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Question Number 46586 by naka3546 last updated on 28/Oct/18

$$Solvex\in \mathbb{R}$$$$x+\frac{x}{\sqrt{x^2+1}}=\frac{35}{12}$$

Answered by Kunal12588 last updated on 29/Oct/18

$$\Rightarrow \frac{x\sqrt{x^2+1}+x}{\sqrt{x^2+1}}=\frac{35}{12}$$$$\Rightarrow 12x\sqrt{x^2+1}+12x=35\sqrt{x^2+1}$$$$\Rightarrow 12x=(35-12x)\sqrt{x^2+1}$$$$\Rightarrow 144x^2=(1225+144x^2-840x)(x^2+1)$$$$\Rightarrow 144x^4-840x^3+1225x^2-840x+1225=0$$$$itsabiquadraticequation,howtosolveit?$$

Commented by Kunal12588 last updated on 29/Oct/18

$$calculator:$$$$realsolutions:3.88510101188and2.02043398548(satisfiesthequestion.)$$

Commented by MJS last updated on 29/Oct/18

$$3.88...\mathrm{is}\mathrm{no}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{eq}.$$$$\mathrm{squaring}\mathrm{leads}\mathrm{to}\mathrm{apparent}\mathrm{solutions}$$

Commented by Kunal12588 last updated on 29/Oct/18

$$yes,onlysolutionbycalculatoris2.02043398548$$

Answered by MJS last updated on 29/Oct/18

$$f\left(x\right)=x+\frac{x}{\sqrt{x^2+1}}-\frac{35}{12}$$$$f\left(0\right)=-\frac{35}{12}<0\wedge f\left(\frac{35}{12}\right)=\frac{\frac{35}{12}}{\sqrt{{\left(\frac{35}{12}\right)}^2+1}}>0$$$$f^{\prime }\left(x\right)=1+\frac{1}{\sqrt{{(x^2+1)}^3}}>0\mathrm{\forall }x\in \mathbb{R}$$$$\Rightarrow f\left(x\right)\mathrm{has}\mathrm{got}\mathrm{exactly}\mathrm{one}\mathrm{zero}$$$$\mathrm{approximation}\mathrm{leads}\mathrm{to}x\approx 2.02043$$$$\mathrm{the}\mathrm{exact}\mathrm{solution}\mathrm{by}\mathrm{solving}\mathrm{the}\mathrm{resulting}$$$$\mathrm{polynome}{x}^4-\frac{35}{6}{x}^3+\frac{1225}{144}{x}^2-\frac{35}{6}x+\frac{1225}{144}=0$$$$\mathrm{is}\mathrm{not}\mathrm{manageable}$$

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