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Question Number 193346 by pascal889 last updated on 11/Jun/23

$$\sqrt{4\mathrm{x}-3}-\sqrt{2\mathrm{x}-5}=2$$$$\mathrm{solve}\mathrm{for}\mathrm{x}$$

Answered by deleteduser1 last updated on 11/Jun/23

$$u-v=2;u^2-2v^2=7$$$$\Rightarrow {(2+v)}^2-2v^2=7\Rightarrow v^2-4v+3=0\Rightarrow v=3or1$$$$v=3\Rightarrow \sqrt{2x-5}=3\Rightarrow x=7;v=1\Rightarrow \sqrt{2x-5}=1\Rightarrow x=3$$$$\Rightarrow x=3or7$$

Answered by cortano12 last updated on 11/Jun/23

$$\mathrm{let}\sqrt{2\mathrm{x}-5}=\mathrm{u}\Rightarrow 2\mathrm{x}={\mathrm{u}}^2+5$$$$4\mathrm{x}-3={2\mathrm{u}}^2+7$$$$\iff \sqrt{{2\mathrm{u}}^2+7}=2+\mathrm{u}$$$$\iff {2\mathrm{u}}^2+7={\mathrm{u}}^2+4\mathrm{u}+4$$$$\iff {\mathrm{u}}^2-4\mathrm{u}+3=0$$$$\iff (\mathrm{u}-1)(\mathrm{u}-3)=0$$$$\iff \{\begin{array}{l}\mathrm{x}=\frac{6}{2}=3\\ \mathrm{x}=\frac{14}{2}=7\end{array}$$$$$$

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