.. meta:: :author: Shiv Shankar Dayal :title: Quadratic Equations Solutions Part 5 :description: Quadratic Equations Solutions Part 5 :keywords: quadratic equations, algebra Quadratic Equations Solutions Part 5 ************************************ 201. Product of roots :math:`= \frac{k + 2}{k} = \frac{c}{a}` :math:`\Rightarrow k = \frac{2a}{c - a}` Sum of roots :math:`= \frac{k + 1}{k} + \frac{k + 2}{k + 1} = -\frac{b}{a}` Substituting for :math:`k` :math:`\frac{c + a}{2a} + \frac{2c}{c + a} = - \frac{b}{a}` :math:`\frac{(a + c)^2 + 4ac}{2a(a + c)} = -\frac{b}{a}` :math:`a(a + c)^2 + 4a^2c = -2abc - 2a^2b` :math:`(a + c)^2 + 4ac = -2bc - 2ab` :math:`(a + b + c)^2 = b^2 - 4ac` 202. Given, :math:`f(x) = ax^2 + bx + c` and that :math:`\alpha,\beta` are the roots of the equation :math:`px^2 + qx + r = 0` :math:`\alpha + \beta = -\frac{q}{p}` and :math:`\alpha\beta = \frac{r}{p}` Now :math:`f(\alpha)f(\beta) = (a\alpha^2 + b\alpha + c)(a\beta^2 + b\beta + c)` :math:`= 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)` :math:`= 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}` :math:`= \frac{1}{p^2}[a^2r^2 + b^2rp + c^2p^2 - abrq + acq^2 - 2acrp - bcqp]` :math:`= \frac{1}{p^2}[(cp - ar)^2 + b^2rp - bcqp - abrq + acq^2]` :math:`= \frac{1}{p^2}[(cp - ar)^2 - (bp - aq)(cq - br)]` Now since :math:`\alpha, \beta` are the roots of the equation :math:`px^2 + qx + r = 0` Therefore, if :math:`ax^2 + bx + c = 0` and :math:`px^2 + qx + r = 0` have to have a common root then it has to be either :math:`\alpha` or :math:`\beta`. :math:`f(\alpha) = 0` or :math:`f(\beta) = 0 \therefore f(\alpha)f(\beta) = 0` :math:`\Rightarrow (cp - ar)^2 - (bp - aq)(cq - br) = 0` :math:`\therefore bp - aq, cp - ar, cq - br` are in G. P. 203. From the given equations it follows that :math:`q` and :math:`r` are roots of the equation :math:`a(p + x)^2 + 2bpx + c = 0 \Rightarrow ax^2 + 2(a + b)px + c = 0` Product of roots :math:`qr = \frac{ap^2 + c}{a} = p^2 + \frac{c}{a}` 204. Since :math:`\alpha, \beta` are the roots of the equation :math:`x^2 - px - (p + c) = 0` :math:`\alpha + \beta = p` and :math:`\alpha + \beta = -(p + c)` Now :math:`(\alpha + 1)(\beta + 1) = -p - c + p + 1 = 1 - c` :math:`\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)}}` :math:`= \frac{(\alpha + 1)^2}{(\alpha + 1)^2 - (\alpha + 1)(\beta + 1) + \frac{(\beta + 1)^2}{(\beta + 1)^2 - (\alpha 1)(\beta + 1)}}` :math:`= \frac{(\alpha + 1)^2}{(\alpha + 1)(\alpha - \beta)} + \frac{(\beta + 1)^2}{(\beta + 1)(\alpha - \beta)} = 1` 205. :math:`\alpha, \beta` are the roots of the equation :math:`x^2 + px + q = 0` :math:`\therefore \alpha + \beta = -p` and :math:`\alpha\beta = q` Since :math:`\alpha, \beta` are the roots of the equation :math:`x^2{2n} + p^nx^n + q^n = 0` Substituting it follows that :math:`\alpha^n, \beta^n` are the roots of the equation :math:`y^2 + p^ny + q^n = 0` :math:`\therefore \alpha^n + \beta^n = -p^n` and :math:`\alpha^n\beta^n = q^n` :math:`(\alpha + \beta)^n = (-p)^n = p^n [\because~\text{n is even}]` Thus, :math:`\alpha^n + \beta^n + (\alpha + \beta)^n = 0` Dividing by :math:`\beta^n` we have :math:`\left(\frac{\alpha}{\beta}\right)^n + 1 + \left(\frac{\alpha}{\beta} + 1\right)^n = 0` Dividing by :math:`\alpha^n` we have :math:`\left(\frac{\beta}{\alpha}\right)^n + 1 + \left(\frac{\beta}{\alpha} + 1\right)^n = 0` From last two equations it is evident that :math:`\frac{\alpha}{\beta}` and :math:`\frac{\beta}{\alpha}` are roots of the equation :math:`x^n + 1 + (x + 1)^n = 0` 206. Following from previous problem: :math:`\alpha^n + \beta^n = -p^n` Also since :math:`\frac{\alpha}{\beta}` and :math:`\frac{\beta}{\alpha}` are roots of the equation :math:`x^n + 1 + (x + 1)^n = 0` :math:`\alpha^n + \beta^n = -(\alpha + \beta)^n = -(-p)^n` From these equations we have :math:`-p^n = -(-p)^n \Rightarrow p^n = (-p)^n` Therefore, :math:`n` must be even. 207. Let :math:`\alpha` and :math:`\beta` are the roots of the given equation. Since roots are real and distinct :math:`D > 0 \Rightarrow a^2 - 4b > 0 \Rightarrow b < \frac{a^2}{4}` Again it is given that :math:`|\alpha - \beta| < c \Rightarrow (\alpha - \beta)^2 < c^2` :math:`(\alpha + \beta)^2 - 4\alpha\beta < c^2 \Rightarrow a^2 - 4b < c^2 \Rightarrow 4b > a^2 - c^2` :math:`\Rightarrow \frac{a^2 - c^2}{4} < b < \frac{a^2}{4}` 208. Given, :math:`ax^2 + bx + c - p = 0` for two integral values of :math:`x` say :math:`\alpha` and :math:`\beta`. Then, :math:`\alpha + \beta = -\frac{b}{a}` and :math:`\alpha\beta = \frac{c - p}{a}` If possible, let :math:`ax^2 + bx + c - 2p = 0` for some integer :math:`k`. :math:`ak^2 + bk + c - p = p \Rightarrow k^2 - (\alpha + \beta)k + \alpha\beta = \frac{p}{a}` :math:`(k - \alpha)(k - \beta) =~\text{an integer}~= \frac{p}{a}` But since :math:`p` is prime this cannot hold true unless :math:`a = p` or :math:`a = 1` :math:`a = p [\because a > 1]` :math:`(k - \alpha)(k - \beta) = 1` which implies that :math:`k - \alpha = k - \beta = 1` which is not possible since :math:`\alpha \ne \beta` Thus, we have a contradiction. Hence, :math:`ax^2 + bx + c \ne 2p` for any integral value of :math:`x`. 209. :math:`\alpha + \beta = -p, \alpha\beta = q, \alpha^4 + \beta^4 = r, \alpha^4\beta^4 = s` Let :math:`D` be the discriminant of :math:`x^2 - 4qx + 2q^2 - r = 0` then :math:`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` :math:`D \ge 0` hence roots of the third equation are always real. 210. :math:`\alpha + \beta = -\frac{b}{a}` and :math:`\alpha\beta = \frac{c}{a}` :math:`\alpha_1 - \beta = -\frac{b_1}{a_1}` and :math:`-\alpha_1\beta = \frac{c_1}{a_1}` :math:`\alpha + \alpha_1 = -\left(\frac{b}{a} + \frac{b_1}{a_1}\right)` Also, dividing :math:`\alpha + \beta` by :math:`\alpha\beta` :math:`\frac{1}{\beta} + \frac{1}{\alpha} = -\frac{b}{c}` Similarly, dividing :math:`\alpha_1 - \beta` by :math:`-\alpha_1\beta` :math:`\frac{1}{\alpha_1} - \frac{1}{\beta} = -\frac{b_1}{c_1}` Thus, :math:`\frac{1}{\alpha} + \frac{1}{\alpha_1} = -\left(\frac{b}{c} + \frac{b_1}{c_1}\right)` Equation whose roots are :math:`\alpha` and :math:`\alpha_1` is :math:`x^2 - (\alpha + \alpha_1)x + \alpha\alpha_1 = 0` :math:`\frac{x^2}{-(\alpha + \alpha_1)} + x - \frac{\alpha\alpha_1}{\alpha + \alpha_1} = 0` :math:`\frac{x^2}{\frac{b}{a} + \frac{b_1}{a_1}} + x + \frac{1}{\frac{b}{c} + \frac{b_1}{c_1}} = 0` 211. Let :math:`\alpha` and :math:`\beta` be roots of such quadratic equation given by :math:`x^2 + px + q = 0` :math:`\alpha + \beta = -p` and :math:`\alpha\beta = q` Now quadratic equation whose roots are :math:`\alpha^2` and :math:`\beta^2` is :math:`x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0` :math:`x^2 - (p^2 - 2q)x + q^2 = 0` But the equation remains unchanged, therefore, :math:`\frac{1}{1} = \frac{p}{p^2 - 2q} = \frac{q}{q^2}` :math:`\Rightarrow q = q^2 \Rightarrow q(q - 1) = 0 \Rightarrow q = 0, 1` If :math:`q = 0 \Rightarrow p = 0, -1` and if :math:`q = 1 \Rightarrow p = -2, 1` Thus, four such quadratic equations are possible. 212. Given :math:`\frac{d}{a}, \frac{e}{b}, \frac{f}{c}` are in A. P. and :math:`a, b, c` are in G. P. Equations :math:`ax^2 + 2bx + c = 0` and :math:`dx^2 + 2ex + f = 0` will have a common root if :math:`\frac{2(bf - ec)}{cd - af} = \frac{cd - af}{2(ae - bd)}` :math:`4(bf - ec)(ae - bd) = (cd - af)^2` :math:`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` :math:`4k.k.b^2 = 4k^2ac` where :math:`k` is the c.d. of the A. P. :math:`b^2 = ac` which is true because :math:`a, b, c` are in G. P. 213. Let :math:`\alpha` be the common root and :math:`\beta_1` another root of :math:`x^2 + ax + 12 = 0, \beta_2` be another root of :math:`x^2 + bx + 15 = 0` and :math:`\beta_3` be a root of :math:`x^2 + (a + b)x + 36 = 0.` :math:`\alpha + \beta_1 = -a` and :math:`\alpha\beta_1 = 12` :math:`\alpha + \beta_2 = -b` and :math:`\alpha\beta_2 = 15` :math:`\alpha + \beta_3 = -(a + b)` and :math:`\alpha\beta_3 = 36` Thus, :math:`2\alpha + \beta_1 + \beta_2 = \alpha + \beta_3 \Rightarrow \alpha = \beta_3 - \beta_1 - \beta_2` and :math:`\alpha(\beta_3 - \beta_1 - \beta_2) = 36 - 12 - 15 = 9` :math:`\Rightarrow \alpha^2 = 9 \Rightarrow \alpha = \pm 3` but :math:`\alpha > 0 \Rightarrow \alpha = 3` :math:`\Rightarrow \beta_1 = 4, \beta_2 = 5, \beta_3 = 12` 214. Given :math:`m(ax^2 + 2bx + c) + px^2 + 1qx + r = n(x + k)^2` Equating coefficients for powers of :math:`x`, we get :math:`ma + p = n, mb + q = nk, mc + r = nk^2` :math:`\Rightarrow m(ak - b) + pk - q = 0 \Rightarrow m = -\frac{pk - q}{ak - b}` :math:`\Rightarrow m(bk - c) + qk - r = 0 \Rightarrow m = -\frac{qk - r}{bk - c}` Equating values for :math:`m` :math:`(ak - b)(qk - r) = (pk - q)(bk - c)` 215. Given equation is :math:`x^3 - x^2 + \beta x + \gamma = 0` Let it roots :math:`x_1, x_2, x_3` be :math:`a - d, a, a + d` respectively. :math:`a - d + a + a + d = 1 \Rightarrow a = \frac{1}{3}` :math:`(a - d)a + a(a + d) + (a - d)(a + d) = \beta \Rightarrow 3a^2 - d^2 = \beta \Rightarrow 1 - 3\beta = 3d^2` :math:`(a - d)a(a + d) = \gamma \Rightarrow a(a^2 - d^2) = \gamma \Rightarrow 1 + 27\gamma = 9d^2` Since :math:`d` is real :math:`\therefore 1 - 3\beta \ge 0 \Rightarrow \beta \le \frac{1}{3}` :math:`1 + 27\gamma \ge 0 \Rightarrow \gamma \ge -\frac{1}{27}` 216. Let :math:`\alpha` be a common root, then :math:`\alpha^3 + 3p\alpha^2 + 3q\alpha + r = 0` ... (1) and :math:`\alpha^2 + 2p\alpha + q = 0` ... (2) :math:`(1) - \alpha (2)` gives us :math:`\Rightarrow p\alpha^2 + 2q\alpha + r = 0` ... (3) By cross multiplication between (2) and (3) :math:`\frac{\alpha^2}{2(pr - q^2)} = \frac{\alpha}{pq - r} = \frac{1}{2(q - p^2)}` Equating for values of :math:`\alpha` we get the desired condition. 