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Question Number 64130 by ANTARES VY last updated on 14/Jul/19

(√(4x+((12)/x)))=((x^2 +7)/(x+1)) x=?

$$\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}+\frac{\mathrm{12}}{\boldsymbol{\mathrm{x}}}}=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{7}}{\boldsymbol{\mathrm{x}}+\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{x}}=? \\ $$

Commented by Prithwish sen last updated on 14/Jul/19

x= 1,1,3 and ((−1±15i)/2)

$$\mathrm{x}=\:\mathrm{1},\mathrm{1},\mathrm{3}\:\mathrm{and}\:\frac{−\mathrm{1}\pm\mathrm{15i}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 14/Jul/19

2(√((x^2 +3)/x))=((x^2 +7)/(x+1)) x=1 and 3 are solutions. ((4(x^2 +3))/x)=((x^4 +14x^2 +49)/(x^2 +2x+1)) x^5 +14x^3 +49x=4x^4 +8x^3 +4x^2 +12x^2 +24x+12 x^5 −4x^4 +6x^3 −16x^2 +25x−12=0 ...... (x−1)^2 (x−3)(x^2 +x+4)=0 ⇒x=1, 3 x^2 +x+4=0 no other real solution.

$$\mathrm{2}\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}}}=\frac{{x}^{\mathrm{2}} +\mathrm{7}}{{x}+\mathrm{1}} \\ $$$${x}=\mathrm{1}\:{and}\:\mathrm{3}\:{are}\:{solutions}. \\ $$$$\frac{\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{{x}}=\frac{{x}^{\mathrm{4}} +\mathrm{14}{x}^{\mathrm{2}} +\mathrm{49}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$$${x}^{\mathrm{5}} +\mathrm{14}{x}^{\mathrm{3}} +\mathrm{49}{x}=\mathrm{4}{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{24}{x}+\mathrm{12} \\ $$$${x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} −\mathrm{16}{x}^{\mathrm{2}} +\mathrm{25}{x}−\mathrm{12}=\mathrm{0} \\ $$$$...... \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1},\:\mathrm{3} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{4}=\mathrm{0}\: \\ $$$${no}\:{other}\:{real}\:{solution}. \\ $$

Answered by ajfour last updated on 14/Jul/19

4(x^2 +3)(x+1)^2 =x(x^2 +7)^2 let 4a^2 (x^2 +3)=x & x+1=a(x^2 +7) ⇒ x=(1/(8a^2 ))±(√((1/(64a^4 ))−3)) x=(1/(2a))±(√((1/(4a^2 ))+(1/a)−7)) let the irrational & rational parts be equal ⇒ a=(1/4) & we see that for this value the irrational parts are both same (=1). hence x= 2±1 . ⇒ real roots are x=1,3 .

$$\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} ={x}\left({x}^{\mathrm{2}} +\mathrm{7}\right)^{\mathrm{2}} \\ $$$${let}\:\:\:\:\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}\right)={x} \\ $$$$\:\:\&\:\:\:{x}+\mathrm{1}={a}\left({x}^{\mathrm{2}} +\mathrm{7}\right) \\ $$$$\Rightarrow\:\:{x}=\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{2}} }\pm\sqrt{\frac{\mathrm{1}}{\mathrm{64}{a}^{\mathrm{4}} }−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}{a}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}}−\mathrm{7}} \\ $$$${let}\:{the}\:{irrational}\:\&\:{rational}\:{parts}\:{be}\:{equal} \\ $$$$\Rightarrow\:\:{a}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\&\:\:{we}\:{see}\:{that}\:{for}\:{this} \\ $$$${value}\:{the}\:{irrational}\:{parts}\:{are} \\ $$$${both}\:{same}\:\left(=\mathrm{1}\right). \\ $$$${hence}\:\:\:\:\:{x}=\:\mathrm{2}\pm\mathrm{1}\:\:. \\ $$$$\Rightarrow\:\:{real}\:{roots}\:{are}\:\:{x}=\mathrm{1},\mathrm{3}\:. \\ $$

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