# 48. Quadratic Equations Solutions Part 5¶

1. Product of roots $$= \frac{k + 2}{k} = \frac{c}{a}$$

$$\Rightarrow k = \frac{2a}{c - a}$$

Sum of roots $$= \frac{k + 1}{k} + \frac{k + 2}{k + 1} = -\frac{b}{a}$$

Substituting for $$k$$

$$\frac{c + a}{2a} + \frac{2c}{c + a} = - \frac{b}{a}$$

$$\frac{(a + c)^2 + 4ac}{2a(a + c)} = -\frac{b}{a}$$

$$a(a + c)^2 + 4a^2c = -2abc - 2a^2b$$

$$(a + c)^2 + 4ac = -2bc - 2ab$$

$$(a + b + c)^2 = b^2 - 4ac$$

2. Given, $$f(x) = ax^2 + bx + c$$ and that $$\alpha,\beta$$ are the roots of the equation $$px^2 + qx + r = 0$$

$$\alpha + \beta = -\frac{q}{p}$$ and $$\alpha\beta = \frac{r}{p}$$

Now $$f(\alpha)f(\beta) = (a\alpha^2 + b\alpha + c)(a\beta^2 + b\beta + c)$$

$$= a^2\alpha^2\beta^2 + b^2\alpha\beta + c^2 + ab\alpha\beta(\alpha + \beta) + ac(\alpha^2 + \beta^2) + bc(\alpha + \beta)$$

$$= a^2\frac{r^2}{p^2} + b^2\frac{r}{p} + c^2 - ab\frac{r}{p}\frac{q}{p} + ac\left(\frac{q^2}{p^2} - \frac{2r}{p}\right) - bc\frac{q}{p}$$

$$= \frac{1}{p^2}[a^2r^2 + b^2rp + c^2p^2 - abrq + acq^2 - 2acrp - bcqp]$$

$$= \frac{1}{p^2}[(cp - ar)^2 + b^2rp - bcqp - abrq + acq^2]$$

$$= \frac{1}{p^2}[(cp - ar)^2 - (bp - aq)(cq - br)]$$

Now since $$\alpha, \beta$$ are the roots of the equation $$px^2 + qx + r = 0$$

Therefore, if $$ax^2 + bx + c = 0$$ and $$px^2 + qx + r = 0$$ have to have a common root then it has to be either $$\alpha$$ or $$\beta$$.

$$f(\alpha) = 0$$ or $$f(\beta) = 0 \therefore f(\alpha)f(\beta) = 0$$

$$\Rightarrow (cp - ar)^2 - (bp - aq)(cq - br) = 0$$

$$\therefore bp - aq, cp - ar, cq - br$$ are in G. P.

3. From the given equations it follows that $$q$$ and $$r$$ are roots of the equation

$$a(p + x)^2 + 2bpx + c = 0 \Rightarrow ax^2 + 2(a + b)px + c = 0$$

Product of roots $$qr = \frac{ap^2 + c}{a} = p^2 + \frac{c}{a}$$

4. Since $$\alpha, \beta$$ are the roots of the equation $$x^2 - px - (p + c) = 0$$

$$\alpha + \beta = p$$ and $$\alpha + \beta = -(p + c)$$

Now $$(\alpha + 1)(\beta + 1) = -p - c + p + 1 = 1 - c$$

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

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

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

5. $$\alpha, \beta$$ are the roots of the equation $$x^2 + px + q = 0$$

$$\therefore \alpha + \beta = -p$$ and $$\alpha\beta = q$$

Since $$\alpha, \beta$$ are the roots of the equation $$x^2{2n} + p^nx^n + q^n = 0$$

Substituting it follows that $$\alpha^n, \beta^n$$ are the roots of the equation $$y^2 + p^ny + q^n = 0$$

$$\therefore \alpha^n + \beta^n = -p^n$$ and $$\alpha^n\beta^n = q^n$$

$$(\alpha + \beta)^n = (-p)^n = p^n [\because~\text{n is even}]$$

Thus, $$\alpha^n + \beta^n + (\alpha + \beta)^n = 0$$

Dividing by $$\beta^n$$ we have

$$\left(\frac{\alpha}{\beta}\right)^n + 1 + \left(\frac{\alpha}{\beta} + 1\right)^n = 0$$

Dividing by $$\alpha^n$$ we have

$$\left(\frac{\beta}{\alpha}\right)^n + 1 + \left(\frac{\beta}{\alpha} + 1\right)^n = 0$$

