \frac{98}{3}` 183. If :math:`x - \alpha` is a factor then it has to satisfy both the equations i.e. :math:`a_1\alpha^2 + b_1\alpha + c_1 = 0` and :math:`a_2\alpha^2 + b_2\alpha + c_1 = 0` By cross-multiplication we have :math:`\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 :math:`\alpha(a_1 - a_2) = b_2 - b_1` 184. Corresponding equation is :math:`6x^2 + 7xy + 2y^2 + 11x + 7y + 3 = 0` :math:`\Rightarrow 6x^2 + (7y + 11)x + 2y^2 + 7y + 3 = 0` :math:`x = \frac{-7y - 11 \pm\sqrt{(7y + 11)^2 - 24(2y^2 + 7y + 3)}}{12}` :math:`x = \frac{-7y - 11 \pm\sqrt{49y^2 + 154y + 121 - 48y^2 - 168y - 72}}{12}` :math:`x = \frac{-7y - 11 \pm\sqrt{y^2 - 14y + 49}}{12}` Therefore, factors are :math:`2x + y + 3` and :math:`3x + 2y + 1` 185. Corresponding equation is :math:`x^2 - 5xy + 4y^2 + x + 2y -2 = 0` :math:`\Rightarrow x^2 + (1 - 5y)x + 4y^2 + 2y - 2 = 0` :math:`x = \frac{5y - 1 \pm \sqrt{(1 - 5y)^2 - 16y^2 - 8y + 8}}{2}` :math:`x = \frac{5y - 1 \pm \sqrt{1 - 10y + 25y^2 - 16y^2 - 8y + 8}}{2}` :math:`x = \frac{5y - 1 \pm \sqrt{9y^2 - 18y + 9}}{2}` Therefore, factors are :math:`x - 4y + 2` and :math:`x - y - 1` 186. Corresponding equation is :math:`2x^2 + 5xy - 3y^2 + x + 17y - 10 = 0` :math:`\Rightarrow 2x^2 + (5y + 1)x - 3y^2 + 17y - 10 = 0` :math:`x = \frac{-5y - 1 \pm \sqrt{25y^2 + 10y + 1 + 24y^2 - 136y + 80}}{4}` :math:`x = \frac{-5y - 1 \pm \sqrt{(7y - 9)^2}}{4}` Therefore, factors are :math:`2x - y + 5` and :math:`x + 3y - 2` 187. Corresponding equation is :math:`3x^2 + 5xy - 2y^2 - 3x + 8y - 6 = 0` Proceeding as previous problems factors can be found as :math:`3x - y + 3` and :math:`x + 2y - 2` 188. We know that :math:`ax^2 + 2hxy + by^2 + 2gx + 2fy + c` can be resolved into two linear factors if and only if :math:`abc + 2fgh - af^2 - bg^2 - ch^2 = 0` Given expression is :math:`3x^2 + 2\alpha xy + 2y^2 + 2ax - 4y + 1` Comparing we find coefficients as :math:`a = 3, h = \alpha, b = 2, g = a, f = -2, c = 1` :math:`\therefore 6 - 4a\alpha - 12 - 2a^2 - \alpha^2 = 0` Clearly, :math:`\alpha` has to be a root of the equation :math:`x^2 + 4ax + 2a^2 + 6 = 0` 189. Let :math:`D_1` be discriminant of the equation :math:`x^2 + px + q = 0` and :math:`D_2` be discriminant of the equation :math:`x^2 + rx + s = 0` :math:`D_1 + D_2 = p^2 - 4q + r^2 - 4s > p^2 + r^2 - pr [\because 4(q + s) < pr]` If both :math:`p` and :math:`r` are zero, then :math:`D_1 + D_2 > 0` In case one of :math:`p` or :math:`r` is non-zero then :math:`D_1 + D_2 > r^2\left\{\left(\frac{p}{r}\right)^2 - \frac{p}{r} + 1\right\}` if :math:`r \ne 0` :math:`D_1 + D_2 > p^2\left\{\left(\frac{r}{p}\right)^2 - \frac{r}{p} + 1\right\}` if :math:`p \ne 0` Since for real :math:`x, x^2 - x + 1 > 0` as corresponding equation has imaginary roots. Thus, in all cases :math:`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. 