.. meta:: :author: Shiv Shankar Dayal :title: Quadratic Equations Solutions Part 3 :description: Quadratic Equations Solutions Part 3 :keywords: quadratic equations, algebra Quadratic Equations Solutions Part 3 ************************************ 101. Since :math:`\alpha, \beta` are roots of the equation :math:`x^2 - px + q = 0`, :math:`\alpha + \beta = p` and :math:`\alpha\beta = q` Let us assume that :math:`\alpha + \frac{1}{\beta}` is a root of :math:`qx^2 - p(1 + q)x + (1 + q)^2 = 0` then it must satisfy the equation. Substituting the values we have :math:`\alpha\beta\frac{(\alpha\beta + 1)^2}{\beta^2} - \frac{(\alpha + \beta)(1 + \alpha\beta)(\alpha\beta + 1)}{\beta} + (1 + \alpha\beta)^2 = 0` :math:`(\alpha\beta + 1)^2[\alpha\beta - (\alpha + \beta)\beta - \beta^2] = 0` :math:`\because L. H. S. = R. H. S.` it is proven that :math:`\alpha + \frac{1}{\beta}` is a root of the given equation. 102. Let :math:`\alpha` be a common root. Then, :math:`\frac{\alpha^2}{-8m - 6} = \frac{\alpha}{4 + 6} = \frac{1}{9 - 8m}` Solving for :math:`m` we obtain :math:`\frac{7}{4}` and :math:`-\frac{11}{8}` as two values. 103. Proceeding as previous part we find :math:`a = 24` 104. The condition for having common roots is :math:`(ba - c^2)(ca - b^2) = (a^2 - bc)^2` :math:`a^2bc - ab^3 - ac^3 + b^2c^2 = a^4 - 2a^2bc + b^2c^2` :math:`3a^2bc - ab^3 -ac^3 - a^4 = 0` :math:`a(3abc - b^3 - c^3 - a^3) = 0` :math:`\because a\ne = 0 \Rightarrow a^3 + b^3 + c^3 - 3abc = 0` :math:`(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0` :math:`\Rightarrow a + b + c = 0` or :math:`a = b = c` 105. Proceeding as in last example, condition for common root is :math:`(10m - 189)(9 - 10) = (21 - m)^2` :math:`189 - 10m = 441 - 42m + m^2 \Rightarrow m^2 - 32m + 252 = 0 \Rightarrow m = 18, 14` Roots of :math:`x^2 + 10x + 21 = 0` are :math:`-3, -7` When :math:`m = 18` roots of :math:`x^2 + 9x + 18 = 0` are :math:`-3, -6` In that case equation formed with :math:`-7` and :math:`-6` is :math:`x^2 + 13x + 42 = 0` When :math:`m= 14` roots of :math:`x^2 + 9x + 14 = 0` are :math:`-2, -7` In that case equation formed with :math:`-3` and :math:`-2` are :math:`x^2 + 5x + 6 = 0` 106. Following condition for common roots, we have :math:`(-3 + 120)(10 + 3) = (3 + 36)^2` :math:`117 * 13 = 39^2` which is true and thus equations have a common root. Roots of :math:`x^2 - x - 12 = 0` are :math:`4, -3` and roots of :math:`3x^2 + 10x + 3 = 0` are :math:`-3, -\frac{1}{3}` and thus common root is :math:`-3` 107. Again condition for common root is given below: :math:`(p - q)(3q - 2p) = (3 - 2)^2` :math:`(2p - 3q)(p - q) + 1 = 0` :math:`2p^2 + 3q^2 - 5pq + 1 = 0` 108. Again repeating the condition for common root we have :math:`(b - c)(a - b) = (a - c)^2` :math:`ab - ac - b^2 + bc = a^2 + c^2 - 2ac` :math:`a^2 + b^2 + c^2 - ab - ac - bc = 0` :math:`\frac{1}{2}(a - b)^2(b - c)^2(c - a)^2 = 0` :math:`\Rightarrow a = b = c` 109. Let :math:`\alpha` be the common root then :math:`\frac{\alpha^2}{pq_1 - p_1q} = \frac{\alpha}{q - q_1} = \frac{1}{p_1 - p}` Clearly, the root is either :math:`\frac{pq_1 - p_1q}{q - q_1}` or :math:`\frac{q - q_1}{p_1 - p}` 110. Condition for having common root is: :math:`(-4b + 3c)(-6a - 2b) = (4a - 2c)^2` Solving this we arrive at the given condition. 