From last two equations it is evident that $$\frac{\alpha}{\beta}$$ and $$\frac{\beta}{\alpha}$$ are roots of the equation $$x^n + 1 + (x + 1)^n = 0$$

6. Following from previous problem: $$\alpha^n + \beta^n = -p^n$$

Also since $$\frac{\alpha}{\beta}$$ and $$\frac{\beta}{\alpha}$$ are roots of the equation $$x^n + 1 + (x + 1)^n = 0$$

$$\alpha^n + \beta^n = -(\alpha + \beta)^n = -(-p)^n$$

From these equations we have $$-p^n = -(-p)^n \Rightarrow p^n = (-p)^n$$

Therefore, $$n$$ must be even.

7. Let $$\alpha$$ and $$\beta$$ are the roots of the given equation.

Since roots are real and distinct $$D > 0 \Rightarrow a^2 - 4b > 0 \Rightarrow b < \frac{a^2}{4}$$

Again it is given that $$|\alpha - \beta| < c \Rightarrow (\alpha - \beta)^2 < c^2$$

$$(\alpha + \beta)^2 - 4\alpha\beta < c^2 \Rightarrow a^2 - 4b < c^2 \Rightarrow 4b > a^2 - c^2$$

$$\Rightarrow \frac{a^2 - c^2}{4} < b < \frac{a^2}{4}$$

8. Given, $$ax^2 + bx + c - p = 0$$ for two integral values of $$x$$ say $$\alpha$$ and $$\beta$$.

Then, $$\alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c - p}{a}$$

If possible, let $$ax^2 + bx + c - 2p = 0$$ for some integer $$k$$.

$$ak^2 + bk + c - p = p \Rightarrow k^2 - (\alpha + \beta)k + \alpha\beta = \frac{p}{a}$$

$$(k - \alpha)(k - \beta) =~\text{an integer}~= \frac{p}{a}$$

But since $$p$$ is prime this cannot hold true unless $$a = p$$ or $$a = 1$$

$$a = p [\because a > 1]$$

$$(k - \alpha)(k - \beta) = 1$$ which implies that $$k - \alpha = k - \beta = 1$$

which is not possible since $$\alpha \ne \beta$$

Thus, we have a contradiction. Hence, $$ax^2 + bx + c \ne 2p$$ for any integral value of $$x$$.

9. $$\alpha + \beta = -p, \alpha\beta = q, \alpha^4 + \beta^4 = r, \alpha^4\beta^4 = s$$

Let $$D$$ be the discriminant of $$x^2 - 4qx + 2q^2 - r = 0$$ then

$$D = 16q^2 - 4(2q^2 - r) = 8q^2 + 4r = 8\alpha^2\beta^2 + 4(\alpha^4 + \beta^4) = 4(\alpha^2 + \beta^2)^2$$

$$D \ge 0$$ hence roots of the third equation are always real.

10. $$\alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c}{a}$$

$$\alpha_1 - \beta = -\frac{b_1}{a_1}$$ and $$-\alpha_1\beta = \frac{c_1}{a_1}$$

$$\alpha + \alpha_1 = -\left(\frac{b}{a} + \frac{b_1}{a_1}\right)$$

Also, dividing $$\alpha + \beta$$ by $$\alpha\beta$$

$$\frac{1}{\beta} + \frac{1}{\alpha} = -\frac{b}{c}$$

Similarly, dividing $$\alpha_1 - \beta$$ by $$-\alpha_1\beta$$

$$\frac{1}{\alpha_1} - \frac{1}{\beta} = -\frac{b_1}{c_1}$$

Thus, $$\frac{1}{\alpha} + \frac{1}{\alpha_1} = -\left(\frac{b}{c} + \frac{b_1}{c_1}\right)$$

Equation whose roots are $$\alpha$$ and $$\alpha_1$$ is

$$x^2 - (\alpha + \alpha_1)x + \alpha\alpha_1 = 0$$

$$\frac{x^2}{-(\alpha + \alpha_1)} + x - \frac{\alpha\alpha_1}{\alpha + \alpha_1} = 0$$

$$\frac{x^2}{\frac{b}{a} + \frac{b_1}{a_1}} + x + \frac{1}{\frac{b}{c} + \frac{b_1}{c_1}} = 0$$

11. Let $$\alpha$$ and $$\beta$$ be roots of such quadratic equation given by $$x^2 + px + q = 0$$

$$\alpha + \beta = -p$$ and $$\alpha\beta = q$$

Now quadratic equation whose roots are $$\alpha^2$$ and $$\beta^2$$ is

$$x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0$$

$$x^2 - (p^2 - 2q)x + q^2 = 0$$

But the equation remains unchanged, therefore,

$$\frac{1}{1} = \frac{p}{p^2 - 2q} = \frac{q}{q^2}$$

$$\Rightarrow q = q^2 \Rightarrow q(q - 1) = 0 \Rightarrow q = 0, 1$$

If $$q = 0 \Rightarrow p = 0, -1$$

and if $$q = 1 \Rightarrow p = -2, 1$$

Thus, four such quadratic equations are possible.