217. Let :math:`\alpha` be a common root, then :math:`\alpha^3 + 2a\alpha^2 + 3b\alpha + c = 0` ... (1) and :math:`\alpha^3 + a\alpha^2 + 2b\alpha = 0` ... (2) Since :math:`c \ne 0,` therefore :math:`\alpha = 0` cannot be a common root. Therefore, from (2) :math:`\alpha^2 + a\alpha + 2b = 0` ... (3) :math:`(1) - \alpha (2) \Rightarrow a\alpha^2 + b\alpha + c = 0` ... (4) Solving (3) and (4) by cross-multiplication yields the desired result. 218. Given equation is :math:`x^3 + ax + b = 0` and :math:`\alpha, \beta, \gamma` be its real roots. Then we have :math:`\alpha + \beta + \gamma = 0` ... (1) :math:`\alpha\beta + \beta\gamma + \alpha\gamma = a` ... (2) :math:`\alpha\beta\gamma = -b` Let :math:`y = (\alpha - \beta)^2,` then :math:`y = (\alpha + \beta)^2 - 4\alpha\beta` :math:`y = \gamma^2 + \frac{4b}{\gamma}` :math:`\Rightarrow \gamma^3 - y\gamma + 4b = 0` Also, :math:`\gamma` is a root of the original equation. :math:`\gamma^3 + a\gamma + b = 0` :math:`(a + y)\gamma - 3b = 0 \Rightarrow \gamma = \frac{3b}{a + y}` :math:`\Rightarrow \frac{27b^3}{(a + y)^3} + a\left(\frac{3b}{a + y}\right) + b = 0` :math:`y^3 + 6ay^2 + 9a^2y + 4a^3 + 27b^2 = 0` We will get same equation if we would have chosen :math:`y = (\beta - \alpha)^2` or :math:`y = (\gamma - \alpha)^2` Hence, product of roots :math:`-(4a^3 + 27b^2) = (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 \ge 0` :math:`\therefore 4a^3 + 27b^2 \le 0` 219. :math:`\alpha` is a root of the equation :math:`ax^2 + bx + c = 0` :math:`\therefore a\alpha^2 + b\alpha + c = 0` Similarly, :math:`-a\beta^2 + b\beta + c = 0` Let :math:`f(x) = \frac{a}{2}x^2 + bx + c = 0` :math:`f(\alpha) = -\frac{a}{2}\alpha^2` :math:`f(\beta) = \frac{3}{2}\beta^2` :math:`\therefore f(\alpha)f(\beta) = -\frac{3}{4}a^2\alpha^2\beta^2 < 0 [\because \alpha,\beta \ne 0]` :math:`\therefore f(\alpha)` and :math:`f(\beta)` have opposite signs. Therefore, :math:`f(x)` will have exactly one root between :math:`\alpha` and :math:`\beta`. 220. Let :math:`f(x) = ax^2 + bx + c = 0` Since equation :math:`ax^2 + bx + c = 0` i.e. equation :math:`f(x) = 0` has no real root, therefore, :math:`f(x)` will have same sign for real values of :math:`x`. :math:`\therefore f(1)f(0) > 0 \Rightarrow (a + b + c)c > 0` 221. Let :math:`f(x) = (x - a)(x - c) + \lambda (x - b)(x - d)` Given :math:`a > b > c > d` Now :math:`f(b) = (b - a)(b - c) < 0` and :math:`f(d) = (d - a)(d - c) > 0` Since :math:`f(b)` and :math:`f(d)` have opposite signs, therefore equation :math:`f(x) = 0` will have one real root between :math:`b` and :math:`d`. Since one root is real and :math:`a, b, c, d, \lambda` are all real the other root will also be real. 222. Let :math:`f'(x) = ax^2 + bx + c,` then :math:`f(x) = a\frac{x^3}{3} + b\frac{x^2}{2} + cx + k = \frac{2ax^3 + 3bx^2 + 4cx + 6k}{6}` :math:`f(1) = \frac{2a + 3b + 6c + 6k}{6} = k` Again, :math:`f(0) = k` Thus, :math:`f(0) = f(1)` hence equation will have at least one root between :math:`0` and :math:`1` which implies that it will have a real root between :math:`0` and :math:`2`. 