111. Condition for having a common root is: :math:`[(r - p)(q - r) - (p - q)^2][(p - q)(q - r) - (r - p)^2] = [(q - r)^2 - (p - q)(r - p)]^2` Solving this leads to an equality which means the equations have a common root. 112. Let :math:`\alpha` be a common root then :math:`\frac{\alpha^2}{ab^2 - ac^2} = \frac{1}{b - c} = \frac{1}{ac - ab}` :math:`\alpha = -a(b + c)` or :math:`\alpha = -\frac{1}{a}` Let :math:`\alpha, \beta` be roots of first and :math:`\alpha, \gamma` be roots of the second equation. Then, :math:`\alpha + \beta = -ab` and :math:`\alpha\beta = c` also, :math:`\alpha + \gamma = -ac` and :math:`\alpha\gamma = b` :math:`\Rightarrow 2\alpha + \beta + \gamma = -a(b + c)` and :math:`\alpha^2\beta\gamma = bc` Equation formed by :math:`\beta` and :math:`\gamma` would be :math:`x^2 - (\beta + \gamma)x + \beta\gamma = 0` For either values of :math:`\alpha` equation is :math:`x^2 - a(b + c)x + a^2bc = 0` 113. This question has been left as an exercise. 114. It is a quadratic equation but satisfied by three values of :math:`x = 1, 2, 3` therefore it is an identity. 115. It is a quadratic equation but satisfied by three values of :math:`x = a, b, c` therefore it is an identity. 116. Let :math:`x^5 = y` then equation becomes :math:`3y^2 - 2y - 8 = 0` Since it is satisfied by two distinct values and it is a quadratic equation therefore it is an equation. 117. :math:`\frac{(x + 2)^2 - (x - 2)^2}{x^2 - 4} = \frac{5}{6}` :math:`\frac{8x}{x^2 - 4} = \frac{5}{6}` :math:`5x^2 - 20 - 48x = 0` :math:`x = 10 , -\frac{2}{5}` 118. Let :math:`x = y^2` :math:`\Rightarrow \frac{2y + 1}{3 - y} = \frac{11 - 3y}{5y - 9}` :math:`10y^2 - 13y - 9 = 33 - 20y + 3y^2` :math:`7y^2 + 7y - 42 = 0` :math:`y = 2, -3` :math:`x = 4, 9` but :math:`x = 9` does not apply to the equation and is an impossible solution. 119. :math:`(x + 1)(x - 3)(x + 2)(x - 4) = 336` :math:`(x^2 - 2x - 3)(x^2 - 2x - 8) = 336` Let :math:`x^2 - 2x - 3 = y` :math:`y(y - 5) = 336` :math:`y^2 - 5y - 336 = 0` :math:`y = 21, -16` :math:`\Rightarrow x = -4, 6, 1 \pm 2\sqrt{3}i` 120 and 121 are left as an exercise. 122. Let the speed be :math:`x` km/hour. Then, from the statement :math:`\frac{800}{x} = \frac{800}{x + 40} + \frac{2}{3}` Solving we get :math:`x = 200` km/hour 123. Let width be :math:`w` meter. Thus, :math:`(w + 8)(w - 2) = 119` :math:`w^2 + 6w - 135 = 0` :math:`w = 9, -15` but width cannot be negative. Length is :math:`11` m. 124. Equivalent equation is :math:`-x^2 + 3x + 4 = 0` and roots are :math:`-1, 4`. Since coefficient of :math:`x^2` is -ve the expression will be +ve if :math:`x` lies between the root. Therefore, for :math:`-x^2 + 3x + 4 > 0` the range is :math:`]-1, 4[` which is fully open interval. 125. :math:`5x - 1 < (x + 1)^2 \Rightarrow x^2 - 3x + 2 > 0` Roots of equivalent equation :math:`x^2 - 3x + 2 = 0` are :math:`x = 2, 1` Since coefficient of :math:`x^2` is positive, :math:`x` must lie outside the range of :math:`[1, 2]` for the expression to be positive. Now considering, :math:`(x + 1)^2 < 7x - 3` :math:`x^2 - 5x + 4 < 0` Roots of the equivalent equation :math:`x^2 - 5x + 4 = 0` are :math:`x = 1, 4` and for expression to be negative :math:`x` must lie inside the open interval :math:`]1, 4[`. Therefore, the only integral value satisfying the original expression is :math:`3`. 