12. Given $$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$$ are in A. P. and $$a, b, c$$ are in G. P.

Equations $$ax^2 + 2bx + c = 0$$ and $$dx^2 + 2ex + f = 0$$ will have a common root if

$$\frac{2(bf - ec)}{cd - af} = \frac{cd - af}{2(ae - bd)}$$

$$4(bf - ec)(ae - bd) = (cd - af)^2$$

$$4\left[\left(\frac{f}{c} - \frac{e}{b}\right)bc\right]\left[\left(\frac{e}{b} - \frac{d}{a}\right)ab\right] = \left(\frac{d}{a} - \frac{a}{f}\right)^2a^2c^2$$

$$4k.k.b^2 = 4k^2ac$$ where $$k$$ is the c.d. of the A. P.

$$b^2 = ac$$ which is true because $$a, b, c$$ are in G. P.

13. Let $$\alpha$$ be the common root and $$\beta_1$$ another root of $$x^2 + ax + 12 = 0, \beta_2$$ be another root of $$x^2 + bx + 15 = 0$$ and $$\beta_3$$ be a root of $$x^2 + (a + b)x + 36 = 0.$$

$$\alpha + \beta_1 = -a$$ and $$\alpha\beta_1 = 12$$

$$\alpha + \beta_2 = -b$$ and $$\alpha\beta_2 = 15$$

$$\alpha + \beta_3 = -(a + b)$$ and $$\alpha\beta_3 = 36$$

Thus, $$2\alpha + \beta_1 + \beta_2 = \alpha + \beta_3 \Rightarrow \alpha = \beta_3 - \beta_1 - \beta_2$$

and $$\alpha(\beta_3 - \beta_1 - \beta_2) = 36 - 12 - 15 = 9$$

$$\Rightarrow \alpha^2 = 9 \Rightarrow \alpha = \pm 3$$ but $$\alpha > 0 \Rightarrow \alpha = 3$$

$$\Rightarrow \beta_1 = 4, \beta_2 = 5, \beta_3 = 12$$

14. Given $$m(ax^2 + 2bx + c) + px^2 + 1qx + r = n(x + k)^2$$

Equating coefficients for powers of $$x$$, we get

$$ma + p = n, mb + q = nk, mc + r = nk^2$$

$$\Rightarrow m(ak - b) + pk - q = 0 \Rightarrow m = -\frac{pk - q}{ak - b}$$

$$\Rightarrow m(bk - c) + qk - r = 0 \Rightarrow m = -\frac{qk - r}{bk - c}$$

Equating values for $$m$$

$$(ak - b)(qk - r) = (pk - q)(bk - c)$$

15. Given equation is $$x^3 - x^2 + \beta x + \gamma = 0$$

Let it roots $$x_1, x_2, x_3$$ be $$a - d, a, a + d$$ respectively.

$$a - d + a + a + d = 1 \Rightarrow a = \frac{1}{3}$$

$$(a - d)a + a(a + d) + (a - d)(a + d) = \beta \Rightarrow 3a^2 - d^2 = \beta \Rightarrow 1 - 3\beta = 3d^2$$

$$(a - d)a(a + d) = \gamma \Rightarrow a(a^2 - d^2) = \gamma \Rightarrow 1 + 27\gamma = 9d^2$$

Since $$d$$ is real $$\therefore 1 - 3\beta \ge 0 \Rightarrow \beta \le \frac{1}{3}$$

$$1 + 27\gamma \ge 0 \Rightarrow \gamma \ge -\frac{1}{27}$$

16. Let $$\alpha$$ be a common root, then

$$\alpha^3 + 3p\alpha^2 + 3q\alpha + r = 0$$ … (1) and $$\alpha^2 + 2p\alpha + q = 0$$ … (2)

$$(1) - \alpha (2)$$ gives us $$\Rightarrow p\alpha^2 + 2q\alpha + r = 0$$ … (3)

By cross multiplication between (2) and (3)

$$\frac{\alpha^2}{2(pr - q^2)} = \frac{\alpha}{pq - r} = \frac{1}{2(q - p^2)}$$

Equating for values of $$\alpha$$ we get the desired condition.

