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Question Number 194756 by horsebrand11 last updated on 15/Jul/23

$$\underbrace{\mathit{b}}$$

Answered by cortano12 last updated on 15/Jul/23

$$\frac{\mathrm{x}-\mathrm{a}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}}=\frac{\mathrm{x}-\mathrm{a}}{3(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}})}+2\sqrt{\mathrm{a}}$$$$\frac{3(\mathrm{x}-\mathrm{a})}{3(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}})}-\frac{(\mathrm{x}-\mathrm{a})}{3(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}})}=2\sqrt{\mathrm{a}}$$$$\frac{2(\mathrm{x}-\mathrm{a})}{3(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}})}=2\sqrt{\mathrm{a}}$$$$\Rightarrow 2\mathrm{x}-2\mathrm{a}=6\sqrt{\mathrm{ax}}+6\mathrm{a}$$$$\Rightarrow 2\mathrm{x}-8\mathrm{a}=6\sqrt{\mathrm{ax}}$$$$\Rightarrow \mathrm{x}-4\mathrm{a}=3\sqrt{\mathrm{ax}}$$$$\Rightarrow {\mathrm{x}}^2-8\mathrm{ax}+{16\mathrm{a}}^2=9\mathrm{ax}$$$$\Rightarrow {\mathrm{x}}^2-17\mathrm{ax}+{16\mathrm{a}}^2=0$$$$\Rightarrow (\mathrm{x}-16\mathrm{a})(\mathrm{x}-\mathrm{a})=0,\mathrm{x}=\mathrm{a}\left(\mathrm{reject}\right)$$$$\Rightarrow \overline{)\begin{array}{c}\mathrm{x}=16\mathrm{a}\end{array}}$$$$$$

Commented by Rasheed.Sindhi last updated on 15/Jul/23

$$\mathrm{x}=\mathrm{a}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{satisfy}\mathrm{in}\mathrm{general}.\mathrm{It}$$$$\mathrm{satisfies}\mathrm{only}\mathrm{when}\mathrm{x}=\mathrm{a}=0.$$

Commented by cortano12 last updated on 15/Jul/23

$$\mathrm{yes}\mathrm{sir}.\mathrm{right}$$

Commented by Frix last updated on 15/Jul/23

$$x=a=0\mathrm{is}\mathrm{not}\mathrm{allowed}\mathrm{because}\frac{...}{\sqrt{x}+\sqrt{a}}\Rightarrow$$$$\sqrt{x}+\sqrt{a}\ne 0$$

Commented by Rasheed.Sindhi last updated on 15/Jul/23

$$SorryformistakeSir!$$

Answered by Rasheed.Sindhi last updated on 15/Jul/23

$$\underset{―}{⋮\cancel{\underbrace{}}}\mathrm{x}-\mathrm{a}$$

Answered by Frix last updated on 15/Jul/23

$$x>0\wedge a>0$$$$\sqrt{x}+\sqrt{a}=t>0\iff t-\sqrt{a}>0\iff x={(t-\sqrt{a})}^2$$$$\mathrm{Inserting}:$$$$t-2\sqrt{a}=\frac{t+4\sqrt{a}}{3}$$$$t=5\sqrt{a}$$$$x={(5\sqrt{a}-\sqrt{a})}^2=16a$$

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