Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 61479 by Tawa1 last updated on 03/Jun/19

$$\mathrm{Solve}\mathrm{for}\mathrm{x}:\frac{6\sqrt{2\mathrm{x}}}{\mathrm{x}-1}+\frac{5\sqrt{\mathrm{x}-1}}{2\mathrm{x}}=13$$

Commented by MJS last updated on 03/Jun/19

$$\mathrm{we}\mathrm{cannot}\mathrm{generally}\mathrm{solve}\mathrm{this}...$$

Answered by ajfour last updated on 03/Jun/19

$$letx=\sec {}^2\theta$$$$\Rightarrow 6\sqrt{2}\sec \theta \cot {}^2\theta +\frac{5\tan \theta \cos {}^2\theta }{2}=13$$$$\frac{12\sqrt{2}\cos \theta }{\sin {}^2\theta }+5\sin \theta \cos \theta =26$$$${\left(\frac{12\sqrt{2}+5\sin {}^3\theta }{\sin {}^2\theta }\right)}^2(1-\sin {}^2\theta )=676$$$$\Rightarrow {(a+{bt}^3)}^2(1-t^2)={ct}^4$$$$wherea=12\sqrt{2},b=5,c=676,$$$$andt=\sin \theta$$$$\left(Apolynomialofdegree8\right)$$

Commented by Tawa1 last updated on 03/Jun/19

$$\mathrm{Ohh}.$$

Answered by MJS last updated on 03/Jun/19

$$x\approx 2.028522\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{solution}\in \mathbb{R}$$$$\left(\mathrm{found}\mathrm{no}\mathrm{exact}\mathrm{value}\right)$$

Commented by Tawa1 last updated on 03/Jun/19

$$\mathrm{Any}\mathrm{workings}\mathrm{sir}$$

Commented by MJS last updated on 03/Jun/19

$$\mathrm{we}\mathrm{can}\mathrm{square}\mathrm{it}2\mathrm{times}\mathrm{but}\mathrm{we}\mathrm{end}\mathrm{up}\mathrm{with}$$$$\mathrm{a}\mathrm{polynome}\mathrm{of}\mathrm{degree}8\mathrm{which}\mathrm{has}\mathrm{got}\mathrm{no}$$$$\mathrm{trivial}\mathrm{solution}$$$$[\mathrm{squaring}\mathrm{like}\mathrm{this}:$$$$\sqrt{a}+\sqrt{b}=c$$$$a+\sqrt{a}\sqrt{b}+b=c^2$$$$ab={(c^2-a-b)}^2]$$$$$$$$\mathrm{approximated}\mathrm{using}\mathrm{a}\mathrm{calculator}$$$$\mathrm{the}\mathrm{equation}\mathrm{is}\mathrm{defined}\mathrm{for}x>1\mathrm{so}\mathrm{I}\mathrm{started}$$$$\mathrm{with}x=1.5,x=2,x=2.5$$

Answered by behi83417@gmail.com last updated on 03/Jun/19

$$\sqrt{2\mathrm{x}}=\mathrm{a},\sqrt{\mathrm{x}-1}=\mathrm{b}\Rightarrow \mathrm{x}=\frac{{\mathrm{a}}^2}{2}={\mathrm{b}}^2+1$$$$\frac{6\mathrm{a}}{{\mathrm{b}}^2}+\frac{5\mathrm{b}}{{\mathrm{a}}^2}=13,{\mathrm{a}}^2={2\mathrm{b}}^2+2$$$$\Rightarrow \{\begin{array}{l}{6\mathrm{a}}^3+{5\mathrm{b}}^3={13\mathrm{a}}^2{\mathrm{b}}^2\\ {\mathrm{a}}^2-{2\mathrm{b}}^2=2\Rightarrow {\mathrm{a}}^3=2\mathrm{a}({\mathrm{b}}^2+1)\end{array}$$$$\Rightarrow \{\begin{array}{l}5(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2)+{\mathrm{a}}^3={13\mathrm{a}}^2{\mathrm{b}}^2\\ (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})={\mathrm{b}}^2+2\end{array}$$$$\underset{\mathrm{ab}=\mathrm{q}}{\stackrel{\mathrm{a}+\mathrm{b}=\mathrm{p}}{\Rightarrow }}\{\begin{array}{l}5\mathrm{p}({\mathrm{p}}^2-3\mathrm{q})={13\mathrm{a}}^2{\mathrm{b}}^2-2\mathrm{a}({\mathrm{b}}^2+1)\\ {\mathrm{p}}^2({\mathrm{p}}^2-4\mathrm{q})={({\mathrm{b}}^2+2)}^2\end{array}$$$$\Rightarrow \{\begin{array}{l}{5\mathrm{p}}^3-15\mathrm{pq}={13\mathrm{a}}^2(\frac{{\mathrm{a}}^2}{2}-1)-2\mathrm{a}\left(\frac{{\mathrm{a}}^2}{2}\right)\\ {\mathrm{p}}^4-{4\mathrm{p}}^2\mathrm{q}={(\frac{{\mathrm{a}}^2}{2}+1)}^2\end{array}$$$$\Rightarrow \{\begin{array}{l}{80\mathrm{p}}^3-240\mathrm{pq}=8\mathrm{p}({13\mathrm{a}}^4-{2\mathrm{a}}^3-{26\mathrm{a}}^2)\\ {60\mathrm{p}}^4-{240\mathrm{p}}^2\mathrm{q}=15({\mathrm{a}}^4+{4\mathrm{a}}^2+4)\end{array}$$$${140\mathrm{p}}^4-{8\mathrm{pa}}^2({13\mathrm{a}}^2-2\mathrm{a}-26)-15{({\mathrm{a}}^2+2)}^2=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com