# 53. Quadratic Equations Solutions Part 4

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.