44. Quadratic Equations Solutions Part 3ΒΆ

  1. Since \(\alpha, \beta\) are roots of the equation \(x^2 - px + q = 0\), \(\alpha + \beta = p\) and \(\alpha\beta = q\)

    Let us assume that \(\alpha + \frac{1}{\beta}\) is a root of \(qx^2 - p(1 + q)x + (1 + q)^2 = 0\) then it must satisfy the equation. Substituting the values we have

    \(\alpha\beta\frac{(\alpha\beta + 1)^2}{\beta^2} - \frac{(\alpha + \beta)(1 + \alpha\beta)(\alpha\beta + 1)}{\beta} + (1 + \alpha\beta)^2 = 0\)

    \((\alpha\beta + 1)^2[\alpha\beta - (\alpha + \beta)\beta - \beta^2] = 0\)

    \(\because L. H. S. = R. H. S.\) it is proven that \(\alpha + \frac{1}{\beta}\) is a root of the given equation.

  2. Let \(\alpha\) be a common root. Then,

    \(\frac{\alpha^2}{-8m - 6} = \frac{\alpha}{4 + 6} = \frac{1}{9 - 8m}\)

    Solving for \(m\) we obtain \(\frac{7}{4}\) and \(-\frac{11}{8}\) as two values.

  3. Proceeding as previous part we find \(a = 24\)

  4. The condition for having common roots is

    \((ba - c^2)(ca - b^2) = (a^2 - bc)^2\)

    \(a^2bc - ab^3 - ac^3 + b^2c^2 = a^4 - 2a^2bc + b^2c^2\)

    \(3a^2bc - ab^3 -ac^3 - a^4 = 0\)

    \(a(3abc - b^3 - c^3 - a^3) = 0\)

    \(\because a\ne = 0 \Rightarrow a^3 + b^3 + c^3 - 3abc = 0\)

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

    \(\Rightarrow a + b + c = 0\) or \(a = b = c\)

  5. Proceeding as in last example, condition for common root is

    \((10m - 189)(9 - 10) = (21 - m)^2\)

    \(189 - 10m = 441 - 42m + m^2 \Rightarrow m^2 - 32m + 252 = 0 \Rightarrow m = 18, 14\)

    Roots of \(x^2 + 10x + 21 = 0\) are \(-3, -7\)

    When \(m = 18\) roots of \(x^2 + 9x + 18 = 0\) are \(-3, -6\)

    In that case equation formed with \(-7\) and \(-6\) is \(x^2 + 13x + 42 = 0\)

    When \(m= 14\) roots of \(x^2 + 9x + 14 = 0\) are \(-2, -7\)

    In that case equation formed with \(-3\) and \(-2\) are \(x^2 + 5x + 6 = 0\)

  6. Following condition for common roots, we have

    \((-3 + 120)(10 + 3) = (3 + 36)^2\)

    \(117 * 13 = 39^2\) which is true and thus equations have a common root.

    Roots of \(x^2 - x - 12 = 0\) are \(4, -3\) and roots of \(3x^2 + 10x + 3 = 0\) are \(-3, -\frac{1}{3}\) and thus common root is \(-3\)

  7. Again condition for common root is given below:

    \((p - q)(3q - 2p) = (3 - 2)^2\)

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

    \(2p^2 + 3q^2 - 5pq + 1 = 0\)

  8. Again repeating the condition for common root we have

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

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

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

    \(\frac{1}{2}(a - b)^2(b - c)^2(c - a)^2 = 0\)

    \(\Rightarrow a = b = c\)

  9. Let \(\alpha\) be the common root then

    \(\frac{\alpha^2}{pq_1 - p_1q} = \frac{\alpha}{q - q_1} = \frac{1}{p_1 - p}\)

    Clearly, the root is either \(\frac{pq_1 - p_1q}{q - q_1}\) or \(\frac{q - q_1}{p_1 - p}\)

  10. Condition for having common root is:

    \((-4b + 3c)(-6a - 2b) = (4a - 2c)^2\)

    Solving this we arrive at the given condition.

  11. Condition for having a common root is:

    \([(r - p)(q - r) - (p - q)^2][(p - q)(q - r) - (r - p)^2] = [(q - r)^2 - (p - q)(r - p)]^2\)

    Solving this leads to an equality which means the equations have a common root.

