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Question Number 58529 by Tawa1 last updated on 24/Apr/19

$$\mathrm{Solve}\mathrm{for}\mathrm{x}\mathrm{and}\mathrm{y}$$$$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=4.......\left(\mathrm{i}\right)$$$$\frac{{\mathrm{x}}^2}{\mathrm{y}}+\frac{{\mathrm{y}}^2}{\mathrm{x}}=9.......\left(\mathrm{ii}\right)$$

Commented by MJS last updated on 24/Apr/19

$$\mathrm{it}\mathrm{leads}\mathrm{to}\mathrm{an}\mathrm{exactly}\mathrm{solveable}\mathrm{polynome}$$$$\mathrm{of}{4}^{\mathrm{th}}\mathrm{degree}.\mathrm{I}\mathrm{will}\mathrm{show}\mathrm{later}$$

Commented by Tawa1 last updated on 24/Apr/19

$$\mathrm{Waiting}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

Commented by behi83417@gmail.com last updated on 25/Apr/19

$$x+y=4xy$$$$x^3+y^3=9xy\Rightarrow [x+y=p,xy=q]\Rightarrow p^3-3pq=9q$$$$\Rightarrow \{\begin{array}{l}p=4q\\ p^3-3pq=9q\Rightarrow {\left(4q\right)}^3-3\left(4q\right)q=9q\end{array}$$$$\Rightarrow 64q^3-12q^2-9q=0\Rightarrow q(64q^2-12q-9)=0$$$$\Rightarrow 1)q=0\Rightarrow (x=0\vee y=0,notokfororiginal$$$$question)$$$$\Rightarrow 2)64q^2-12q-9=0\Rightarrow q=\frac{6\pm \sqrt{36+9\times 64}}{64}$$$$\Rightarrow [q=\mathbf{xy}=3\times \frac{1\pm \sqrt{17}}{32}\Rightarrow p=\mathbf{x}+\mathbf{y}=3\times \frac{1\pm \sqrt{17}}{8}]$$$${\mathbf{z}}^2-\frac{3}{8}(1\pm \sqrt{17})\mathbf{z}+\frac{3}{32}(1\pm \sqrt{17})=0$$$${\mathbf{z}}_{1,2}=\frac{3}{16}[(1+\sqrt{17})\pm \sqrt{{(1+\sqrt{17})}^2-(1+\sqrt{17})}]$$$$=\frac{3}{16}[(1+\sqrt{17})\pm \sqrt{17+\sqrt{17}}]$$$${\mathbf{z}}_{3,4}=\frac{3}{16}[(1-\sqrt{17})\pm \sqrt{{(1-\sqrt{17})}^2-(1-\sqrt{17})}]$$$$=\frac{3}{16}[(1-\sqrt{17})\pm \sqrt{17-\sqrt{17}}]$$$${\mathbf{z}}_{5,6}=\frac{3}{16}[(1+\sqrt{17})\pm \sqrt{{(1+\sqrt{17})}^2-(1-\sqrt{17})}]$$$$=\frac{3}{16}[(1+\sqrt{17})\pm \sqrt{17+3\sqrt{17}}]$$$${\mathbf{z}}_{7,8}=\frac{3}{16}[(1-\sqrt{17})\pm \sqrt{{(1-\sqrt{17})}^2-(1+\sqrt{17})}]$$$$=\frac{3}{16}[(1-\sqrt{17})\pm \sqrt{17-3\sqrt{17}}]$$

Commented by Tawa1 last updated on 25/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 25/Apr/19

