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

Solve x ∈ R x + (x/(√(x^2 + 1))) = ((35)/(12))

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

Answered by Kunal12588 last updated on 29/Oct/18

⇒((x(√(x^2 +1))+x)/(√(x^2 +1)))=((35)/(12)) ⇒12x(√(x^2 +1))+12x=35(√(x^2 +1)) ⇒12x=(35−12x)(√(x^2 +1)) ⇒144x^2 =(1225+144x^2 −840x)(x^2 +1) ⇒144x^4 −840x^3 +1225x^2 −840x+1225=0 its a biquadratic equation, how to solve it?

$$\Rightarrow\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}=\frac{\mathrm{35}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{12}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{12}{x}=\mathrm{35}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{12}{x}=\left(\mathrm{35}−\mathrm{12}{x}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{144}{x}^{\mathrm{2}} =\left(\mathrm{1225}+\mathrm{144}{x}^{\mathrm{2}} −\mathrm{840}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{144}{x}^{\mathrm{4}} −\mathrm{840}{x}^{\mathrm{3}} +\mathrm{1225}{x}^{\mathrm{2}} −\mathrm{840}{x}+\mathrm{1225}=\mathrm{0} \\ $$$${its}\:{a}\:{biquadratic}\:{equation},\:{how}\:{to}\:{solve}\:{it}? \\ $$

Commented by Kunal12588 last updated on 29/Oct/18

calculator : real solutions : 3.88510101188 and 2.02043398548(satisfies the question.)

$${calculator}\::\: \\ $$$${real}\:{solutions}\::\:\mathrm{3}.\mathrm{88510101188}\:{and}\:\mathrm{2}.\mathrm{02043398548}\left({satisfies}\:{the}\:{question}.\right) \\ $$

Commented by MJS last updated on 29/Oct/18

3.88... is no solution of the given eq. squaring leads to apparent solutions

$$\mathrm{3}.\mathrm{88}...\:\mathrm{is}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{eq}. \\ $$$$\mathrm{squaring}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{apparent}\:\mathrm{solutions} \\ $$

Commented by Kunal12588 last updated on 29/Oct/18

yes, only solution by calculator is 2.02043398548

$${yes},\:{only}\:{solution}\:{by}\:{calculator}\:{is}\:\mathrm{2}.\mathrm{02043398548} \\ $$

Answered by MJS last updated on 29/Oct/18

f(x)=x+(x/(√(x^2 +1)))−((35)/(12)) f(0)=−((35)/(12))<0∧f(((35)/(12)))=(((35)/(12))/(√((((35)/(12)))^2 +1)))>0 f′(x)=1+(1/(√((x^2 +1)^3 )))>0∀x∈R ⇒ f(x) has got exactly one zero approximation leads to x≈2.02043 the exact solution by solving the resulting polynome x^4 −((35)/6)x^3 +((1225)/(144))x^2 −((35)/6)x+((1225)/(144))=0 is not manageable

$${f}\left({x}\right)={x}+\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{35}}{\mathrm{12}} \\ $$$${f}\left(\mathrm{0}\right)=−\frac{\mathrm{35}}{\mathrm{12}}<\mathrm{0}\wedge{f}\left(\frac{\mathrm{35}}{\mathrm{12}}\right)=\frac{\frac{\mathrm{35}}{\mathrm{12}}}{\sqrt{\left(\frac{\mathrm{35}}{\mathrm{12}}\right)^{\mathrm{2}} +\mathrm{1}}}>\mathrm{0} \\ $$$${f}'\left({x}\right)=\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }}>\mathrm{0}\forall{x}\in\mathbb{R} \\ $$$$\Rightarrow\:{f}\left({x}\right)\:\mathrm{has}\:\mathrm{got}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{zero} \\ $$$$\mathrm{approximation}\:\mathrm{leads}\:\mathrm{to}\:{x}\approx\mathrm{2}.\mathrm{02043} \\ $$$$\mathrm{the}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{by}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{resulting} \\ $$$$\mathrm{polynome}\:{x}^{\mathrm{4}} −\frac{\mathrm{35}}{\mathrm{6}}{x}^{\mathrm{3}} +\frac{\mathrm{1225}}{\mathrm{144}}{x}^{\mathrm{2}} −\frac{\mathrm{35}}{\mathrm{6}}{x}+\frac{\mathrm{1225}}{\mathrm{144}}=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{manageable} \\ $$

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