126. :math:`\frac{8x^2 + 16x - 51}{(2x - 3)(x + 4)} > 3` :math:`\Rightarrow \frac{2x^2 + x - 15}{2x^2 + 5x - 12} > 0` :math:`2x^2 + x - 15 = 0` has roots :math:`x = -3 , \frac{5}{2}` :math:`2x^2 + 5x - 12 = 0` has roots :math:`x = -4, \frac{3}{2}` Thus, the inequality will hold true for :math:`x < -4` and :math:`-3 < x < \frac{3}{2}` and :math:`x > \frac{5}{2}` 127. Let :math:`y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}` :math:`(y - 1)x^2 + 3(y + 1)x + 4(y - 1) = 0` Since :math:`x` is real, the discriminant will be greater that :math:`0` :math:`\Rightarrow 9(y + 1)^2 - 16(y - 1)^2 \ge 0` :math:`-7y^2 + 50y - 7 \ge 0` The roots are :math:`7` and :math:`\frac{1}{7}` Since coefficient of :math:`y^2` is negative, for the expression to be positive :math:`y` has to lie between the open interval formed by its roots i.e. :math:`]\frac{1}{7}, 7[` 128. Let :math:`y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}` :math:`(y - 1)x^2 + 2(y - 17)x + (71 - y) = 0` Since :math:`x` is real, the discriminant will be greater that :math:`0` :math:`\Rightarrow 4(y - 17)^2 - 4(y - 1)(71 - 7y) \ge 0` :math:`y^2 - 14y + 45 \ge 0` Its roots are :math:`5` and :math:`9` Since coefficient of :math:`y^2` is positive, therefore for the expression to be positive :math:`y` has to lie outside the open interval formed by its roots. Thus, the expression has no value between :math:`5` and :math:`9`. 129. Let :math:`y = \frac{4x^2 + 36x + 9}{12x^2 + 8x + 1}` :math:`4(3y - 1)x^2 + 4(2y - 9)x + y - 9 = 0` Since :math:`x` is real, the discriminant will be greater that :math:`0` :math:`16(2y - 9)^2 - 16(3y - 1)(y - 1) \ge 0` :math:`y^2 - 8y + 72 \ge 0` Corresponding equation is :math:`y^2 - 8y + 72 = 0` :math:`D = 64 - 288 = -224 < 0` Since coefficient of :math:`y^2` is positive and discriminant is less than :math:`0` therefore :math:`y^2 - 8y + 72 \ge 0` holds true for all value of :math:`y`. Therefore, the expression can take any value. 130. Let :math:`y = \frac{(x - a)(x - c)}{x -b}` :math:`x^2 - (a + c + y)x + ac + yb = 0` Since :math:`x` is real, the discriminant will be greater that :math:`0` :math:`(a + c + y)^2 - 4(ac + yb) \ge 0` :math:`y^2 + 2(a + c - 2b)y + (a - c)^2 \ge 0` Corresponding equation is :math:`y^2 + 2(a + c - 2b)y + (a - c)^2 = 0` Discriminant of above equation is :math:`D = -16(a - b)(b - c)` If :math:`a > b > c` then :math:`D < 0` and if :math:`a < b < c` then also :math:`D < 0` Since coefficient of :math:`y^2` is positive and :math:`D < 0` the expression :math:`y^2 + 2(a + c - 2b)y + (a - c)^2 \ge 0` is true for all real values of :math:`y`. Therefore, the given expression is capable of holding any value for the given conditions. 131. Given :math:`x + y =` constant :math:`= k` (say) Let :math:`z = xy`, then :math:`z = x(k - x) \Rightarrow x^2 - kx + z = 0` Since :math:`x` is real, :math:`D \ge 0` for the above equation. :math:`k^2 - 4z \ge 0 \Rightarrow z \le \frac{k^2}{4}` Hence, the maximum value of :math:`z = \frac{k^2}{4}` Thus, :math:`x^2 - kx + \frac{k^2}{4} = 0 \Rightarrow \left(x - \frac{k}{2}\right)^2 = 0 \Rightarrow x = \frac{k}{2}` :math:`\therefore y = \frac{k}{2}` and thus :math:`xy` is maximum when :math:`x = y` 132. Let :math:`y = 3 - 6x - 8x^2 \Rightarrow 8x^2 + 6x + y - 3 = 0` Since :math:`x` is real, :math:`D \ge 0` for the above equation. :math:`\Rightarrow 