Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 170781 by solomonwells last updated on 30/May/22

$$\sqrt{(2-\mathrm{x})}={(2-\mathrm{x})}^2\mathbf{solve}\mathbf{for}\mathbf{x}$$

Answered by Rasheed.Sindhi last updated on 30/May/22

$$\frac{{(2-x)}^2}{{(2-x)}^{1/2}}=1$$$${(2-x)}^{3/2}=1^{3/2}$$$$2-x=1$$$$x=1$$

Commented by mr W last updated on 30/May/22

$$x=2isalsogood$$

Commented by Rasheed.Sindhi last updated on 30/May/22

$$\mathrm{Of}\mathrm{course}\mathrm{sir}:$$$$\sqrt{2-x}={(2-x)}^2$$$$\sqrt{2-x}-{(2-x)}^2=0$$$$\sqrt{2-x}(1-{(2-x)}^{3/2})=0$$$$\sqrt{2-x}=0\mid {(2-x)}^{3/2}=1$$$$x=2\mid {(2-x)}^{3/2}=1^{3/2}$$$$2-x=1$$$$x=1$$

Answered by MJS_new last updated on 31/May/22

$$t=\sqrt{2-x}⩾0\iff x=2-t^2$$$$t=t^4$$$$t(t^3-1)=0$$$$t=0\vee t=1\Rightarrow x=1\vee x=2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com