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Question Number 66890 by hmamarques1994@gmail.com last updated on 20/Aug/19

$$$$$${\mathbf{x}}^{\mathbf{x}}+{\mathbf{x}}^{2\mathbf{x}}=20$$$$$$$${\mathbf{x}}^{\mathbf{x}}=?$$$$$$

Answered by Kunal12588 last updated on 21/Aug/19

$$let{x}^x=t$$$$t+t^2=20$$$$\Rightarrow t^2+t-20=0$$$$\Rightarrow t=\frac{-1\pm \sqrt{1+80}}{2}=\frac{-1\pm 9}{2}$$$$t_1=4,t_2=-5\left(invalid\right)$$$$\Rightarrow x^x=4$$

Answered by John Kaloki Musau last updated on 20/Aug/19

$$x^x+x^x\times x^x=20$$$$let{x}^{x}bey$$$$y+y^2=20$$$$y^2+y-20=0$$$$y^2+5y-4y-20=0$$$$y(y+5)-4(y+5)=0$$$$(y-4)(y+5)=0$$$$y=4ory=-5$$$$x^x=4or-5$$

Answered by mr W last updated on 20/Aug/19

$$lett=x^x>0$$$$t+t^2=20$$$$(t-4)(t+5)=0$$$$\Rightarrow t=4,-5$$$$sincet>0,onlyt=4isok.$$$$\Rightarrow x^x=4$$$$\Rightarrow x=2$$

Commented by John Kaloki Musau last updated on 21/Aug/19

$$tisanumberandcantakebothpositive$$$$andnegativeforms.I^{\prime }mIcorrect?$$

Commented by mr W last updated on 21/Aug/19

$$buttstandsfor{x}^{x}whichisalways$$$$positive,thereforetispositive.$$

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