.. meta:: :author: Shiv Shankar Dayal :title: Quadratic Equations Solutions Part 6 :description: Quadratic Equations Solutions Part 6 :keywords: quadratic equations, algebra Quadratic Equations Solutions Part 6 ************************************ 251. :math:`D = c^2(3a^2 + b^2)^2 + 4abc^2(6a^2 + ab - 2b^2)` :math:`= c^2(9a^4 + b^4 + 6a^2b^2 + 4a^3b + 4a^2b^2 - 8ab^3)` :math:`= c^2(3a^2 - b^2 + 4ab)^2` which is a perfect square and hence roots are rational. 252. :math:`\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} + \frac{b}{\sqrt{ac}} = 0` :math:`L. H. S. = \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \frac{b}{\sqrt{ac}}` :math:`= \frac{\alpha + \beta}{\sqrt{\alpha\beta}} + \frac{b}{\sqrt{ac}}` :math:`= \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}} + \frac{b}{\sqrt{ac}} = 0` 253. Let :math:`\alpha` be the root, then the second root would be :math:`\alpha^3`. Product of roots :math:`= \alpha^4 = a \Rightarrow \alpha = a^{\frac{1}{4}}` Sum of roots :math:`= \alpha + \alpha^3 = -f(a)` :math:`\Rightarrow f(a) = -a^{\frac{1}{4}} - a^{\frac{3}{4}}` Therefore, the general equation in :math:`x` would be :math:`f(x) = -x^{\frac{1}{4}} - x^{\frac{3}{4}}` 254. Since :math:`\alpha, \beta` are roots of the equation :math:`x^2 - px + q = 0` therefore :math:`\alpha + \beta = p` and :math:`\alpha\beta = q` :math:`(\alpha^2 - \beta^2)(\alpha^3 - \beta^3) = (\alpha - \beta)^2(\alpha + \beta)[(\alpha + \beta^2) - \alpha\beta]` :math:`=(p^2 - 4q)p(p^2 - q)` :math:`\alpha^3\beta^2 + \alpha^2\beta^3 = \alpha^2\beta^2(\alpha + \beta) = pq^2` Therefore, the equation would be :math:`x^2 - p[(p^2 - 4q)(p^2 - q) + q^2]x + p^2q^2(p^2 - 4q)(p^2 - q) = 0` 255. This problem has been left as an exercise. 256. Let :math:`\alpha, \beta` be the roots then :math:`\alpha + \beta = -\frac{b}{a}` and :math:`\alpha\beta = \frac{c}{a}` According to the question :math:`\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}` :math:`-\frac{b}{a} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2\beta^2}` :math:`-\frac{b}{a} = \frac{\frac{b^2}{a^2}}{\frac{c^2}{a^2}} - 2\frac{1}{\frac{c}{a}}` :math:`-\frac{b}{a} = \frac{b^2}{c^2} - 2\frac{a}{c}` :math:`\Rightarrow \frac{b^2}{ac} + \frac{bc}{a^2} = 2` 257. Given, :math:`T = 2\pi \sqrt{\frac{h^2 + k^2}{gh}}` Squaring, :math:`h^2 + k^2 = \frac{T^2gh}{4\pi^2}` :math:`h^2 - \frac{T^2gh}{4\pi^2} + k^2 = 0` Clearly, :math:`h_1` and :math:`h_2` are two possible roots of above equation where :math:`h_1 + h_2 = \frac{T^2g}{4\pi^2}` and :math:`h_1h_2 = k^2` 258. Clearly, :math:`\alpha_1 + \alpha_2 = -p` and :math:`\alpha_1\alpha_2 = q` :math:`\beta_1 + \beta2 = -r` and :math:`\beta_1\beta_2 = s` Solving the two equations in :math:`y` and :math:`z` by elimination we have :math:`\frac{\alpha_1}{\alpha_2} = \frac{\beta_1}{\beta_2} = k` :math:`\frac{p^2}{r^2} = \frac{(\alpha_1 + \alpha_2)^2}{(\beta_1 + \beta_2)^2}` :math:`= \frac{\alpha_1^2(1 + k^2)}{\beta_1(1 + k^2)} = \frac{\frac{\alpha_1\alpha_2}{k}}{\frac{\beta_1\beta_2}{k}} = \frac{q}{s}` 259. :math:`-(1 + \alpha\beta) = -(\frac{a + c}{a})` H. M. of :math:`\alpha` and :math:`\beta = \frac{2\alpha\beta}{\alpha + \beta} = -\frac{2c}{b}` but since :math:`a, b, c` are in H. P. it becomes :math:`= -\frac{2c}{\frac{2ac}{a + c}} = -(\frac{a + c}{a}) = -(1 + \alpha\beta)` 260. Given equation is :math:`x + 1 = \lambda x - \lambda^2x^2` :math:`\lambda^2x^2 + (1 - \lambda)x + 1 = 0` :math:`\Rightarrow \alpha + \beta = \frac{\lambda - 1}{\lambda^2}` and :math:`\alpha\beta = \frac{1}{\lambda^2}` Also given that, :math:`\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = r - 2` :math:`\Rightarrow \alpha^2 + \beta^2 = (r - 2)\alpha\beta` :math:`(\alpha + \beta)^2 = r\alpha\beta` :math:`\frac{(\lambda - 1)^2}{\lambda^4} = \frac{r}{\lambda^2}` :math:`\Rightarrow \lambda_1 + \lambda_2 = \frac{2}{1 - r}` and :math:`\lambda_1\lambda_2 = \frac{1}{1 - r}` Now substituting and solving we get the desired result. 261. Let :math:`\alpha, \beta` be roots of :math:`ax^2 + bx + c = 0` then :math:`\alpha + \beta = -\frac{b}{a}` and :math:`\alpha\beta = \frac{c}{a}` According to question, :math:`\frac{1}{\alpha} + \frac{1}{\beta} = -\frac{m}{l}` and :math:`\frac{1}{\alpha\beta} = \frac{n}{l}` From product of roots, :math:`\frac{c}{a} = \frac{l}{n}` and from sum of roots :math:`\frac{b}{c} = \frac{m}{l}` Hence the relation is established. 262. Let the roots are :math:`l, lm, lm^2, lm^3` which is an increasing G. P. Sum of roots for first equation :math:`= l(1 + m) = 3` Sum of roots for second equation :math:`= lm^2(1 + m) = 12 \Rightarrow m^2 = 4 \Rightarrow m = 2` because G. P. is increasing. :math:`\Rightarrow l = 1` :math:`A = l^2m = 2` and :math:`B = l^2m^5 = 32` 263. For first equation, :math:`p + q = 2` and :math:`pq = A.` For second equation, :math:`r + s = 18` and :math:`rs = B` Let :math:`a` be the first term and :math:`d` be the common difference, then :math:`p = a - 3d, q = a - d, r = a + d, s = a + 3d` Substituting in sums we have :math:`2a - 4d = 2` and :math:`2a + 4d = 18` :math:`\therefore a = 5` and :math:`d = 2` :math:`\therefore p = -1, q = 3, r = 7, s = 11` :math:`\therefore A = -3` and :math:`B = 77` 264. :math:`\alpha + \beta = -a` and :math:`\alpha\beta = -\frac{1}{2a^2}` :math:`\alpha^4 + \beta^4 = ((\alpha + \beta)^2 - 2\alpha\beta)^2 - 2\alpha^2\beta^2` :math:`= 2 + a^4 + \frac{1}{2a^4}` Let :math:`a^4 + \frac{1}{2a^4} = y` :math:`\Rightarrow 2a^8 - 2a^4y - 1 = 0` Since :math:`a` is real. :math:`\therefore y^2 - 2 \ge 0 \Rightarrow y \ge \sqrt{2} [\because a^4 \ge 0]` :math:`\alpha^4 + \beta^4 \ge 2 + \sqrt{2}` 265. :math:`\alpha + \beta = p` and :math:`\alpha\beta = q` :math:`\alpha^{\frac{1}{4}} + \beta^{\frac{1}{4}} = \sqrt[4]{\left(\alpha^{\frac{1}{4}} + \beta^{\frac{1}{4}}\right)^4}` :math:`= \sqrt[4]{\alpha + \beta + 6\sqrt{\alpha\beta} + 4\sqrt[4]{\alpha\beta(\alpha^2 + \beta^2)}}` :math:`= \sqrt[4]{p + 6\sqrt{q} + 4\sqrt[4]{q(p^2 - 2q)}}` 266. Let :math:`\alpha, beta` be roots of first equation and :math:`\gamma, \delta` be that of second equation. :math:`\alpha + \beta = \frac{b}{a}, \alpha\beta = \frac{c}{a}` and :math:`\gamma + \delta = \frac{c}{b}, \gamma\delta = \frac{a}{b}` According to question, :math:`\alpha - \beta = \gamma - \delta` :math:`(\alpha + \beta)^2 - 4\alpha\beta = (\gamma + \delta)^2 - 4\gamma\delta` :math:`\frac{b^2}{a^2} - \frac{4c}{a} = \frac{c^2}{b^2} - \frac{4a}{b}` :math:`\Rightarrow b^4 - a^2c^2 = 4ab(bc - a^2)` 267. A cubic equation whose roots are :math:`\alpha, \beta, \gamma` is given by :math:`f(x) = (x - \alpha)(x - \beta)(x - \gamma)` :math:`\therefore f'(x) = (x - \alpha)(x - \beta) + (x - \beta)(x - \gamma) + (x - \alpha)(x - \gamma)` Now it is trivial to prove that a sign change occurs for the given limits for :math:`f'(x)` and thus a root lies in these limits. 