# 55. Quadratic Equations Solutions Part 5

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 \(k\)

\(\frac{c + a}{2a} + \frac{2c}{c + a} = - \frac{b}{a}\)

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

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

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

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

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.

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 \(qr = \frac{ap^2 + c}{a} = p^2 + \frac{c}{a}\)

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

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

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

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

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

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

\(\alpha, \beta\) 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 \(x^n + 1 + (x + 1)^n = 0\)

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.

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 \(|\alpha - \beta| < c \Rightarrow (\alpha - \beta)^2 < c^2\)

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

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

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 \(x\).

\(\alpha + \beta = -p, \alpha\beta = q, \alpha^4 + \beta^4 = r, \alpha^4\beta^4 = s\)

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.

\(\alpha + \beta = -\frac{b}{a}\) 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

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

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

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

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.

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.

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 \(\alpha > 0 \Rightarrow \alpha = 3\)

\(\Rightarrow \beta_1 = 4, \beta_2 = 5, \beta_3 = 12\)

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 \(m\)

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

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 \(\therefore 1 - 3\beta \ge 0 \Rightarrow \beta \le \frac{1}{3}\)

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

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.

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.

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 \(-(4a^3 + 27b^2) = (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 \ge 0\)

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

\(\alpha\) 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 \(\beta\).

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 \(x\).

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

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.

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 \(2\).

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 \(1 + \cos^8x \ne 0\)

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 \(f(x) - x < 0\)

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]

Proceeding from previous question

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

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

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 \(-2\)

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 \(x\).

\(2 + a + b + 4 > 0 \Rightarrow a + b > - 6\)

\(f(x) = x^3 + 2x^2 + x + 5 = 0\) 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 \(]-\infty, -1]\)

\([\alpha] = -3\)

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 \(y = 2x \Rightarrow x^2 - 2x + 2 = 0 \Rightarrow x = 1 \pm i\)

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 \(y = \frac{x^2}{2} \Rightarrow x = 3 \pm \sqrt{17}\)

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 \(y = 15 - 4\sqrt{14} \Rightarrow t = -1\)

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

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.

\(x^2 - x - 6 = 0 \Rightarrow x = -2, 3\)

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.

\(|x + 2| = 0 \Rightarrow x = -2\) 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.

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.

Proceeding as previous problem \(x = 2\) is the only solution.

\(x = [x] + \{x\},\) given equation is

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

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

\(\therefore \{x\} = \frac{2}{3} \Rightarrow x = \frac{5}{3}\)

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 \(x = y + z = \frac{\sqrt{5} + 1}{2}\)

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 \(x = 7, y = 6\)

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 \(k = \pm \sqrt{7}\)

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 \(v = \pm 5i, \pm 1\)

\(\therefore x = 2 \pm 5i, 3, 1\)

\(y = 2\mp 5i, 1, 3\)

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.

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, \(m = -8, -2\)

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

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

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 \(1 + \sqrt{1 - a} > 0\)

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

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.

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.

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.

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