$$\left(\mathrm{i}\right)\Rightarrow y=\frac{x}{4x-1}\Rightarrow$$$$\Rightarrow \left(\mathrm{ii}\right)x^4-\frac{3}{4}{x}^3-\frac{33}{16}{x}^2+\frac{9}{8}x-\frac{9}{64}=0$$$$\mathrm{put}{x}_{1,2}=\alpha \pm \sqrt{\beta };x_{3,4}=\gamma \pm \sqrt{\delta }$$$$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=0$$$$x^4-2(\alpha +\gamma )x^3+({\alpha }^2+4\alpha \gamma -\beta +{\gamma }^2+\delta )x^2-2(\alpha \gamma (\alpha +\gamma )+\alpha \delta -\beta \gamma )x+({\alpha }^2-\beta )({\gamma }^2+\delta )=0$$$$\mathrm{now}\mathrm{compare}\mathrm{factors}:$$$$\left(1\right)-2(\alpha +\gamma )=-\frac{3}{4}$$$$\left(2\right){\alpha }^2+4\alpha \gamma -\beta +{\gamma }^2+\delta =-\frac{33}{16}$$$$\left(3\right)-2(\alpha \gamma (\alpha +\gamma )+\alpha \delta -\beta \gamma )=\frac{9}{8}$$$$\left(4\right)({\alpha }^2-\beta )({\gamma }^2+\delta )=-\frac{9}{64}$$$$\mathrm{now}\mathrm{solve}\left(1\right)\mathrm{for}\alpha ,\left(2\right)\mathrm{for}\beta ,\left(3\right)\mathrm{for}\gamma$$$$\alpha =\frac{3}{8}-\gamma$$$$\beta =-2{\gamma }^2+\frac{3}{4}\gamma +\delta +\frac{141}{64}$$$$\delta =\frac{32{\gamma }^3-18{\gamma }^2-33\gamma +9}{2(16\gamma -3)}$$$$\Rightarrow \left(4\right)$$$${\gamma }^6-\frac{9}{8}{\gamma }^5-\frac{39}{64}{\gamma }^4+\frac{369}{512}{\gamma }^3+\frac{423}{4096}{\gamma }^2-\frac{3051}{32768}\gamma +\frac{567}{65536}=0$$$$\gamma =r+\frac{3}{16}$$$$r^6-\frac{291}{256}{r}^4+\frac{21267}{65536}{r}^2-\frac{23409}{16777216}=0$$$$r=\sqrt{s}$$$$s^3-\frac{291}{256}{s}^2+\frac{21267}{65536}s-\frac{23409}{16777216}=0$$$$s=t+\frac{97}{256}$$$$t^3-\frac{435}{4096}t+\frac{1673}{131072}=0$$$$\mathrm{trying}\mathrm{factors}\mathrm{of}\frac{1673}{131072}\Rightarrow t_1=\frac{7}{32}$$$$(\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{need}{t}_{2,3}=-\frac{7}{64}\pm \frac{12\sqrt{2}}{64})$$$$t=\frac{7}{32}\Rightarrow s=\frac{153}{256}\Rightarrow r=\frac{3\sqrt{17}}{16}\Rightarrow \gamma =\frac{3}{16}+\frac{3\sqrt{17}}{16}$$$$\mathrm{now}$$$$\alpha =\frac{3}{16}-\frac{3\sqrt{17}}{16}$$$$\beta =\frac{69}{128}+\frac{3\sqrt{17}}{128}$$$$\gamma =\frac{3}{16}+\frac{3\sqrt{17}}{16}$$$$\delta =-\frac{69}{128}+\frac{3\sqrt{17}}{16}$$$$\Rightarrow$$$$x_1=\frac{3}{16}-\frac{3\sqrt{17}}{16}-\frac{\sqrt{6(23+\sqrt{17})}}{16}$$$$x_2=\frac{3}{16}-\frac{3\sqrt{17}}{16}+\frac{\sqrt{6(23+\sqrt{17})}}{16}$$$$x_{3,4}\notin \mathbb{R}$$$$y_1=x_2$$$$y_2=x_1$$

Commented by MJS last updated on 25/Apr/19

$$\mathrm{I}\mathrm{prepared}\mathrm{these}\mathrm{formulas},\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{just}$$$$\mathrm{putting}\mathrm{in}\mathrm{the}\mathrm{constants}$$$$x^4+{ax}^3+{bx}^2+cx+d=0$$$$(x-\alpha -\sqrt{\beta })(x-\alpha +\sqrt{\beta })(x-\gamma -\sqrt{\delta })(x-\gamma +\sqrt{\delta })=0$$$$\left(1\right)-2(\alpha +\gamma )=a$$$$\left(2\right){\alpha }^2+4\alpha \gamma -\beta +{\gamma }^2-\delta =b$$$$\left(3\right)-2(\alpha \gamma (\alpha +\gamma )-\alpha \delta -\beta \gamma )=c$$$$\left(4\right)({\alpha }^2-\beta )({\gamma }^2-\delta )=d$$$$\Rightarrow$$$$\left(1\right)\alpha =-\gamma -\frac{a}{2}$$$$\left(2\right)\beta =-2{\gamma }^2-a\gamma -\delta +\frac{a^2}{4}-b$$$$\left(3\right)\delta =-{\gamma }^2-\frac{a}{2}\gamma +\frac{(a^2-4b)\gamma -2c}{2(4\gamma +a)}$$$$\Rightarrow \beta =-{\gamma }^2-\frac{a}{2}\gamma +\frac{2(a^2-4b)\gamma +a^3-4ab+4c}{4(4\gamma +a)}$$$$\left(4\right)$$$${\gamma }^6+\frac{3a}{2}{\gamma }^5+\frac{3a^2+2b}{4}{\gamma }^4+\frac{a(a^2+4b)}{8}{\gamma }^3+\frac{2a^{2}b+ac+b^2-4d}{16}{\gamma }^2+\frac{a(ac+b^2-4d)}{32}\gamma -\frac{a^{2}d-abc+c^2}{64}=0$$$$\gamma =r-\frac{a}{4}$$$$r^6-\frac{3a^2-8b}{16}{r}^4+\frac{3a^4-16(a^{2}b-ac-b^2+4d)}{256}{r}^2-\frac{{(a^3-4ab+8c)}^2}{4096}=0$$$$r=\sqrt{s}$$$$s^3-\frac{3a^2-8b}{16}{s}^2+\frac{3a^4-16(a^{2}b-ac-b^2+4d)}{256}s-\frac{{(a^3-4ab+8c)}^2}{4096}=0$$$$s=t+\frac{3a^2-8b}{48}$$$$t^3+\frac{3ac-b^2-12d}{48}t-\frac{2b^3+9(3a^{2}d-abc-8bd+3c^2)}{1728}=0$$$$p=\frac{3ac-b^2-12d}{48}$$$$q=-\frac{2b^3+9(3a^{2}d-abc-8bd+3c^2)}{1728}$$$$\mathrm{now}\mathrm{we}\mathrm{have}\mathrm{to}$$$$1.\mathrm{try}\mathrm{factors}\mathrm{of}q\Rightarrow \mathrm{solved}$$$$2.\mathrm{calculate}$$$$D=\frac{p^3}{27}+\frac{q^2}{4}\mathrm{and}\mathrm{decide}\mathrm{which}\mathrm{method}\mathrm{to}\mathrm{use}$$$$D<0\Rightarrow \mathrm{trigonometric}\mathrm{solution}$$$$D⩾0\Rightarrow {\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{solution}$$$$\mathrm{in}\mathrm{most}\mathrm{cases}\mathrm{we}{\mathrm{won}}^{\prime }\mathrm{t}\mathrm{get}\mathrm{useable}\mathrm{solutions}$$$$\mathrm{but}\mathrm{we}\mathrm{always}\mathrm{get}\mathrm{exact}\mathrm{solutions}$$$$\gamma =\sqrt{t+\frac{3a^2-8b}{48}}-\frac{a}{4}$$$$\mathrm{if}t{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{a}‘‘{\mathrm{nice}}^{″}\mathrm{number}\mathrm{we}\mathrm{cannot}\mathrm{handle}$$$$\alpha ,\beta ,\gamma ,\delta \mathrm{nor}x=\{\begin{array}{l}\alpha \pm \sqrt{\beta }\\ \gamma \pm \sqrt{\delta }\end{array}$$