  12. Let \(\alpha\) be a common root then

    \(\frac{\alpha^2}{ab^2 - ac^2} = \frac{1}{b - c} = \frac{1}{ac - ab}\)

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

    Let \(\alpha, \beta\) be roots of first and \(\alpha, \gamma\) be roots of the second equation. Then, \(\alpha + \beta = -ab\) and \(\alpha\beta = c\) also, \(\alpha + \gamma = -ac\) and \(\alpha\gamma = b\)

    \(\Rightarrow 2\alpha + \beta + \gamma = -a(b + c)\) and \(\alpha^2\beta\gamma = bc\)

    Equation formed by \(\beta\) and \(\gamma\) would be

    \(x^2 - (\beta + \gamma)x + \beta\gamma = 0\)

    For either values of \(\alpha\) equation is

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

  13. This question has been left as an exercise.

  14. It is a quadratic equation but satisfied by three values of \(x = 1, 2, 3\) therefore it is an identity.

  15. It is a quadratic equation but satisfied by three values of \(x = a, b, c\) therefore it is an identity.

  16. Let \(x^5 = y\) then equation becomes \(3y^2 - 2y - 8 = 0\)

    Since it is satisfied by two distinct values and it is a quadratic equation therefore it is an equation.

  17. \(\frac{(x + 2)^2 - (x - 2)^2}{x^2 - 4} = \frac{5}{6}\)

    \(\frac{8x}{x^2 - 4} = \frac{5}{6}\)

    \(5x^2 - 20 - 48x = 0\)

    \(x = 10 , -\frac{2}{5}\)

  18. Let \(x = y^2\)

    \(\Rightarrow \frac{2y + 1}{3 - y} = \frac{11 - 3y}{5y - 9}\)

    \(10y^2 - 13y - 9 = 33 - 20y + 3y^2\)

    \(7y^2 + 7y - 42 = 0\)

    \(y = 2, -3\)

    \(x = 4, 9\) but \(x = 9\) does not apply to the equation and is an impossible solution.

  19. \((x + 1)(x - 3)(x + 2)(x - 4) = 336\)

    \((x^2 - 2x - 3)(x^2 - 2x - 8) = 336\)

    Let \(x^2 - 2x - 3 = y\)

    \(y(y - 5) = 336\)

    \(y^2 - 5y - 336 = 0\)

    \(y = 21, -16\)

    \(\Rightarrow x = -4, 6, 1 \pm 2\sqrt{3}i\)

    120 and 121 are left as an exercise.

  1. Let the speed be \(x\) km/hour. Then, from the statement

    \(\frac{800}{x} = \frac{800}{x + 40} + \frac{2}{3}\)

    Solving we get \(x = 200\) km/hour

  2. Let width be \(w\) meter. Thus,

    \((w + 8)(w - 2) = 119\)

    \(w^2 + 6w - 135 = 0\)

    \(w = 9, -15\) but width cannot be negative.

    Length is \(11\) m.

  3. Equivalent equation is \(-x^2 + 3x + 4 = 0\) and roots are \(-1, 4\).

    Since coefficient of \(x^2\) is -ve the expression will be +ve if \(x\) lies between the root.

    Therefore, for \(-x^2 + 3x + 4 > 0\) the range is \(]-1, 4[\) which is fully open interval.

  4. \(5x - 1 < (x + 1)^2 \Rightarrow x^2 - 3x + 2 > 0\)

    Roots of equivalent equation \(x^2 - 3x + 2 = 0\) are \(x = 2, 1\)

    Since coefficient of \(x^2\) is positive, \(x\) must lie outside the range of \([1, 2]\) for the expression to be positive.

    Now considering, \((x + 1)^2 < 7x - 3\)

    \(x^2 - 5x + 4 < 0\)

    Roots of the equivalent equation \(x^2 - 5x + 4 = 0\) are \(x = 1, 4\) and for expression to be negative \(x\) must lie inside the open interval \(]1, 4[\).

    Therefore, the only integral value satisfying the original expression is \(3\).