17. Let $$\alpha$$ be a common root, then

$$\alpha^3 + 2a\alpha^2 + 3b\alpha + c = 0$$ … (1) and $$\alpha^3 + a\alpha^2 + 2b\alpha = 0$$ … (2)

Since $$c \ne 0,$$ therefore $$\alpha = 0$$ cannot be a common root. Therefore, from (2)

$$\alpha^2 + a\alpha + 2b = 0$$ … (3)

$$(1) - \alpha (2) \Rightarrow a\alpha^2 + b\alpha + c = 0$$ … (4)

Solving (3) and (4) by cross-multiplication yields the desired result.

18. Given equation is $$x^3 + ax + b = 0$$ and $$\alpha, \beta, \gamma$$ be its real roots. Then we have

$$\alpha + \beta + \gamma = 0$$ … (1) $$\alpha\beta + \beta\gamma + \alpha\gamma = a$$ … (2) $$\alpha\beta\gamma = -b$$

Let $$y = (\alpha - \beta)^2,$$ then $$y = (\alpha + \beta)^2 - 4\alpha\beta$$

$$y = \gamma^2 + \frac{4b}{\gamma}$$ $$\Rightarrow \gamma^3 - y\gamma + 4b = 0$$

Also, $$\gamma$$ is a root of the original equation.

$$\gamma^3 + a\gamma + b = 0$$

$$(a + y)\gamma - 3b = 0 \Rightarrow \gamma = \frac{3b}{a + y}$$

$$\Rightarrow \frac{27b^3}{(a + y)^3} + a\left(\frac{3b}{a + y}\right) + b = 0$$

$$y^3 + 6ay^2 + 9a^2y + 4a^3 + 27b^2 = 0$$

We will get same equation if we would have chosen $$y = (\beta - \alpha)^2$$ or $$y = (\gamma - \alpha)^2$$

Hence, product of roots $$-(4a^3 + 27b^2) = (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 \ge 0$$

$$\therefore 4a^3 + 27b^2 \le 0$$

19. $$\alpha$$ is a root of the equation $$ax^2 + bx + c = 0$$

$$\therefore a\alpha^2 + b\alpha + c = 0$$

Similarly, $$-a\beta^2 + b\beta + c = 0$$

Let $$f(x) = \frac{a}{2}x^2 + bx + c = 0$$

$$f(\alpha) = -\frac{a}{2}\alpha^2$$

$$f(\beta) = \frac{3}{2}\beta^2$$

$$\therefore f(\alpha)f(\beta) = -\frac{3}{4}a^2\alpha^2\beta^2 < 0 [\because \alpha,\beta \ne 0]$$

$$\therefore f(\alpha)$$ and $$f(\beta)$$ have opposite signs. Therefore, $$f(x)$$ will have exactly one root between $$\alpha$$ and $$\beta$$.

20. Let $$f(x) = ax^2 + bx + c = 0$$

Since equation $$ax^2 + bx + c = 0$$ i.e. equation $$f(x) = 0$$ has no real root, therefore, $$f(x)$$ will have same sign for real values of $$x$$.

$$\therefore f(1)f(0) > 0 \Rightarrow (a + b + c)c > 0$$

21. Let $$f(x) = (x - a)(x - c) + \lambda (x - b)(x - d)$$

Given $$a > b > c > d$$

Now $$f(b) = (b - a)(b - c) < 0$$

and $$f(d) = (d - a)(d - c) > 0$$

Since $$f(b)$$ and $$f(d)$$ have opposite signs, therefore equation $$f(x) = 0$$ will have one real root between $$b$$ and $$d$$.

Since one root is real and $$a, b, c, d, \lambda$$ are all real the other root will also be real.

22. Let $$f'(x) = ax^2 + bx + c,$$ then

$$f(x) = a\frac{x^3}{3} + b\frac{x^2}{2} + cx + k = \frac{2ax^3 + 3bx^2 + 4cx + 6k}{6}$$

$$f(1) = \frac{2a + 3b + 6c + 6k}{6} = k$$

Again, $$f(0) = k$$

Thus, $$f(0) = f(1)$$ hence equation will have at least one root between $$0$$ and $$1$$ which implies that it will have a real root between $$0$$ and $$2$$.

