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

Commented by Frix last updated on 17/Nov/24

$$\mathrm{There}\mathrm{is}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{complex}\mathrm{solutions}\mathrm{but}\mathrm{the}$$$$\mathrm{exact}\mathrm{form}\mathrm{is}\mathrm{not}\mathrm{useable}.$$$$x\approx 1.32848492\pm .570204126\mathrm{i}$$

Commented by Ghisom last updated on 17/Nov/24

$$\mathrm{maybe}\mathrm{a}\mathrm{misprint}?$$$$\mathrm{we}\mathrm{get}\mathrm{one}\mathrm{solution}\in \mathbb{Z}\mathrm{for}n⩾0\mathrm{with}$$$$3x^2-x-(3x-1)\sqrt{x+3}-(n^2-n-3)(3n^2-10)=0$$$$\Rightarrow x_1=n^2-3$$$$\mathrm{with}n=2\mathrm{we}\mathrm{have}$$$$3x^2-x-(3x-1)\sqrt{x+3}+2=0$$$$\Rightarrow x_1=1$$$$x_2=\frac{2}{9}(1+{46}^{1/2}\sin \frac{\pi +\arcsin \frac{349}{2^{5/2}{23}^{3/2}}}{3})\approx$$$$\approx 1.65014793569$$

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

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

Answered by A5T last updated on 17/Nov/24

$${(3x^2-x+3)}^2={(3x-1)}^2(x+3)$$$$\Rightarrow 9x^4+x^2+9-6x^3-6x+18x^2=(9x^2-6x+1)(x+3)$$$$\Rightarrow 9x^4-6x^3+19x^2-6x+9=9x^3+21x^2-17x+3$$$$\Rightarrow 9x^4-15x^3-2x^2+11x+6=0$$$$x\approx 1.3285\underset{-}{+}0.5702i$$

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