  5. \(\frac{8x^2 + 16x - 51}{(2x - 3)(x + 4)} > 3\)

    \(\Rightarrow \frac{2x^2 + x - 15}{2x^2 + 5x - 12} > 0\)

    \(2x^2 + x - 15 = 0\) has roots \(x = -3 , \frac{5}{2}\)

    \(2x^2 + 5x - 12 = 0\) has roots \(x = -4, \frac{3}{2}\)

    Thus, the inequality will hold true for \(x < -4\) and \(-3 < x < \frac{3}{2}\) and \(x > \frac{5}{2}\)

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

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

    Since \(x\) is real, the discriminant will be greater that \(0\)

    \(\Rightarrow 9(y + 1)^2 - 16(y - 1)^2 \ge 0\)

    \(-7y^2 + 50y - 7 \ge 0\)

    The roots are \(7\) and \(\frac{1}{7}\)

    Since coefficient of \(y^2\) is negative, for the expression to be positive \(y\) has to lie between the open interval formed by its roots i.e. \(]\frac{1}{7}, 7[\)

  7. Let \(y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}\)

    \((y - 1)x^2 + 2(y - 17)x + (71 - y) = 0\)

    Since \(x\) is real, the discriminant will be greater that \(0\)

    \(\Rightarrow 4(y - 17)^2 - 4(y - 1)(71 - 7y) \ge 0\)

    \(y^2 - 14y + 45 \ge 0\)

    Its roots are \(5\) and \(9\)

    Since coefficient of \(y^2\) is positive, therefore for the expression to be positive \(y\) has to lie outside the open interval formed by its roots. Thus, the expression has no value between \(5\) and \(9\).

  8. Let \(y = \frac{4x^2 + 36x + 9}{12x^2 + 8x + 1}\)

    \(4(3y - 1)x^2 + 4(2y - 9)x + y - 9 = 0\)

    Since \(x\) is real, the discriminant will be greater that \(0\)

    \(16(2y - 9)^2 - 16(3y - 1)(y - 1) \ge 0\)

    \(y^2 - 8y + 72 \ge 0\)

    Corresponding equation is \(y^2 - 8y + 72 = 0\)

    \(D = 64 - 288 = -224 < 0\)

    Since coefficient of \(y^2\) is positive and discriminant is less than \(0\) therefore \(y^2 - 8y + 72 \ge 0\) holds true for all value of \(y\). Therefore, the expression can take any value.

  9. Let \(y = \frac{(x - a)(x - c)}{x -b}\)

    \(x^2 - (a + c + y)x + ac + yb = 0\)

    Since \(x\) is real, the discriminant will be greater that \(0\)

    \((a + c + y)^2 - 4(ac + yb) \ge 0\)

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

    Corresponding equation is \(y^2 + 2(a + c - 2b)y + (a - c)^2 = 0\)

    Discriminant of above equation is \(D = -16(a - b)(b - c)\)

    If \(a > b > c\) then \(D < 0\) and if \(a < b < c\) then also \(D < 0\)

    Since coefficient of \(y^2\) is positive and \(D < 0\) the expression \(y^2 + 2(a + c - 2b)y + (a - c)^2 \ge 0\) is true for all real values of \(y\).

    Therefore, the given expression is capable of holding any value for the given conditions.

  10. Given \(x + y =\) constant \(= k\) (say)

    Let \(z = xy\), then

    \(z = x(k - x) \Rightarrow x^2 - kx + z = 0\)

    Since \(x\) is real, \(D \ge 0\) for the above equation.

    \(k^2 - 4z \ge 0 \Rightarrow z \le \frac{k^2}{4}\)

    Hence, the maximum value of \(z = \frac{k^2}{4}\)

    Thus, \(x^2 - kx + \frac{k^2}{4} = 0 \Rightarrow \left(x - \frac{k}{2}\right)^2 = 0 \Rightarrow x = \frac{k}{2}\)

    \(\therefore y = \frac{k}{2}\) and thus \(xy\) is maximum when \(x = y\)

  11. Let \(y = 3 - 6x - 8x^2 \Rightarrow 8x^2 + 6x + y - 3 = 0\)

    Since \(x\) is real, \(D \ge 0\) for the above equation.

    \(\Rightarrow 36 - 32(y - 3) \ge 0\)

    \(y \le \frac{33}{8}\). Hence, maximum value of \(y = \frac{33}{8}\)

    \(\Rightarrow 64x^2 + 48x + 9 = 0\)

    \((8x + 3)^2 = 0 \Rightarrow x = -\frac{3}{8}\)

  12. Let \(y = \frac{12x}{4x^2 + 9}\)

    \(\Rightarrow 4yx^2 - 12x + 9y = 0\)

    Since \(x\) is real, \(D \ge 0\) for the above equation.