223. Let :math:`f(x) = \int (1 + \cos^8x)(ax^2 + bx + c)dx` then :math:`f'(x) = (1 + \cos^8x)(ax^2 + bx + c)` Given, :math:`\int_0^1 (1 + \cos^8 x)(ax^2 + bx + c)dx = \int_0^2 (1 + \cos^8 x)(ax^2 + bx + c)dx` :math:`f(1) - f(0) = f(2) - f(0) \Rightarrow f(1) = f(2)` Therefore, equation :math:`f(x) = 0` has at least one root between :math:`1` and :math:`2` which implies that :math:`ax^2 + bx + c` has a root between these two limits as :math:`1 + \cos^8x \ne 0` 224. Given equation :math:`f(x) - x = 0` has non-real roots where :math:`f(x) = ax^2 + bx + c` is a continuous function. :math:`\therefore f(x) - x` has same sign for all :math:`x \in R` Let :math:`f(x) - x > 0~\forall~x \in R` :math:`\Rightarrow f(f(x)) - f(x) > 0~\forall~x\in R` :math:`\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 :math:`f(x) - x < 0` 225. Let :math:`f(x) = ax^2 - bx + c = 0` and that :math:`\alpha, \beta` be its roots. Then, :math:`f(x) = a(x - \alpha)(x - \beta)` Given :math:`\alpha \ne \beta, 0 < \alpha < 1, 0 < \beta < 1` and :math:`a, b, c \in N` Since quadratic equation has both roots between :math:`0` and :math:`1`, therefore :math:`f(0)f(1) > 0` :math:`f(0)f(1) = c(a - b + c) =` an integer Thus, :math:`f(0)f(1) \ge 1 \Rightarrow a\alpha(1 - \alpha)a\beta(1 - \beta) = a^2\alpha\beta(1 - \alpha)(1 - \beta)` Let :math:`y = \alpha(1 - \alpha) \Rightarrow \alpha^2 - \alpha + y = 0` Since :math:`\alpha` is real :math:`\therefore 1 - 4y \ge 0 \Rightarrow y \le \frac{1}{4} \Rightarrow \alpha = \frac{1}{2}~\text{max value}` Similarly, maximum value of :math:`\beta = \frac{1}{2}` Maximum value of :math:`\therefore f(0)f(1) < \frac{a^2}{16} > 1` :math:`\Rightarrow a > 4 \Rightarrow a = 5` [least integral value] 226. Proceeding from previous question :math:`b^2 - 4ac > 0 \Rightarrow b^2 > 4.5.1 [\because c \ge 1] \Rightarrow b = 5` :math:`\log_5(abc) \ge 2` 227. Given equation is :math:`ax^2 + bx + 6 = 0`. Let :math:`f(x) = ax^2 + bx + 6` Since the equation has imaginary roots or real and equal roots its sign will never change. :math:`f(0) = 6 > 0` :math:`\therefore f(x) \ge 0` for all real :math:`x` :math:`f(3) \ge 0 \Rightarrow 9a + 3b + 6 \ge 0` :math:`3a + b \ge -2` and hence least value is :math:`-2` 228. Let :math:`\alpha, \beta, \gamma` be the roots of the equation. Then, :math:`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 :math:`a, b > 0` :math:`f(0) = 4 > 0 \therefore f(1) > 0` because sign of :math:`f(x)` will not change for all :math:`x`. :math:`2 + a + b + 4 > 0 \Rightarrow a + b > - 6` 229. :math:`f(x) = x^3 + 2x^2 + x + 5 = 0` and :math:`f'(x) = 3x^2 + 4x + 1` which has roots :math:`-1` and :math:`-\frac{1}{3}`. :math:`f(0) = 5` and :math:`f(x)` is increasing in :math:`(0, \infty)` therefore it will have no root in :math:`[0, \infty[` :math:`f(-2) = 