268. Since :math:`\alpha, \beta, \gamma, \delta` are in A. P. let :math:`\alpha = l - 3m, \beta = l - m, \gamma = l + m, \delta = l + 3m` where :math:`l` is the first term and :math:`m` is the common difference of A. P. :math:`\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}` and :math:`\gamma + \delta = -\frac{q}{p}, \gamma\delta = \frac{r}{p}` :math:`\frac{D_1}{D_2} = \frac{b^2 - 4ac}{q^2 - 4pr} = \frac{\frac{b^2}{a^2} - \frac{4c}{a}}{\frac{a^2}{p^2} - \frac{4r}{p}}\frac{a^2}{p^2}` :math:`= \frac{(\alpha - \beta)^2}{(\gamma - \delta)^2}\frac{a^2}{p^2} = \frac{4d^2}{4d^2}\frac{a^2}{p^2}` 269. This is similar to 268 and has been left as an exercise. 270. This is a very easy problem and has been left as an exercise. 271. :math:`\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}` and :math:`\alpha^4 + \beta^4 = -\frac{m}{l}, \alpha^4\beta^4 = \frac{n}{l}` Discriminant of given quadratic equation, :math:`D = 16a^2c^2l^2 - 4a2^l(2c^2l + a^2m)` :math:`= 8a^2c^2l^2 - 4a^4lm` :math:`= 4a^4l^2\left(2\frac{c^2}{a^2} - \frac{m}{l}\right)` :math:`= 4a^4l^2(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 2a^4l^2(\alpha^2 + \beta^2)^2` Therefore, roots of the given equation can be computed which are found to be :math:`(\alpha + \beta)^2, -(\alpha + \beta)^2` which are equal and opposite in sign. 272. :math:`\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}` and :math:`\gamma + \delta = -\frac{m}{l}, \gamma\delta = \frac{n}{l}` Equation whose roots are :math:`\alpha\gamma + \beta\delta` and :math:`\alpha\delta + \beta\gamma` is :math:`x^2 - (\alpha\gamma + \beta\delta + \alpha\delta + \beta\gamma)x + (\alpha\gamma + \beta\delta)(\alpha\delta + \beta\gamma) = 0` :math:`x^ - (\alpha + \beta)(\gamma + \delta)x + ((\alpha^2 + \beta^2)\gamma\delta + (\gamma^2 + \delta^2)\alpha\beta) = 0` :math:`a^2l^2x^2 - ablmx + (b^2 - 2ac)ln + (m^2 - 2ln)ac = 0` 273. :math:`p` and :math:`q` are roots of the equation :math:`3x^2 - 5x - 2 = 0` :math:`\Rightarrow p + q = \frac{5}{3}` and :math:`pq = -\frac{2}{3}` Equation whose roots are :math:`3p - 2q` and :math:`3q - 2p` is :math:`x^2 - (p + q)x - 6p^2 - 6q^2 + 13pq = 0` :math:`3x^2 - 5x - 100 = 0` 274. Since :math:`p` and :math:`q` are roots of the equation :math:`x^2 + bx + c = 0` therefore :math:`p + q = -b` and :math:`pq = c` Equation whose roots are :math:`b` and :math:`c` is :math:`x^2 - (b + c)x + bc = 0` :math:`x^2 +(p + q - pq)x - pq(p + q) = 0` 275. Sum of roots :math:`= 2\alpha = -p` and product of roots :math:`= \alpha^2 - \beta = q \Rightarrow \beta = \frac{p^2 - 4q}{4}` Equation whose roots are :math:`\frac{1}{\alpha}\pm \frac{1}{\sqrt{\beta}}` is :math:`x^2 - \frac{2}{\alpha}x + \frac{1}{\alpha^2} - \frac{1}{\beta} = 0` :math:`x^2 + \frac{2}{p}x + \frac{1}{p^2} - \frac{4}{p^2 - 4q} = 0` :math:`(p^2 - 4q)(p^2x^2 + 4px) = 16q` 276. Proceeding like previous problem the equation can be found as :math:`qx^2 - p(p^2 - q)(p^2 - 4q)x - p^2q^2(p^2 - 4q) = 0` Rest of the problems have been left as exercises.