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Question Number 186241 by a.lgnaoui last updated on 02/Feb/23

$$x+\frac{1}{x}=\frac{-1+\sqrt{5}}{2}$$$$x?$$

Commented by MJS_new last updated on 02/Feb/23

$$\mathrm{are}\mathrm{you}\mathrm{unable}\mathrm{to}\mathrm{solve}\mathrm{a}\mathrm{simple}\mathrm{quadratic}?$$

Commented by a.lgnaoui last updated on 02/Feb/23

$$Thevalue{\left(5x\right)}^5=?,withoutcalculofx.$$

Commented by a.lgnaoui last updated on 02/Feb/23

$$Iwantbythissimpleequationto$$$$evaluate{\left(5x\right)}^{5}withoutcalculofx?$$$$$$

Answered by Frix last updated on 02/Feb/23

$$x=\frac{-1+\sqrt{5}}{4}\pm \frac{\sqrt{10+2\sqrt{5}}}{4}\mathrm{i}$$

Answered by MJS_new last updated on 02/Feb/23

$$x+\frac{1}{x}=\frac{-1+\sqrt{5}}{2}$$$$\iff$$$$x^2+\left(\frac{1-\sqrt{5}}{2}\right)x+1=0$$$$\mathrm{we}\mathrm{know}$$$$(x^2+\left(\frac{1-\sqrt{5}}{2}\right)x+1)(x^2+\left(\frac{1+\sqrt{5}}{2}\right)x+1)(x-1)=x^5-1=0$$$$\Rightarrow x^5=1\Rightarrow {\left(5x\right)}^5=3125$$$$$$

Commented by a.lgnaoui last updated on 02/Feb/23

$$good$$

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