# 44. Quadratic Equations Solutions Part 3¶

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 $$7$$.