3 > 0` and :math:`f(-3) = -7 < 0` Since :math:`f(-2)` and :math:`f(-3)` are of opposite sign therefore equation :math:`f(x) = 0` will have one root between :math:`-2` and :math:`-3` and this will be only one root as :math:`f(x)` is increasing in :math:`]-\infty, -1]` :math:`[\alpha] = -3` 230. Given equation is :math:`(x^2 + 2)^2 + 8x^2 = 6x(x^2 + 2)` Let :math:`y = x^2 + 2` then above equation becomes :math:`y^2 + 8x^2 = 6xy` :math:`\Rightarrow y = 4x, 2x` If :math:`y = 4x \Rightarrow x^2 - 4x + 2 = 0 \Rightarrow x = 2 \pm \sqrt{2}` If :math:`y = 2x \Rightarrow x^2 - 2x + 2 = 0 \Rightarrow x = 1 \pm i` 231. Given equation is :math:`3x^3 = (x^2 + \sqrt{18}x + \sqrt{32})(x^2 - \sqrt{18}x - \sqrt{32}) - 4x^2` :math:`3x^3 = x^4 - (\sqrt{18}x + \sqrt{32})^2 - 4x^2` :math:`x^2(3x + 4) = x^4 - 2(3x + 4)^2` :math:`x^2y = x^4 - 2y^2` where :math:`y = 3x + 4` :math:`\Rightarrow y = -x^2, \frac{x^2}{2}` If :math:`y = -x^2 \Rightarrow x = \frac{-3 \pm \sqrt{7}i}{2}` and if :math:`y = \frac{x^2}{2} \Rightarrow x = 3 \pm \sqrt{17}` 232. Clearly, :math:`(15 + 4\sqrt{14})^t(15 - 4\sqrt{14})^t = (225 - 224)^t = 1` Let :math:`(15 + 4\sqrt{14})^t = y,` then :math:`(15 - 4\sqrt{14})^t = \frac{1}{y}` Substituting for the given equation :math:`y + \frac{1}{y} = 30 \Rightarrow y^2 - 30y + 1 = 0` :math:`y = 15 \pm 4\sqrt{14}` If :math:`y = 15 + 4\sqrt{14} \Rightarrow t = 1` :math:`\therefore x^2 - 2|x| = 1 \Rightarrow |x|^2 - 2|x| - 1 = 0` :math:`\Rightarrow |x| = 1 + \sqrt{2} \therefore x = \pm(1 + \sqrt{2})` If :math:`y = 15 - 4\sqrt{14} \Rightarrow t = -1` :math:`\Rightarrow |x|^2 - 2|x| + 1 = 0 \Rightarrow |x| = 1 \Rightarrow x = \pm 1` 233. Given equation is :math:`x^2 - 2a|x - a| - 3a^2 = 0` When :math:`a = 0` equation becomes :math:`x^2 = 0 \Rightarrow x = 0` Let :math:`a < 0`. Case I: When :math:`x < a` then equation becomes :math:`x^2 + 2a(x - a) - 3a^2 = 0 \Rightarrow x^2 + 2ax - 5a^2 = 0 \Rightarrow x = -a \pm \sqrt{6}a` Since :math:`x < a, x = -a - \sqrt{6}a` is not acceptable. Case II: When :math:`x > a` the equation becomes :math:`x^2 - 2ax - a^2 = 0 \Rightarrow x = a \pm \sqrt{2}a` Since :math:`x > a, x = a + \sqrt{2}a` is not acceptable. Clearly, :math:`x = a` does not satisfy the equation. 234. :math:`x^2 - x - 6 = 0 \Rightarrow x = -2, 3` Case I: When :math:`x < -2` or :math:`x > 3` then :math:`x^2 - x - 6 > 0` Then equation becomes :math:`x^2 - x - 6 = x + 2 \Rightarrow x^2 - 2x - 8 = 0` :math:`x = -2, 4` but :math:`x = -2` is not acceptable as :math:`x < -2` Case II: When :math:`-2 < x < 3` :math:`x^2 - x - 6 < 0` Then equation becomes :math:`-(x^2 - x - 6) = x + 2 \Rightarrow x^2 - 4 = 0 \Rightarrow x = 2` because :math:`x = -2` is not acceptable. Case III: Clearly :math:`x = -2` satisfies the equation by :math:`x = 3` does not. 235. :math:`|x + 2| = 0 \Rightarrow x = -2` and :math:`|2^{x + 1} - 1| = 0 \Rightarrow 2^{x + 1} = 1 \Rightarrow x = -1` Case I: When :math:`x < -2` then :math:`x + 2 < 0` and :math:`2^{x + 1} - 1 < 0` Equation becomes :math:`2^{-(x + 2)} - [-(2^{x + 1} - 1)] = 2^{x + 1} + 1` :math:`\Rightarrow x = 3` Case II: