If frac{sqrt{3+x}+sqrt{3-x}}{sqrt{3+x}-sqrt{3 | vedclass

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(B) Given equation: $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2$Rationalizing the denominator:$\frac{(\sqrt{3+x}+\sqrt{3-x})^2}{(\sqrt{3+x}

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$\frac{12}{5}$

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(B) Given equation: $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2$Rationalizing the denominator:$\frac{(\sqrt{3+x}+\sqrt{3-x})^2}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})} = 2$$\frac{(3+x) + (3-x) + 2\sqrt{(3+x)(3-x)}}{(3+x) - (3-x)} = 2$$\frac{6 + 2\sqrt{9-x^2}}{2x} = 2$$\frac{2(3 + \sqrt{9-x^2})}{2x} = 2$$3 + \sqrt{9-x^2} = 2x$$\sqrt{9-x^2} = 2x - 3$Squaring both sides:$9 - x^2 = (2x - 3)^2$$9 - x^2 = 4x^2 - 12x + 9$$5x^2 - 12x = 0$$x(5x - 12) = 0$Since $x$ must satisfy the original equation,$x = \frac{12}{5}$ (as $x=0$ leads to $1=2$ which is false).

If $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2,$ then $x$ is equal to

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