23. Let $$f(x) = \int (1 + \cos^8x)(ax^2 + bx + c)dx$$ then $$f'(x) = (1 + \cos^8x)(ax^2 + bx + c)$$

Given, $$\int_0^1 (1 + \cos^8 x)(ax^2 + bx + c)dx = \int_0^2 (1 + \cos^8 x)(ax^2 + bx + c)dx$$

$$f(1) - f(0) = f(2) - f(0) \Rightarrow f(1) = f(2)$$

Therefore, equation $$f(x) = 0$$ has at least one root between $$1$$ and $$2$$ which implies that $$ax^2 + bx + c$$ has a root between these two limits as $$1 + \cos^8x \ne 0$$

24. Given equation $$f(x) - x = 0$$ has non-real roots where $$f(x) = ax^2 + bx + c$$ is a continuous function.

$$\therefore f(x) - x$$ has same sign for all $$x \in R$$

Let $$f(x) - x > 0~\forall~x \in R$$

$$\Rightarrow f(f(x)) - f(x) > 0~\forall~x\in R$$

$$\Rightarrow f(f(x)) - x = f(f(x)) - f(x) + f(x) - x > 0~\forall~x\in R$$

Hence it has no real roots. Similarly, it can be proven for $$f(x) - x < 0$$

25. Let $$f(x) = ax^2 - bx + c = 0$$ and that $$\alpha, \beta$$ be its roots.

Then, $$f(x) = a(x - \alpha)(x - \beta)$$

Given $$\alpha \ne \beta, 0 < \alpha < 1, 0 < \beta < 1$$ and $$a, b, c \in N$$

Since quadratic equation has both roots between $$0$$ and $$1$$, therefore

$$f(0)f(1) > 0$$

$$f(0)f(1) = c(a - b + c) =$$ an integer

Thus, $$f(0)f(1) \ge 1 \Rightarrow a\alpha(1 - \alpha)a\beta(1 - \beta) = a^2\alpha\beta(1 - \alpha)(1 - \beta)$$

Let $$y = \alpha(1 - \alpha) \Rightarrow \alpha^2 - \alpha + y = 0$$

Since $$\alpha$$ is real $$\therefore 1 - 4y \ge 0 \Rightarrow y \le \frac{1}{4} \Rightarrow \alpha = \frac{1}{2}~\text{max value}$$

Similarly, maximum value of $$\beta = \frac{1}{2}$$

Maximum value of $$\therefore f(0)f(1) < \frac{a^2}{16} > 1$$

$$\Rightarrow a > 4 \Rightarrow a = 5$$ [least integral value]

26. Proceeding from previous question

$$b^2 - 4ac > 0 \Rightarrow b^2 > 4.5.1 [\because c \ge 1] \Rightarrow b = 5$$

$$\log_5(abc) \ge 2$$

27. Given equation is $$ax^2 + bx + 6 = 0$$. Let $$f(x) = ax^2 + bx + 6$$

Since the equation has imaginary roots or real and equal roots its sign will never change.

$$f(0) = 6 > 0$$

$$\therefore f(x) \ge 0$$ for all real $$x$$

$$f(3) \ge 0 \Rightarrow 9a + 3b + 6 \ge 0$$

$$3a + b \ge -2$$ and hence least value is $$-2$$

28. Let $$\alpha, \beta, \gamma$$ be the roots of the equation. Then,

$$f(x) = 2x^3 - \frac{\alpha + \beta + \gamma}{2}x^2 + \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{2}x - \frac{\alpha\beta\gamma}{2} = 0$$

Clearly, all roots have to be negative for signs to be satisfied as $$a, b > 0$$

$$f(0) = 4 > 0 \therefore f(1) > 0$$ because sign of $$f(x)$$ will not change for all $$x$$.

$$2 + a + b + 4 > 0 \Rightarrow a + b > - 6$$

29. $$f(x) = x^3 + 2x^2 + x + 5 = 0$$ and $$f'(x) = 3x^2 + 4x + 1$$ which has roots $$-1$$ and $$-\frac{1}{3}$$.