190. Roots of equation :math:`P(x)Q(x) = 0` will be the roots of equation :math:`P(x) = 0` i.e. :math:`ax^2 + bx + c = 0` and :math:`Q(x) = -ax^2 + bx + c = 0` Let :math:`D_1` and :math:`D_2` be the discriminants of two equations, then :math:`D_1 + D_2 = b^2 - 4ax + b^2 + 4ac = 2b^2 > 0` Hence, at least one of :math:`D_1` and :math:`D_2` will be zero. Hence, :math:`P(x)Q(x) = 0` has at least two real roots. 191. Let :math:`D_1` be the discriminant of :math:`bx^2 + (b - c)x + b - c - a = 0` and :math:`D_2` be discriminant of :math:`ax^2 + 2bx + b = 0`, then :math:`D_1 + D_2 = (b - c)^2 - 4b(b - c - a) + 4b^2 - 4ab = (b + c)^2 \ge 0` Hence, if :math:`D_2 < 0`, then :math:`D_1 > 0`. Therefore, roots of :math:`bx^2 + (b - c)x + b - c - a = 0` will be real if roots of :math:`ax^2 + 2bx + b = 0` are imaginary and vice versa. 192. Let :math:`a = 2m + 1, b = 2n + 1, c = 2r + 1` Now :math:`D = (2n + 1)^2 - 4(2m + 1)(2r + 1)` :math:`= (\text{an odd number}) - (\text{an even number}) =` an odd number If possible let :math:`D` be a perfect square then it has to be square of an odd number. :math:`(2k + 1)^2 = (2n + 1)^2 - 4(2m + 1)(2r + 1)` :math:`(2m + 1)(2r + 1) = (n + k + 1)(n - k)` If :math:`n` and :math:`k` are both odd or even then :math:`n - k` will be even or zero. However, if one is odd and one is even then :math:`(n + k + 1)` will be even. So, R. H. S. is an even while L. H. S. is an odd number. Thus, :math:`D` cannot be a perfect square. Hence, roots cannot be a rational numbers. 193. Let :math:`D_1` be discriminant of :math:`ax^2 + 2bx + c = 0` then :math:`D_1 = 4b^2 - 4ac = 4k,` where :math:`k = b^2 - ac` Let :math:`D_2` is discriminant of :math:`(a + c)(ax^2 + 2bx + c) = 2(ac - b^2)(x^2 + 1)` :math:`D_2 = 4(a + c)^2b^2 - 4(a^2 + b^2 + k)(b^2 + c^2 + k)` :math:`= -D1[4b^2 + (a - c)^2] \Rightarrow D_2 < 0 \because D_1 > 0` Therefore, roots of second equation are non-real complex numbers. 194. :math:`D = 4[(^nC_r)^2 - ^nC_{r - 1}^nC_{r + 1}]` :math:`= 4(a - b),` where :math:`a = ^(nC_r)^2, b = ^nC_{r - 1}^nC_{r + 1}` :math:`\frac{a}{b} = \left(1 + \frac{1}{r}\right)\left(1 + \frac{1}{n - r}\right) > 1` :math:`a > b \Rightarrow D > 0` Thus, roots of given equation are real and distinct. 195. Let :math:`D` be the discriminant of the given equation. :math:`D = (2m - 1)^2 - 4m(m - 2)` :math:`= 4m + 1 =` an odd number. For roots to be rational discriminant must be a perfect square. :math:`(2k + 1)^2 = 4m + 1,` where :math:`k \in I` :math:`m = k(k + 1)` 196. Let :math:`y = e^{\sin x}` then given equation becomes :math:`y - \frac{1}{y} - 4 = 0` :math:`y = 2\pm \sqrt{5} \therefore e^{\sin x} = 2 \pm \sqrt{5}` :math:`\sin x = \log_e (2 - \sqrt{5})` is