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Question Number 213803 by muallimRiyoziyot last updated on 17/Nov/24

Commented by Frix last updated on 17/Nov/24

There is a pair of complex solutions but the exact form is not useable. x≈1.32848492±.570204126i

$$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{exact}\:\mathrm{form}\:\mathrm{is}\:\mathrm{not}\:\mathrm{useable}. \\ $$$${x}\approx\mathrm{1}.\mathrm{32848492}\pm.\mathrm{570204126i} \\ $$

Commented by Ghisom last updated on 17/Nov/24

maybe a misprint? we get one solution ∈Z for n≥0 with 3x^2 −x−(3x−1)(√(x+3))−(n^2 −n−3)(3n^2 −10)=0 ⇒ x_1 =n^2 −3 with n=2 we have 3x^2 −x−(3x−1)(√(x+3))+2=0 ⇒ x_1 =1 x_2 =(2/9)(1+46^(1/2) sin ((π+arcsin ((349)/(2^(5/2) 23^(3/2) )))/3))≈ ≈1.65014793569

$$\mathrm{maybe}\:\mathrm{a}\:\mathrm{misprint}? \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{one}\:\mathrm{solution}\:\in\mathbb{Z}\:\mathrm{for}\:{n}\geqslant\mathrm{0}\:\mathrm{with} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −{x}−\left(\mathrm{3}{x}−\mathrm{1}\right)\sqrt{{x}+\mathrm{3}}−\left({n}^{\mathrm{2}} −{n}−\mathrm{3}\right)\left(\mathrm{3}{n}^{\mathrm{2}} −\mathrm{10}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} ={n}^{\mathrm{2}} −\mathrm{3} \\ $$$$\mathrm{with}\:{n}=\mathrm{2}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −{x}−\left(\mathrm{3}{x}−\mathrm{1}\right)\sqrt{{x}+\mathrm{3}}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{1}+\mathrm{46}^{\mathrm{1}/\mathrm{2}} \mathrm{sin}\:\frac{\pi+\mathrm{arcsin}\:\frac{\mathrm{349}}{\mathrm{2}^{\mathrm{5}/\mathrm{2}} \mathrm{23}^{\mathrm{3}/\mathrm{2}} }}{\mathrm{3}}\right)\approx \\ $$$$\:\:\:\:\:\approx\mathrm{1}.\mathrm{65014793569} \\ $$

Answered by Rasheed.Sindhi last updated on 17/Nov/24

(√(x+3)) =y⇒x=y^2 −3 3(y^2 −3)^2 −(y^2 −3)−( 3(y^2 −3)−1)y+3=0 3(y^4 −6y^2 +9)−y^2 +3−((3y^2 −9)−1)y+3=0 3y^4 −18y^2 +27−y^2 +3−3y^3 +10y+3=0 3y^4 −3y^3 −19y^2 +10y+33=0

$$\sqrt{{x}+\mathrm{3}}\:={y}\Rightarrow{x}={y}^{\mathrm{2}} −\mathrm{3} \\ $$$$\mathrm{3}\left({y}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} −\left({y}^{\mathrm{2}} −\mathrm{3}\right)−\left(\:\mathrm{3}\left({y}^{\mathrm{2}} −\mathrm{3}\right)−\mathrm{1}\right){y}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}\left({y}^{\mathrm{4}} −\mathrm{6}{y}^{\mathrm{2}} +\mathrm{9}\right)−{y}^{\mathrm{2}} +\mathrm{3}−\left(\left(\mathrm{3}{y}^{\mathrm{2}} −\mathrm{9}\right)−\mathrm{1}\right){y}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}{y}^{\mathrm{4}} −\mathrm{18}{y}^{\mathrm{2}} +\mathrm{27}−{y}^{\mathrm{2}} +\mathrm{3}−\mathrm{3}{y}^{\mathrm{3}} +\mathrm{10}{y}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}{y}^{\mathrm{4}} −\mathrm{3}{y}^{\mathrm{3}} −\mathrm{19}{y}^{\mathrm{2}} +\mathrm{10}{y}+\mathrm{33}=\mathrm{0} \\ $$

Answered by A5T last updated on 17/Nov/24

(3x^2 −x+3)^2 =(3x−1)^2 (x+3) ⇒9x^4 +x^2 +9−6x^3 −6x+18x^2 =(9x^2 −6x+1)(x+3) ⇒9x^4 −6x^3 +19x^2 −6x+9=9x^3 +21x^2 −17x+3 ⇒9x^4 −15x^3 −2x^2 +11x+6=0 x≈1.3285+_− 0.5702i

$$\left(\mathrm{3}{x}^{\mathrm{2}} −{x}+\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{9}{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{9}−\mathrm{6}{x}^{\mathrm{3}} −\mathrm{6}{x}+\mathrm{18}{x}^{\mathrm{2}} =\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{9}{x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{19}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}=\mathrm{9}{x}^{\mathrm{3}} +\mathrm{21}{x}^{\mathrm{2}} −\mathrm{17}{x}+\mathrm{3} \\ $$$$\Rightarrow\mathrm{9}{x}^{\mathrm{4}} −\mathrm{15}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{6}=\mathrm{0} \\ $$$${x}\approx\mathrm{1}.\mathrm{3285}\underset{−} {+}\mathrm{0}.\mathrm{5702}{i} \\ $$

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