$$f(0) = 5$$ and $$f(x)$$ is increasing in $$(0, \infty)$$ therefore it will have no root in $$[0, \infty[$$

$$f(-2) = 3 > 0$$ and $$f(-3) = -7 < 0$$

Since $$f(-2)$$ and $$f(-3)$$ are of opposite sign therefore equation $$f(x) = 0$$ will have one root between $$-2$$ and $$-3$$ and this will be only one root as $$f(x)$$ is increasing in $$]-\infty, -1]$$

$$[\alpha] = -3$$

30. Given equation is $$(x^2 + 2)^2 + 8x^2 = 6x(x^2 + 2)$$

Let $$y = x^2 + 2$$ then above equation becomes $$y^2 + 8x^2 = 6xy$$

$$\Rightarrow y = 4x, 2x$$

If $$y = 4x \Rightarrow x^2 - 4x + 2 = 0 \Rightarrow x = 2 \pm \sqrt{2}$$

If $$y = 2x \Rightarrow x^2 - 2x + 2 = 0 \Rightarrow x = 1 \pm i$$

31. Given equation is $$3x^3 = (x^2 + \sqrt{18}x + \sqrt{32})(x^2 - \sqrt{18}x - \sqrt{32}) - 4x^2$$

$$3x^3 = x^4 - (\sqrt{18}x + \sqrt{32})^2 - 4x^2$$

$$x^2(3x + 4) = x^4 - 2(3x + 4)^2$$

$$x^2y = x^4 - 2y^2$$ where $$y = 3x + 4$$

$$\Rightarrow y = -x^2, \frac{x^2}{2}$$

If $$y = -x^2 \Rightarrow x = \frac{-3 \pm \sqrt{7}i}{2}$$

and if $$y = \frac{x^2}{2} \Rightarrow x = 3 \pm \sqrt{17}$$

32. Clearly, $$(15 + 4\sqrt{14})^t(15 - 4\sqrt{14})^t = (225 - 224)^t = 1$$

Let $$(15 + 4\sqrt{14})^t = y,$$ then $$(15 - 4\sqrt{14})^t = \frac{1}{y}$$

Substituting for the given equation

$$y + \frac{1}{y} = 30 \Rightarrow y^2 - 30y + 1 = 0$$

$$y = 15 \pm 4\sqrt{14}$$

If $$y = 15 + 4\sqrt{14} \Rightarrow t = 1$$

$$\therefore x^2 - 2|x| = 1 \Rightarrow |x|^2 - 2|x| - 1 = 0$$

$$\Rightarrow |x| = 1 + \sqrt{2} \therefore x = \pm(1 + \sqrt{2})$$

If $$y = 15 - 4\sqrt{14} \Rightarrow t = -1$$

$$\Rightarrow |x|^2 - 2|x| + 1 = 0 \Rightarrow |x| = 1 \Rightarrow x = \pm 1$$

33. Given equation is $$x^2 - 2a|x - a| - 3a^2 = 0$$

When $$a = 0$$ equation becomes $$x^2 = 0 \Rightarrow x = 0$$

Let $$a < 0$$.

Case I: When $$x < a$$ then equation becomes

$$x^2 + 2a(x - a) - 3a^2 = 0 \Rightarrow x^2 + 2ax - 5a^2 = 0 \Rightarrow x = -a \pm \sqrt{6}a$$

Since $$x < a, x = -a - \sqrt{6}a$$ is not acceptable.

Case II: When $$x > a$$ the equation becomes

$$x^2 - 2ax - a^2 = 0 \Rightarrow x = a \pm \sqrt{2}a$$

Since $$x > a, x = a + \sqrt{2}a$$ is not acceptable.

Clearly, $$x = a$$ does not satisfy the equation.

34. $$x^2 - x - 6 = 0 \Rightarrow x = -2, 3$$

Case I: When $$x < -2$$ or $$x > 3$$ then $$x^2 - x - 6 > 0$$

Then equation becomes $$x^2 - x - 6 = x + 2 \Rightarrow x^2 - 2x - 8 = 0$$

$$x = -2, 4$$ but $$x = -2$$ is not acceptable as $$x < -2$$

Case II: When $$-2 < x < 3$$ $$x^2 - x - 6 < 0$$

Then equation becomes $$-(x^2 - x - 6) = x + 2 \Rightarrow x^2 - 4 = 0 \Rightarrow x = 2$$ because $$x = -2$$ is not acceptable.

Case III: Clearly $$x = -2$$ satisfies the equation by $$x = 3$$ does not.