    \(\Rightarrow 144 - 144y^2 \ge 0\)

    \(\Rightarrow y^2 \le 1\)

    \(\Rightarrow -1 \le y \le 1 \Leftrightarrow |y| \le 1 \Leftrightarrow\left|\frac{12x}{4x^2 + 9}\right| \le 1\)

    Now, \(\left|\frac{12x}{4x^2 + 9}\right| = 1 \Leftrightarrow 4|x|^2 - 12|x| + 9 = 0\)

    \((2|x| - 3)^2 = 0 \Rightarrow |x| = \frac{3}{2}\)

  13. \(x^2 + 9y^2 - 4x + 3 = 0\)

    Since \(x\) is real, \(D \ge 0\) for the above equation.

    \(\Rightarrow (-4)^2 - 4(9y^2 + 3) \ge 0\)

    \(\Rightarrow 9y^2 - 1 \le 0 \Leftrightarrow y^2 \le \frac{1}{9}\)

    \(\Rightarrow -\frac{1}{3} \le y \le \frac{1}{3}\)

    The given equation can also be written as

    \(9y^2 + x^2 - 4x + 3 = 0\)

    Since \(y\) is real, \(D \ge 0\) for the above equation.

    \(-36(x^2 - 4x + 3) \ge 0\)

    \(x^2 - 4x + 3 \le 0\)

    Since coefficient of \(x^2\) is positive, it must lie between its root for the above expression to be negative. Therefore, \(x\) must lie between \(1\) and \(3\).

  14. Given expression is \(x^2 - ax + 1 - 2a^2 > 0\)

    Since \(x\) is real the discriminant of the corresponding equation has to be negative for it to be positive for all values of \(x\).

    \(a^2 - 4(1 - 2a^2) < 0 \Leftrightarrow 9a^2 \le 4\)

    \(-\frac{2}{3} < a < \frac{2}{3}\)

  15. Let \(\alpha\) be a common factor, therefore it will satisfy both the equations.

    \(\alpha^2 - 11\alpha + a = 0\) and \(\alpha^2 - 14\alpha + 2a = 0\)

    By cross-multiplication

    \(\frac{\alpha^2}{-22a + 14x} = \frac{\alpha}{a - 2a} = \frac{1}{-14 + 11}\)

    \(\frac{\alpha^2}{-8a} = \frac{\alpha}{-a} = -\frac{1}{3}\)

    From first two we have \(\alpha = 8\) and from last two we have \(\alpha = \frac{a}{3}\)

    \(\therefore a = 24\)

  16. \(y = mx\) is a factor of \(ax^2 + bxy + cy^2\) means \(ax^2 + bxy + cy^2\) will be zero when \(y = mx\)

    \(ax^2 + bx.mx + cm^2x^2 = 0 \Rightarrow cm^2 + bm + a = 0\)

    Similarly, \(a_1m^2 + b_1m + c_1 = 0\) since \(my - x\) is a factor of \(a_1x^2 + b_1xy + c_1y^2\)

    Solving these two equations in \(m\) by cross-multiplication

    \(\frac{m^2}{bc_1 - ab_1} = \frac{m}{aa_1 - cc_1} = \frac{1}{cb_1 - ba_1}\)

    From first two we get, \(m = \frac{bc_1 - ab_1}{aa_1 - cc_1}\)

    and from last two we get, \(m = \frac{aa_1 - cc_1}{cb_1 - ba_1}\)

    Equating the two values of \(m\) obtained we get

    \((bc_1 - ab_1)(cb_1 - ba_1) = (aa_1 - cc_1)^2\)

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

    \(abc + 2fgh - af^2 - bg^2 - ch^2 = 0\) and \(h^2 - ab > 0\)

    Given expression is \(2x^2 + mxy + 3y^2 - 5y - 2\)

    Here, \(a = 2, h = \frac{m}{2}, b = 3, g = 0, f = \frac{-5}{2}, c = -2\)

    \(h^2 - ab = \frac{m^2}{4} - 6 > 0\Rightarrow m^2 > 24\)

    Applying the second condition

    \(-12 - \frac{25}{2} + \frac{m^2}{2} = 0\)

    \(m^2 = 49 \therefore m = \pm 7\)

