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Question Number 62489 by Tawa1 last updated on 21/Jun/19

$$\mathrm{solve}\mathrm{for}\mathrm{x}:\frac{\sqrt{2-\mathrm{x}}+\sqrt{2+\mathrm{x}}}{\sqrt{2-\mathrm{x}}-\sqrt{2+\mathrm{x}}}=3$$

Answered by som(math1967) last updated on 22/Jun/19

$$\frac{\sqrt{2-x}+\sqrt{2+x}+\sqrt{2-x}-\sqrt{2+x}}{\sqrt{2-x}+\sqrt{2+x}-\sqrt{2-x}+\sqrt{2+x}}=\frac{3+1}{3-1}★$$$$\frac{2\sqrt{2-x}}{2\sqrt{2+x}}=\frac{4}{2}$$$$\frac{2-x}{2+x}=4★★$$$$8+4x=2-x\Rightarrow x=-\frac{6}{5}ans$$$$★usingcomponendoanddividendo$$$$★★squaringbothside$$

Commented by MJS last updated on 22/Jun/19

$$\mathrm{you}\mathrm{should}\mathrm{make}\mathrm{your}\mathrm{first}\mathrm{step}\mathrm{clearer}.$$$$\mathrm{in}\mathrm{fact}\mathrm{there}\mathrm{are}\mathrm{a}\mathrm{few}\mathrm{steps}:$$$$\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$$$$\sqrt{2-x}+\sqrt{2+x}=3\sqrt{2-x}-3\sqrt{2+x}$$$$4\sqrt{2+x}=2\sqrt{2-x}$$$$2\sqrt{2+x}=\sqrt{2-x}$$$$4(2+x)=2-x$$$$6+5x=0$$$$x=-\frac{6}{5}$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}$$

Commented by peter frank last updated on 22/Jun/19

$$nicetoo$$

Answered by MJS last updated on 21/Jun/19

$$\mathrm{defined}\mathrm{for}$$$$2-x⩾0\wedge 2+x⩾0\wedge \sqrt{2-x}\ne \sqrt{2+x}$$$$\Rightarrow -2⩽x⩽2\wedge x\ne 0$$$$\frac{{(\sqrt{2-x}+\sqrt{2+x})}^2}{(\sqrt{2-x}-\sqrt{2+x})(\sqrt{2-x}+\sqrt{2+x})}=3$$$$-\frac{2+\sqrt{2-x}\sqrt{2+x}}{x}=3$$$$\sqrt{2-x}\sqrt{2+x}=-3x-2$$$$4-x^2=9x^2+12x+4$$$$10x^2+12x=0$$$$x(5x+6)=0$$$$x_1=0\left[\mathrm{not}\mathrm{valid}\right]$$$$x_2=-\frac{6}{5}$$

Commented by Tawa1 last updated on 21/Jun/19

$$\mathrm{Good}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by MJS last updated on 22/Jun/19

$$\mathrm{the}\mathrm{path}\mathrm{of}som\left(math67\right)\mathrm{is}\mathrm{better},\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}$$$$\mathrm{realize}\mathrm{this}\mathrm{possibility}\mathrm{at}\mathrm{first}(\mathrm{always}\mathrm{looking}$$$$\mathrm{for}\mathrm{trouble},\mathrm{my}\mathrm{thinking}\mathrm{is}\mathrm{too}\mathrm{complicated}$$$$\mathrm{sometimes})$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{Hahahaha}$$

Commented by peter frank last updated on 22/Jun/19

$$thankyou$$

Answered by Rasheed.Sindhi last updated on 22/Jun/19

$$\mathbb{A}\mathrm{n}\mathbb{O}\mathrm{ther}\mathbb{W}\mathrm{ay}$$$$\frac{\sqrt{2-\mathrm{x}}+\sqrt{2+\mathrm{x}}}{\sqrt{2-\mathrm{x}}-\sqrt{2+\mathrm{x}}}=3$$$$\Rightarrow \frac{\frac{\sqrt{2-x}}{\sqrt{2+x}}+1}{\frac{\sqrt{2-x}}{\sqrt{2+x}}-1}=3$$$$\Rightarrow \frac{\frac{\sqrt{2-x}}{\sqrt{2+x}}+1}{\frac{\sqrt{2-x}}{\sqrt{2+x}}-1}=\frac{3}{1}=\frac{2+1}{2-1}$$$$\frac{\sqrt{2-x}}{\sqrt{2+x}}=2$$$$\frac{2-x}{2+x}=4$$$$8+4x=2-x$$$$5x=-6$$$$x=-6/5$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by peter frank last updated on 22/Jun/19

$$verynice$$$$$$

Commented by Rasheed.Sindhi last updated on 22/Jun/19

$$\mathrm{\top }\nparallel \mathrm{\forall }\cap \mathbb{k}\lessgtr !$$

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