When :math:`-2 < x < 1` then :math:`x + 2 > 0` and :math:`2^{x + 1} - 1 < 0` Equation becomes :math:`2^{x + 2} - [-(2^{x + 1} - 1)] = 2^{x + 1} + 1` :math:`\Rightarrow x = 1` Case III: When :math:`x > -1` then :math:`x + 2 > 0` and :math:`2^{x + 1} - 1 > 0` Equation becomes :math:`2^{x + 2} - (2^{x + 1} - 1) = 2^{x + 1} + 1` :math:`\Rightarrow x + 2 = x + 2` which is true for all :math:`x` but only values for :math:`x > -1` are acceptable. Case IV: Clearly, :math:`x = -2` does not satisfy the equation but :math:`x = -1` satisfies it. 236. Given equation is :math:`3^x + 4^x + 5^x = 6^x` :math:`\left(\frac{3}{6}\right)^x + \left(\frac{4}{6}\right)^x + \left(\frac{5}{6}\right)^x = 1` Clearly, :math:`x = 3` satisfies the equation. When :math:`x > 3` :math:`\left(\frac{3}{6}\right)^x + \left(\frac{4}{6}\right)^x + \left(\frac{5}{6}\right)^x < 1` When :math:`x < 3` :math:`\left(\frac{3}{6}\right)^x + \left(\frac{4}{6}\right)^x + \left(\frac{5}{6}\right)^x > 1` Therefore, :math:`x = 3` is the only solution. 237. Proceeding as previous problem :math:`x = 2` is the only solution. 238. :math:`x = [x] + \{x\},` given equation is :math:`4\{x\} = x + [x] \Rightarrow \{x\} = \frac{2}{3}[x]` :math:`\because 0 < \{x\} < 1 \therefore 0 < \frac{2}{3}[x] < 1 \Rightarrow 0 < [x] < \frac{3}{2} \Rightarrow [x] = 1` :math:`\therefore \{x\} = \frac{2}{3} \Rightarrow x = \frac{5}{3}` 239. Given, :math:`[x]^2 = x(x - [x])` :math:`\Rightarrow [x]^2 = ([x] + \{x\})\{x\} [\because x = [x] + \{x\}]` :math:`y^2 = (y + z)z,` where :math:`y = [x]` and :math:`z = \{x\}` :math:`z^2 + yz - y^2 = 0 \Rightarrow z = \frac{-y \pm \sqrt{5}y}{2}` Since :math:`0 < z < 1` If :math:`z = -\frac{\sqrt{5} + 1}{2}y` then :math:`0 > y > -\frac{2}{\sqrt{5} + 1}` :math:`-\frac{\sqrt{5} - 1}{2} < y < 0` is not possible as :math:`y` is an integer. If :math:`z = \frac{\sqrt{5} - 1}{2}y` then :math:`0 < y < \frac{2}{\sqrt{5} - 1} \Rightarrow y = 1` :math:`z = \frac{\sqrt{5} - 1}{2}` and :math:`x = y + z = \frac{\sqrt{5} + 1}{2}` 240. Let :math:`y = mx` the equations become :math:`x^3(1 - m^3) = 127` and :math:`x^3(m - m^2) = 42` Dividing we get :math:`\frac{1 - m^3}{m - m^2} = \frac{127}{42}` :math:`\Rightarrow \frac{1 + m + m^2}{m} = \frac{127}{42} [\because m = 1]` does not satisfy the equations. :math:`\Rightarrow m = \frac{7}{6}, \frac{6}{7}` Substituting we get :math:`x = -6, y = -7` and :math:`x = 7, y = 6` 241. Solving first two equations by cross-multiplication :math:`\frac{x}{7} = \frac{y}{7} = \frac{z}{7}` or :math:`x = y = z = k` Substituting in third equation :math:`k = \pm \sqrt{7}` 242. Let :math:`x = u + v` and :math:`y = u - v` then first equation becomes :math:`(u + v)^4 + (u - v)^4 = 82` :math:`\Rightarrow u^4 + 6u^2v^2 + v^4 = 41` Second equation becomes :math:`2u = 4 \Rightarrow u = 2` Substituting in above equation :math:`v = \pm 5i, \pm 1` :math:`\therefore x = 2 \pm 5i, 3, 1` :math:`y = 2\mp 5i, 1, 3` 243. Let :math:`y = 2^x > 0` then give equation becomes :math:`\sqrt{a(y - 2) + 1} = 1 - y` :math:`\Rightarrow y^2 - (a + 2)y + 2a = 0` :math:`y = 2, a` but :math:`y = 2` does not satisfy the equation. When :math:`y = a` then :math:`\sqrt{a(a - 2) + 1} = 1 - a \Rightarrow a \le 1` :math:`\therefore 0 < a \le 1 [\because y > 0]` :math:`y = a \Rightarrow x =log_2 a,` where :math:`0 < a \le 1` When :math:`a > 1,` given equation has no solution. 