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Question Number 154497 by mathdanisur last updated on 18/Sep/21

$$\mathrm{Solve}\mathrm{the}\mathrm{equations}:$$$$\mathbf{a})2\sqrt{{2\mathrm{x}}^3-\mathrm{x}}={3\mathrm{x}}^2-3\mathrm{x}+2$$$$\mathbf{b})\sqrt{\frac{{\mathrm{x}}^4+16}{2}}+\sqrt{2({\mathrm{x}}^2+4)}=3\mathrm{x}+2$$

Answered by ARUNG_Brandon_MBU last updated on 18/Sep/21

$$\mathrm{a}.x=1\mathrm{is}\mathrm{a}\mathrm{solution}$$

Answered by Fridunatjan08 last updated on 19/Sep/21

$$$$$$\mathrm{Solve}\mathrm{the}\mathrm{equations}:$$$$\mathbf{a})2\sqrt{{2\mathrm{x}}^3-\mathrm{x}}={3\mathrm{x}}^2-3\mathrm{x}+2$$$${\left(2\sqrt{2x^3-x}\right)}^2={(3x^2-3x+2)}^2$$$$8x^3-4x=9x^4-18x^3+21x^2-12x+4$$$$9x^4-26x^3+21x^2-8x+4=0$$$$p\left(x\right)=9x^4-26x^3+21x^2-8x+4$$$$p\left(1\right)=0$$$$(x-1)(9x^3-17x^2+4x-4)=0$$$$x-1=0\mid 9x^3-17x^2+4x-4=0$$$$x_1=1.$$

Commented by mathdanisur last updated on 19/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{Ser}\mathrm{thanks}$$

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