not defined. :math:`\sin x = \log_e (2 + \sqrt{5}) > 1` is not possible. Hence, roots of given equation cannot be real. 197. Given equation is :math:`az^2 + bz + c + i = 0` :math:`z = \frac{-b \pm \sqrt{b^2 - 4a(c + i)}}{2a} = \frac{-b \pm(p + iq)}{2a}` where :math:`\sqrt{b^2 - 4a(c + i)} = p + iq` :math:`b^2 - 4ac = p^2 - q^2` and :math:`-4a = 2qp` Since :math:`z` is purely imaginary :math:`\frac{-b \pm p}{2a} = 0 \Rightarrow \pm p = b` :math:`-4a = 2(\pm)q \Rightarrow q = \pm \frac{2a}{b}` Then, :math:`b^2 - 4ac = b^2 - \frac{4a^2}{b^2}` :math:`\Rightarrow c = \frac{a}{b^2} \Rightarrow a = b^2c` 198. :math:`D = a^2 - 4b` Let :math:`a` be an odd number then :math:`D` is an odd number and a perfect square as roots are rational. Let :math:`D = (2n + 1)^2,` and :math:`a = 2m + 1` where :math:`m, n \in I` Now roots :math:`= \frac{-(2m + 1)\pm (2n + 1)}{2} = \frac{\text{an even no.}}{2} = \text{an integer}` Similarly, it can be proven when :math:`a` is an even no. then roots are integers. 199. Let :math:`\alpha, \beta` be integral roots of the given equation. :math:`\alpha + \beta = -7` and :math:`\alpha\beta = 14(q^2 + 1)` :math:`\frac{\alpha\beta}{7} = 2(q^2 + 1) = \text{an integer}` :math:`\therefore \alpha\beta` is divisible by :math:`7` and :math:`7` is a prime number. :math:`\therefore` at least one of :math:`\alpha` and :math:`\beta` must be a multiple of :math:`7`. Let :math:`\alpha = 7k,` where :math:`k \in I` :math:`\beta = -7(k + 1)` Thus, :math:`-\frac{2(q^2 + 1)}{7} = k(k + 1) = \text{an integer}` Let :math:`f(q) = q^2 + 1` then it can be shown that :math:`f(1), f(2), ..., f(7)` are not divisible by :math:`7`. :math:`f(q + 7) = q^2 + 1 + 14q + 49` which is not divisible by :math:`7` as :math:`q^2 + 1` is not divisible by :math:`7`. Hence, :math:`\alpha, \beta` cannot be integers. 200. Given equation is :math:`[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 :math:`a^3(b - c) + b^3(c - a) + c^3(a - b) = -(a - b)(b - c)(c - a)(a + b + c)` and :math:`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 :math:`a^2(b - c) + b^2(c - a) + c^2(a - b) = -(a - b)(b - c)(c - a)` the above equation becomes :math:`(a + b + c)x^2 - (ab + bc + ca)x + abc = 0` Roots are :math:`\frac{(ab + bc + ca \pm \sqrt{(ab + bc + ca)^2 - 4abc(a + b + c)})}{2(a + b + c)}` Roots will be equal if :math:`D = 0` If :math:`\frac{1}{\sqrt{a}}\pm \frac{1}{\sqrt{b}}\pm \frac{1}{c} = 0` :math:`\frac{\sqrt{bc} \pm \sqrt{ca} \pm \sqrt{ab}}{\sqrt{abc}} = 0` :math:`\Rightarrow \sqrt{bc} \pm \sqrt{ca} \pm \sqrt{ab} = 0` Squaring :math:`bc + ca + ab \pm 2\sqrt{abc}(\sqrt{a}\pm \sqrt{b} \pm \sqrt{c}) = 0` :math:`(bc + ca + ab)^2 = 4abc(a + b + c + \sqrt{bc} \pm \sqrt{ca} \pm \sqrt{ab})` :math:`\Rightarrow D = 0` i.e. roots are equal.