Commented by Tawa1 last updated on 24/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by tanmay last updated on 25/Apr/19

$$withoutusingcalculatorhowtosolvethisproblem?$$$$isitfeasibletocaculatesuchbignumber$$$$65536,16777216,33768etctocalculatebysimple$$$$+,-,\times ,/etc$$$$$$

Commented by tanmay last updated on 25/Apr/19

$$oksir$$

Answered by tanmay last updated on 24/Apr/19

$$x+y=4xy$$$$x^3+y^3=9xy$$$${(x+y)}^3-3xy(x+y)=9xy$$$$64x^3{y}^3-12x^2{y}^2-9xy=0$$$$xy\ne 0$$$$64x^2{y}^2-12xy-9=0$$$$xy=\frac{12\pm \sqrt{144+64\times 36}}{2\times 64}=a$$$$$$$$x+y=4a$$$$xy=a$$$$x(4a-x)=a$$$$-x^2+4ax-a=0$$$$x^2-4ax+a=0$$$$x=\frac{4a\pm \sqrt{16a^2-4a}}{2}$$$$y=4a-\left(\frac{4a\pm \sqrt{16a^2-4a}}{2}\right)$$$$plscalculatevalueofa...ihaveapathytocaculate$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by Tawa1 last updated on 24/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 24/Apr/19

$$\left(\mathrm{i}\right)\Rightarrow \frac{\mathrm{x}+\mathrm{y}}{\mathrm{xy}}=4.......\left(\mathrm{iii}\right)$$$$\left(\mathrm{ii}\right)\Rightarrow \frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{\mathrm{xy}}=9......\left(\mathrm{iv}\right)$$$$\left(\mathrm{iv}\right)\mathbf{÷}\left(\mathrm{iii}\right):\frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{\mathrm{x}+\mathrm{y}}=\frac{9}{4}$$$${\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{xy}=\frac{9}{4}$$$${(\mathrm{x}+\mathrm{y})}^2-3\mathrm{xy}=\frac{9}{4}$$$$\mathrm{From}\left(\mathrm{iii}\right)\mathrm{xy}=\frac{\mathrm{x}+\mathrm{y}}{4}$$$$\mathrm{So}{(\mathrm{x}+\mathrm{y})}^2-3\left(\frac{\mathrm{x}+\mathrm{y}}{4}\right)-\frac{9}{4}=0$$$$4{(\mathrm{x}+\mathrm{y})}^2-3(\mathrm{x}+\mathrm{y})-9=0$$$$\mathrm{x}+\mathrm{y}=\frac{3\pm \sqrt{9-4\left(4\right)(-9)}}{2\left(4\right)}$$$$=\frac{3\pm 3\sqrt{17}}{8}$$$$Continue$$

Commented by Tawa1 last updated on 24/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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