35. $$|x + 2| = 0 \Rightarrow x = -2$$ and $$|2^{x + 1} - 1| = 0 \Rightarrow 2^{x + 1} = 1 \Rightarrow x = -1$$

Case I: When $$x < -2$$ then $$x + 2 < 0$$ and $$2^{x + 1} - 1 < 0$$

Equation becomes $$2^{-(x + 2)} - [-(2^{x + 1} - 1)] = 2^{x + 1} + 1$$

$$\Rightarrow x = 3$$

Case II: When $$-2 < x < 1$$ then $$x + 2 > 0$$ and $$2^{x + 1} - 1 < 0$$

Equation becomes $$2^{x + 2} - [-(2^{x + 1} - 1)] = 2^{x + 1} + 1$$

$$\Rightarrow x = 1$$

Case III: When $$x > -1$$ then $$x + 2 > 0$$ and $$2^{x + 1} - 1 > 0$$

Equation becomes $$2^{x + 2} - (2^{x + 1} - 1) = 2^{x + 1} + 1$$

$$\Rightarrow x + 2 = x + 2$$

which is true for all $$x$$ but only values for $$x > -1$$ are acceptable.

Case IV: Clearly, $$x = -2$$ does not satisfy the equation but $$x = -1$$ satisfies it.

36. Given equation is $$3^x + 4^x + 5^x = 6^x$$

$$\left(\frac{3}{6}\right)^x + \left(\frac{4}{6}\right)^x + \left(\frac{5}{6}\right)^x = 1$$

Clearly, $$x = 3$$ satisfies the equation.

When $$x > 3$$

$$\left(\frac{3}{6}\right)^x + \left(\frac{4}{6}\right)^x + \left(\frac{5}{6}\right)^x < 1$$

When $$x < 3$$

$$\left(\frac{3}{6}\right)^x + \left(\frac{4}{6}\right)^x + \left(\frac{5}{6}\right)^x > 1$$

Therefore, $$x = 3$$ is the only solution.

37. Proceeding as previous problem $$x = 2$$ is the only solution.

38. $$x = [x] + \{x\},$$ given equation is

$$4\{x\} = x + [x] \Rightarrow \{x\} = \frac{2}{3}[x]$$

$$\because 0 < \{x\} < 1 \therefore 0 < \frac{2}{3}[x] < 1 \Rightarrow 0 < [x] < \frac{3}{2} \Rightarrow [x] = 1$$

$$\therefore \{x\} = \frac{2}{3} \Rightarrow x = \frac{5}{3}$$

39. Given, $$[x]^2 = x(x - [x])$$

$$\Rightarrow [x]^2 = ([x] + \{x\})\{x\} [\because x = [x] + \{x\}]$$

$$y^2 = (y + z)z,$$ where $$y = [x]$$ and $$z = \{x\}$$

$$z^2 + yz - y^2 = 0 \Rightarrow z = \frac{-y \pm \sqrt{5}y}{2}$$

Since $$0 < z < 1$$

If $$z = -\frac{\sqrt{5} + 1}{2}y$$ then

$$0 > y > -\frac{2}{\sqrt{5} + 1}$$

$$-\frac{\sqrt{5} - 1}{2} < y < 0$$ is not possible as $$y$$ is an integer.

If $$z = \frac{\sqrt{5} - 1}{2}y$$ then $$0 < y < \frac{2}{\sqrt{5} - 1} \Rightarrow y = 1$$

$$z = \frac{\sqrt{5} - 1}{2}$$ and $$x = y + z = \frac{\sqrt{5} + 1}{2}$$

40. Let $$y = mx$$ the equations become

$$x^3(1 - m^3) = 127$$ and $$x^3(m - m^2) = 42$$

Dividing we get

$$\frac{1 - m^3}{m - m^2} = \frac{127}{42}$$

$$\Rightarrow \frac{1 + m + m^2}{m} = \frac{127}{42} [\because m = 1]$$ does not satisfy the equations.

$$\Rightarrow m = \frac{7}{6}, \frac{6}{7}$$

Substituting we get $$x = -6, y = -7$$ and $$x = 7, y = 6$$

41. Solving first two equations by cross-multiplication

$$\frac{x}{7} = \frac{y}{7} = \frac{z}{7}$$ or $$x = y = z = k$$

Substituting in third equation $$k = \pm \sqrt{7}$$

42. Let $$x = u + v$$ and $$y = u - v$$ then first equation becomes

$$(u + v)^4 + (u - v)^4 = 82$$

$$\Rightarrow u^4 + 6u^2v^2 + v^4 = 41$$

Second equation becomes $$2u = 4 \Rightarrow u = 2$$

Substituting in above equation $$v = \pm 5i, \pm 1$$

$$\therefore x = 2 \pm 5i, 3, 1$$

$$y = 2\mp 5i, 1, 3$$

43. Let $$y = 2^x > 0$$ then give equation becomes

$$\sqrt{a(y - 2) + 1} = 1 - y$$

$$\Rightarrow y^2 - (a + 2)y + 2a = 0$$

$$y = 2, a$$ but $$y = 2$$ does not satisfy the equation.