  18. Given expression is \(ax^2 + by^2 + cz^2 + 2ayz + 2bzx + 2cxy\)

    \(= z^2\left[a\left(\frac{x}{z}\right)^2 + b\left(\frac{y}{z}\right)^2 + c + 2a\frac{y}{z} + 2b\frac{x}{z} + 2c\frac{xy}{z^2}\right]\)

    \(= z^2(aX^2 + bY^2 + c + 2aY + 2bX + 2cXY)\) where \(X = \frac{x}{z}, Y = \frac{y}{z}\)

    Now this will resolve in linear factors if

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

    \(a^3 + b^3 + c^3 = 3abc\)

  19. Given expression is \(2x^2 - y^2 - x + xy + 2y -1\)

    Corresponding equation is \(2x^2 - y^2 - x + xy + 2y -1 = 0\)

    \(x = \frac{1 - y \pm \sqrt{(1 - y)^2 + 8(y^2 - 2y + 1)}}{4}\)

    \(x = 1 - y, -\frac{1 - y}{2}\)

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

  20. Corresponding quadratic equation is \(x^2 + 2(a + b + c)x + 3(ab + bc + ca) = 0\)

    It will be a perfect square if its discriminant is zero.

    \(\Rightarrow 4(a + b + c)^2 - 4.3(ab + bc + ca) = 0\)

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

    \(\Rightarrow \frac{1}{2}(a - b)^2(b - c)^2(c - a)^2 = 0\)

    \(\Rightarrow a = b = c\)

  21. Discriminant of the given equation is \(D = 36 - 72 < 0\)

    Now since coefficient of \(x^2\) is less than zero the expression is always positive.

  22. \(8x - 15 - x^2 > 0\)

    \(\Rightarrow x^2 - 8x + 15 < 0\)

    \((x - 3)(x - 5) < 0\)

    The above is true if \(x\) lies in the open interval \(]3, 5[\).

  23. \(-x^2 + 5x - 4 > 0\)

    \(\Rightarrow x^2 - 5x + 4 < 0\)

    \((x - 4)(x - 1) < 0\)

    The above is true if \(x\) lies in the open interval \(]1, 4[\).

  24. \(x^2 + 6x - 27 > 0\)

    \(\Rightarrow (x + 9)(x - 3) > 0\)

    This is true if \(x < -9\) or \(x > 3\)

  25. \(\frac{4x}{x^2 + 3} \le 1\)

    \(\Rightarrow x^2 + 3 \le 4x\)

    \(\Rightarrow x^2 - 4x + 3 \le 0\)

    \((x - 3)(x - 1) le 0\)

    This is true for closed interval \([1, 3]\).

  26. \(x^2 - 3x + 2 > 0\)

    \((x - 2)(x - 1) > 0\)

    This is true for \(x > 2\) or \(x < 1\)

    \(x^2 - 3x - 4 \le 0\)

    \((x - 4)(x + 1) \le 0\)

    This is true for \(-1 \le x \le 4\)

    Thus values of \(x\) which satisfy both are \(-1 \le x < 1\) and \(2 < x \le 4\).

  27. Since roots of \(ax^2 + bx + c\) are imaginary, therefore discriminant is negative. \(\Rightarrow b^2 - 4ac < 0\)

    Discriminant of \(a^2x^2 + abx + ac\) is:

    \(D = a^2b^2 - 4a^3c = a^2(b^2 - 4ac) < 0\)

    But coefficient of the expression is positive hence it will be always positive.

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

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

    Since \(x\) is real discriminant will be greater or equal to zero.

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

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

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

    Roots of corresponding equation are \(\frac{1}{3}, 3\). Since coefficient of \(y^2\) is negative, for above to be true \(y\) must lie between \(\frac{1}{3}\) and \(3\).

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

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

    Since \(x\) is real discriminant will be greater or equal to zero.

    \(9(y + 1)^2 - 16(y - 1)^2 \ge 0\)

    \(9y^2 + 18y + 9 - 16y^2 + 32y - 16 \ge 0\)

    \(-7y^2 + 50y - 7 \ge 0\)

    Roots of the corresponding equation are \(\frac{1}{7}, 7\). Since coefficients of \(y^2\) is negative, for the above to be true \(y\) must lie between \(\frac{1}{7}\) and \(7\).