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Question Number 91491 by jagoll last updated on 01/May/20

$$x^3+1=2\sqrt[3]{2x-1}$$$$x=?$$

Answered by MJS last updated on 01/May/20

$$\mathrm{obviously}x=1\mathrm{is}\mathrm{a}\mathrm{solution}$$$${(x^3+1)}^3=8(2x-1)$$$$x^9+3x^6+3x^3-16x+9=0$$$$(x-1)(x^8+x^7+x^6+4x^5+4x^4+4x^3+7x^2+7x-9)=0$$$$\mathrm{approximately}\mathrm{solving}\mathrm{leads}\mathrm{to}$$$$x\approx -1.61803\vee x\approx .618034$$$$\mathrm{these}\mathrm{seem}\mathrm{to}\mathrm{be}$$$$x=-\frac{1}{2}-\frac{\sqrt{5}}{2}\vee x=-\frac{1}{2}+\frac{\sqrt{5}}{2}$$$$\mathrm{yes}!$$$$(x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$$$\mathrm{now}\mathrm{we}\mathrm{can}\mathrm{only}\mathrm{use}\mathrm{approximation}\mathrm{by}\mathrm{an}$$$$\mathrm{electronic}\mathrm{calculator}$$$$x\approx -1.08891\pm .824866\mathrm{i}$$$$x\approx .123133\pm 1.43020\mathrm{i}$$$$x\approx .965776\pm 1.18647\mathrm{i}$$$$\mathrm{only}\mathrm{these}\mathrm{solve}\mathrm{the}\mathrm{given}\mathrm{equation}:$$$$x=1$$$$x=-\frac{1}{2}\pm \frac{\sqrt{5}}{2}$$$$x\approx -1.08891\pm .824866\mathrm{i}$$

Commented by jagoll last updated on 01/May/20

$$isthereanelectroniccalculator$$$$intheplaystore?$$

Commented by MJS last updated on 01/May/20

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know},{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{got}\mathrm{an}\mathrm{old}\mathrm{Texas}\mathrm{Instruments}$$$$\mathrm{TI}-89\mathrm{device}$$

Commented by jagoll last updated on 01/May/20

$$thankalotsir$$

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