Question

Rationalise the denominator:
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Answer 1

Step-by-step explanation:

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}}}Ifba=dc⟹a−ba+b=c−dc+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}⟹a+x+a−x−a+x+a−xa+x+a−x+a+x−a−x=b−1b+1

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

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

/* On squaring both sides, we get */

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

\implies \frac{2a}{2x} = \frac{2(b^{2}+1^{2})}{4b}⟹2x2a=4b2(b2+12)

\implies \frac{a}{x} = \frac{(b^{2}+1)}{2b}⟹xa=2b(b2+1)

\implies \frac{x}{a} = \frac{2b}{b^{2}+1}⟹ax=b2+12b

\implies x= \frac{2ab}{b^{2}+1}⟹x=b2+12ab

Therefore.,

\red{Value \:of \:x} \green {= \frac{2ab}{b^{2}+1}}Valueofx=b2+12ab