36 - 32(y - 3) \ge 0` :math:`y \le \frac{33}{8}`. Hence, maximum value of :math:`y = \frac{33}{8}` :math:`\Rightarrow 64x^2 + 48x + 9 = 0` :math:`(8x + 3)^2 = 0 \Rightarrow x = -\frac{3}{8}` 133. Let :math:`y = \frac{12x}{4x^2 + 9}` :math:`\Rightarrow 4yx^2 - 12x + 9y = 0` Since :math:`x` is real, :math:`D \ge 0` for the above equation. :math:`\Rightarrow 144 - 144y^2 \ge 0` :math:`\Rightarrow y^2 \le 1` :math:`\Rightarrow -1 \le y \le 1 \Leftrightarrow |y| \le 1 \Leftrightarrow\left|\frac{12x}{4x^2 + 9}\right| \le 1` Now, :math:`\left|\frac{12x}{4x^2 + 9}\right| = 1 \Leftrightarrow 4|x|^2 - 12|x| + 9 = 0` :math:`(2|x| - 3)^2 = 0 \Rightarrow |x| = \frac{3}{2}` 134. :math:`x^2 + 9y^2 - 4x + 3 = 0` Since :math:`x` is real, :math:`D \ge 0` for the above equation. :math:`\Rightarrow (-4)^2 - 4(9y^2 + 3) \ge 0` :math:`\Rightarrow 9y^2 - 1 \le 0 \Leftrightarrow y^2 \le \frac{1}{9}` :math:`\Rightarrow -\frac{1}{3} \le y \le \frac{1}{3}` The given equation can also be written as :math:`9y^2 + x^2 - 4x + 3 = 0` Since :math:`y` is real, :math:`D \ge 0` for the above equation. :math:`-36(x^2 - 4x + 3) \ge 0` :math:`x^2 - 4x + 3 \le 0` Since coefficient of :math:`x^2` is positive, it must lie between its root for the above expression to be negative. Therefore, :math:`x` must lie between :math:`1` and :math:`3`. 135. Given expression is :math:`x^2 - ax + 1 - 2a^2 > 0` Since :math:`x` is real the discriminant of the corresponding equation has to be negative for it to be positive for all values of :math:`x`. :math:`a^2 - 4(1 - 2a^2) < 0 \Leftrightarrow 9a^2 \le 4` :math:`-\frac{2}{3} < a < \frac{2}{3}` 136. Let :math:`\alpha` be a common factor, therefore it will satisfy both the equations. :math:`\alpha^2 - 11\alpha + a = 0` and :math:`\alpha^2 - 14\alpha + 2a = 0` By cross-multiplication :math:`\frac{\alpha^2}{-22a + 14x} = \frac{\alpha}{a - 2a} = \frac{1}{-14 + 11}` :math:`\frac{\alpha^2}{-8a} = \frac{\alpha}{-a} = -\frac{1}{3}` From first two we have :math:`\alpha = 8` and from last two we have :math:`\alpha = \frac{a}{3}` :math:`\therefore a = 24` 137. :math:`y = mx` is a factor of :math:`ax^2 + bxy + cy^2` means :math:`ax^2 + bxy + cy^2` will be zero when :math:`y = mx` :math:`ax^2 + bx.mx + cm^2x^2 = 0 \Rightarrow cm^2 + bm + a = 0` Similarly, :math:`a_1m^2 + b_1m + c_1 = 0` since :math:`my - x` is a factor of :math:`a_1x^2 + b_1xy + c_1y^2` Solving these two equations in :math:`m` by cross-multiplication :math:`\frac{m^2}{bc_1 - ab_1} = \frac{m}{aa_1 - cc_1} = \frac{1}{cb_1 - ba_1}` From first two we get, :math:`m = \frac{bc_1 - ab_1}{aa_1 - cc_1}` and from last two we get, :math:`m = \frac{aa_1 - cc_1}{cb_1 - ba_1}` Equating the two values of :math:`m` obtained we get :math:`(bc_1 - ab_1)(cb_1 - ba_1) = (aa_1 - cc_1)^2` 138. 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` and :math:`h^2 - ab > 0` Given expression is :math:`2x^2 + mxy + 3y^2 - 5y - 2` Here, :math:`a = 2, h = \frac{m}{2}, b = 3, g = 0, f = \frac{-5}{2}, c = -2` :math:`h^2 - ab = \frac{m^2}{4} - 6 > 0\Rightarrow m^2 > 24` Applying the second condition :math:`-12 - \frac{25}{2} + \frac{m^2}{2} = 0` :math:`m^2 = 49 \therefore m = \pm 7` 139. Given expression is :math:`ax^2 + by^2 + cz^2 + 2ayz + 2bzx + 2cxy` :math:`= z^2\left[a\left(\frac{x}{z}\right)^2 + b\left(\frac{y}{z}\right)^2 + c + 2a\frac{y}{z} + 2b\frac{x}{z} + 2c\frac{xy}{z^2}\right]` :math:`= z^2(aX^2 + bY^2 + c + 2aY + 2bX + 2cXY)` where :math:`X = \frac{x}{z}, Y = \frac{y}{z}` Now this will resolve in linear factors if :math:`abc + 2abc - a.a^2 - b.b^2 -c.c^2 = 0` :math:`a^3 + b^3 + c^3 = 3abc` 140. Given expression is :math:`2x^2 - y^2 - x + xy + 2y -1` Corresponding equation is :math:`2x^2 - y^2 - x + xy + 2y -1 = 0` :math:`x = \frac{1 - y \pm \sqrt{(1 - y)^2 + 8(y^2 - 2y + 1)}}{4}` :math:`x = 1 - y, -\frac{1 - y}{2}` Therefore, required linear factors are :math:`x + y - 1` and :math:`2x - y + 1` 141. Corresponding quadratic equation is :math:`x^2 + 2(a + b + c)x + 3(ab + bc + ca) = 0` It will be a perfect square if its discriminant is zero. :math:`\Rightarrow 4(a + b + c)^2 - 4.3(ab + bc + ca) = 0` :math:`\Rightarrow a^2 + b^2 + c^2 - ab - bc - ca = 0` :math:`\Rightarrow \frac{1}{2}(a - b)^2(b - c)^2(c - a)^2 = 0` :math:`\Rightarrow a = b = c` 142. Discriminant of the given equation is :math:`D = 36 - 72 < 0` Now since coefficient of :math:`x^2` is less than zero the expression is always positive. 143. :math:`8x - 15 - x^2 > 0` :math:`\Rightarrow x^2 - 8x + 15 < 0` :math:`(x - 3)(x - 5) < 0` The above is true if :math:`x` lies in the open interval :math:`]3, 5[`. 144. :math:`-x^2 + 5x - 4 > 0` :math:`\Rightarrow x^2 - 5x + 4 < 0` :math:`(x - 4)(x - 1) < 0` The above is true if :math:`x` lies in the open interval :math:`]1, 4[`. 145. :math:`x^2 + 6x - 27 > 0` :math:`\Rightarrow (x + 9)(x - 3) > 0` This is true if :math:`x < -9` or :math:`x > 3` 146. :math:`\frac{4x}{x^2 + 3} \le 1` :math:`\Rightarrow x^2 + 3 \le 4x` :math:`\Rightarrow x^2 - 4x + 3 \le 0` :math:`(x - 3)(x - 1) le 0` This is true for closed interval :math:`[1, 3]`. 147. :math:`x^2 - 3x + 2 > 0` :math:`(x - 2)(x - 1) > 0` This is true for :math:`x > 2` or :math:`x < 1` :math:`x^2 - 3x - 4 \le 0` :math:`(x - 4)(x + 1) \le 0` This is true for :math:`-1 \le x \le 4` Thus values of :math:`x` which satisfy both are :math:`-1 \le x < 1` and :math:`2 < x \le 4`. 148. Since roots of :math:`ax^2 + bx + c` are imaginary, therefore discriminant is negative. :math:`\Rightarrow b^2 - 4ac < 0` Discriminant of :math:`a^2x^2 + abx + ac` is: :math:`D = a^2b^2 - 4a^3c = a^2(b^2 - 4ac) < 0` But coefficient of the expression is positive hence it will be always positive. 149. Let :math:`y = \frac{x^2 - 2x + 4}{x^2 + 2x + 4}` :math:`(y - 1)x^2 + 2(y + 1)x + 4(y - 1) = 0` Since :math:`x` is real discriminant will be greater or equal to zero. :math:`4(y + 1)^2 - 16(y - 1)^2 \ge 0` :math:`y^2 + 2y + 1 - 4y^2 + 8y - 4 \ge 0` :math:`-3y^2 + 10y - 3 \ge 0` Roots of corresponding equation are :math:`\frac{1}{3}, 3`. Since coefficient of :math:`y^2` is negative, for above to be true :math:`y` must lie between :math:`\frac{1}{3}` and :math:`3`. 150. Let :math:`y = \frac{2x^2 - 3x + 2}{2x^2 + 3x + 2}` :math:`2(y - 1)x^2 + 3(y + 1)x + 2(y - 1) = 0` Since :math:`x` is real discriminant will be greater or equal to zero. :math:`9(y + 1)^2 - 16(y - 1)^2 \ge 0` :math:`9y^2 + 18y + 9 - 16y^2 + 32y - 16 \ge 0` :math:`-7y^2 + 50y - 7 \ge 0` Roots of the corresponding equation are :math:`\frac{1}{7}, 7`. Since coefficients of :math:`y^2` is negative, for the above to be true :math:`y` must lie between :math:`\frac{1}{7}` and :math:`7`.