Question

√(a+x) + √(a-x)/ √(a+x) -√(a-x)=b solve for x with the properties of proportion

Answer 1

$\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}= b$

Here, we are using Componendo and Dividendo Rule:

$\boxed{\pink { If \: \frac{a}{b} = \frac{c}{d}\implies \frac{a+b}{a-b} = \frac{c+d}{c-d}}}$

$\implies \frac{\sqrt{a+x}+\sqrt{a-x} + \sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x} -\sqrt{a+x}+\sqrt{a-x}}= \frac{b+1}{b-1}$

$\implies \frac{2(\sqrt{a+x})}{2(\sqrt{a-x})} = \frac{b+1}{b-1}$

$\implies \frac{(\sqrt{a+x})}{(\sqrt{a-x})} = \frac{b+1}{b-1}$

/* On squaring both sides, we get */

$\implies \frac{a+x}{a-x} = \frac{(b+1)^{2}}{(b-1)^{2}}$

/* Applying above rule again */

$\implies \frac{a+x+a-x}{a+x-a+x} = \frac{(b+1)^{2}+(b-1)^{2}}{(b+1)^{2}-(b-1)^{2}}$

$\implies \frac{2a}{2x} = \frac{2(b^{2}+1^{2})}{4b}$

$\implies \frac{a}{x} = \frac{(b^{2}+1)}{2b}$

$\implies \frac{x}{a} = \frac{2b}{b^{2}+1}$

$\implies x= \frac{2ab}{b^{2}+1}$

Therefore.,

$\red{Value \:of \:x} \green {= \frac{2ab}{b^{2}+1}}$

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Answer 2

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