244. Given :math:`(x - 5)(x + m) = -2` Since :math:`x` and :math:`m` are both integers, therefore, :math:`x - 5` and :math:`x + m` are also integers. So we have following combination of solutions: :math:`x - 5 = 1` and :math:`x + m = 2` then :math:`x = 6, m = -8` :math:`x - 5 = 2` and :math:`x + m = -1` then :math:`x = 7, m = -8` :math:`x - 5 = -1` and :math:`x + m = 2` then :math:`x = 4, m = -2` :math:`x - 5 = -2` and :math:`x + m = 1` then :math:`x = 3, m = -2` Thus, :math:`m = -8, -2` 245. Multiplying the equations we get :math:`(xy)^{x + y} = (xy)^{2n} \therefore x + y = 2n` where :math:`xy \ne 1` :math:`\Rightarrow x^2 = y` then :math:`x + x^2 = 2n` :math:`x = \frac{-1 \pm \sqrt{1 + 8n}}{2}` But :math:`x > 0` :math:`\therefore x = \frac{-1 + \sqrt{1 + 8n}}{2}` :math:`y = x^2 = \frac{1 + 4n - \sqrt{1 + 8n}}{2}` 246. Let :math:`y = 12^{|x|},` then given equation becomes :math:`y^2 - 2y + a = 0` :math:`y = 1 \pm \sqrt{1 - a}` :math:`|x| = \log_{12}(1 + \sqrt{1 - a})` as :math:`y = 1 - \sqrt{1 - a}` has to be rejected as :math:`y > 1` But for :math:`\sqrt{1 - a}` has to be real :math:`1 - a \ge 0 \Rightarrow a \le 1` For :math:`\log_{12}(1 + \sqrt{1 - a})` to be defined :math:`1 + \sqrt{1 - a} > 0` :math:`\therefore x = \pm \log_{12}(1 + \sqrt{1 - a})` 247. Let :math:`m = 2p + 1` and :math:`n = 2n + 1` the :math:`D = 4(2p + 1)^2 - 8(2q + 1) =` an even no. Let :math:`D` be a perfect square then it has to be perfect square of an even no. Let that no. be :math:`2r` then :math:`4r^2 = 4(2p + 1)^2 - 8(2q + 1) \Rightarrow 2(2q + 1) = (2p + 1 - r)(2p + 1 + r)` Clearly, if :math:`r` is an even no. then L. H. S. is an even and R. H. S. is even no which is not possible. Let :math:`r` is an odd no. then R. H. S. is product of 2 even numbers. Let :math:`2p + 1 - r = 2k` and :math:`2p + 1 + r = 2l` :math:`2(2q + 1) = 4kl` which is an odd no. :math:`2q + 1` having equality to even no. :math:`2kl` which is again not possible. Thus, under the given conditions equation cannot have rational roots. 248. Equation representing points of local extrema is :math:`f'(x) = 3ax^2 + 2bx + c = 0` Let one of these points is :math:`\alpha` and then second would be :math:`-\alpha` Sum of these roots :math:`= \alpha - \alpha = -\frac{2b}{3a} \Rightarrow b = 0` Product of roots :math:`= -\alpha^2 = \frac{c}{3a}` but since roots are opposite in equation it implies that :math:`a` and :math:`c` have opposite signs. :math:`\therefore b^2 - 4ac = -4ac > 0` therefore roots of :math:`ax^2 + bx + c` will have real and distinct roots. 249. Given equation is :math:`\frac{(x - a)(ax - 1)}{x^2 - 1} = b` :math:`ax^2 - (1 + a^2)x + a = bx^2 - b \Rightarrow (a - b)x^2 - (1 + a^2)x + a + b = 0` Discriminant :math:`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. 250. This question has been left as an exercise as it is trivial.