When $$y = a$$ then $$\sqrt{a(a - 2) + 1} = 1 - a \Rightarrow a \le 1$$

$$\therefore 0 < a \le 1 [\because y > 0]$$

$$y = a \Rightarrow x =log_2 a,$$ where $$0 < a \le 1$$

When $$a > 1,$$ given equation has no solution.

44. Given $$(x - 5)(x + m) = -2$$

Since $$x$$ and $$m$$ are both integers, therefore, $$x - 5$$ and $$x + m$$ are also integers.

So we have following combination of solutions:

$$x - 5 = 1$$ and $$x + m = 2$$ then $$x = 6, m = -8$$

$$x - 5 = 2$$ and $$x + m = -1$$ then $$x = 7, m = -8$$

$$x - 5 = -1$$ and $$x + m = 2$$ then $$x = 4, m = -2$$

$$x - 5 = -2$$ and $$x + m = 1$$ then $$x = 3, m = -2$$

Thus, $$m = -8, -2$$

45. Multiplying the equations we get

$$(xy)^{x + y} = (xy)^{2n} \therefore x + y = 2n$$ where $$xy \ne 1$$

$$\Rightarrow x^2 = y$$ then $$x + x^2 = 2n$$

$$x = \frac{-1 \pm \sqrt{1 + 8n}}{2}$$

But $$x > 0$$ $$\therefore x = \frac{-1 + \sqrt{1 + 8n}}{2}$$

$$y = x^2 = \frac{1 + 4n - \sqrt{1 + 8n}}{2}$$

46. Let $$y = 12^{|x|},$$ then given equation becomes $$y^2 - 2y + a = 0$$

$$y = 1 \pm \sqrt{1 - a}$$

$$|x| = \log_{12}(1 + \sqrt{1 - a})$$ as $$y = 1 - \sqrt{1 - a}$$ has to be rejected as $$y > 1$$

But for $$\sqrt{1 - a}$$ has to be real $$1 - a \ge 0 \Rightarrow a \le 1$$

For $$\log_{12}(1 + \sqrt{1 - a})$$ to be defined $$1 + \sqrt{1 - a} > 0$$

$$\therefore x = \pm \log_{12}(1 + \sqrt{1 - a})$$

47. Let $$m = 2p + 1$$ and $$n = 2n + 1$$ the $$D = 4(2p + 1)^2 - 8(2q + 1) =$$ an even no.

Let $$D$$ be a perfect square then it has to be perfect square of an even no. Let that no. be $$2r$$ then

$$4r^2 = 4(2p + 1)^2 - 8(2q + 1) \Rightarrow 2(2q + 1) = (2p + 1 - r)(2p + 1 + r)$$

Clearly, if $$r$$ is an even no. then L. H. S. is an even and R. H. S. is even no which is not possible.

Let $$r$$ is an odd no. then R. H. S. is product of 2 even numbers. Let $$2p + 1 - r = 2k$$ and $$2p + 1 + r = 2l$$

$$2(2q + 1) = 4kl$$ which is an odd no. $$2q + 1$$ having equality to even no. $$2kl$$ which is again not possible. Thus, under the given conditions equation cannot have rational roots.

48. Equation representing points of local extrema is $$f'(x) = 3ax^2 + 2bx + c = 0$$

Let one of these points is $$\alpha$$ and then second would be $$-\alpha$$

Sum of these roots $$= \alpha - \alpha = -\frac{2b}{3a} \Rightarrow b = 0$$

Product of roots $$= -\alpha^2 = \frac{c}{3a}$$ but since roots are opposite in equation it implies that $$a$$ and $$c$$ have opposite signs.

$$\therefore b^2 - 4ac = -4ac > 0$$ therefore roots of $$ax^2 + bx + c$$ will have real and distinct roots.

49. Given equation is $$\frac{(x - a)(ax - 1)}{x^2 - 1} = b$$

$$ax^2 - (1 + a^2)x + a = bx^2 - b \Rightarrow (a - b)x^2 - (1 + a^2)x + a + b = 0$$

Discriminant $$D^2 = (1 + a^2)^2 - a^2 + b^2 = 1 + a^2 + a^4 + b^2 > 0 [\because b \ne 0]$$

Therefore, roots can never be equal.

50. This question has been left as an exercise as it is trivial.