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Question Number 192958 by Mingma last updated on 31/May/23

Answered by Frix last updated on 01/Jun/23

$$\mathrm{Question}191675$$$$x=\frac{1+\sqrt{5}}{2}$$$${(2x-1)}^4={\left(\sqrt{5}\right)}^4=25$$

Answered by BaliramKumar last updated on 01/Jun/23

$$\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}=x$$$$\sqrt{\frac{x^2-1}{x}}+\sqrt{\frac{x-1}{x}}=x$$$${(\sqrt{x^2-1}+\sqrt{x-1})}^2={\left(x\sqrt{x}\right)}^2$$$$x^2-1+x-1+2\sqrt{(x^2-1)(x-1)}=x^3$$$$2\sqrt{x^3-x^2-x+1}=x^3-x^2-x+1+1$$$$put{x}^3-x^2-x+1=t^2$$$$2\sqrt{t^2}=t^2+1$$$$t^2-2t+1=0\Rightarrow (\mathrm{t}-1)(\mathrm{t}-1)=0$$$$t=1\Rightarrow {\mathrm{t}}^2=1^2$$$$\Rightarrow {\mathrm{t}}^2=1$$$$x^3-x^2-x+\cancel{1}=\cancel{1}$$$$x^3-x^2-x=0$$$$x^2-x-1=0x\ne 0[\because x\ge 1]$$$$x=\frac{1\pm \sqrt{5}}{2}x\ne \frac{1-\sqrt{5}}{2}[\because \sqrt{x-1}\ge 0]$$$$x-1\ge 0\mathrm{or}x\ge 1$$$$\therefore x=\frac{1+\sqrt{5}}{2}\Rightarrow 2x-1=\sqrt{5}$$$${(2x-1)}^4={\left(\sqrt{5}\right)}^4=25$$$$$$

Commented by Mingma last updated on 01/Jun/23

Perfect ��

Answered by ajfour last updated on 01/Jun/23

$$p+q=x$$$$p-q=t$$$$p^2-q^2=x-1$$$$p^2+q^2=x+1-\frac{2}{x}$$$$tx=x-1$$$$x^2+t^2=2x+2-\frac{4}{x}$$$$x^2+{(1-\frac{1}{x})}^2=2x+2-\frac{4}{x}$$$$x^2+\frac{1}{x^2}=1+2(x-\frac{1}{x})$$$${(x-\frac{1}{x})}^2-2(x-\frac{1}{x})+1=0$$$$x-\frac{1}{x}-1=0$$$$x^2-x-1=0$$$$x=\frac{1+\sqrt{5}}{2}$$$$2x-1=\sqrt{5}$$$${(2x-1)}^4=25$$

Commented by York12 last updated on 01/Jun/23

$$canyoumakeitmoreclear$$

Commented by Mingma last updated on 01/Jun/23

Perfect ��

Answered by York12 last updated on 01/Jun/23

$$\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}=x$$$$\sqrt{x^3-x}+\sqrt{x^2-x}=x^2.......\left(i\right)$$$$\Rightarrow \sqrt{x^3-x}-\sqrt{x^2-x}=(x-1)......\left(ii\right)$$$$\left(i\right)-\left(ii\right)$$$$\Rightarrow 2\sqrt{\underset{\lambda }{\underbrace{x^2-x}}}=\underset{\lambda }{\underbrace{x^2-x}}+1$$$$\Rightarrow 2\sqrt{\lambda }=\lambda +1\Rightarrow 4\lambda ={\lambda }^2+2\lambda +1$$$${\lambda }^2-2\lambda +1=0\Rightarrow {(\lambda -1)}^2=0,\lambda =1$$$$x^2-x-1=0\Rightarrow x=\frac{1\mp \sqrt{5}}{2}$$$$but\frac{1-\sqrt{5}}{2}{doesn}^{\prime }tsatisfythegiven$$$$equation\Rightarrow x=\frac{1+\sqrt{5}}{2}★$$$$(2x-1)=\sqrt{5}\Rightarrow {(2x-1)}^4=25★$$$$$$

Commented by Mingma last updated on 01/Jun/23

Perfect ��

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