25th 


$25. If x = \frac{\sqrt{a+2b} + \sqrt{a -2b}}{\sqrt{a +2b - \sqrt{a -2b}}}, show that b{x}^2-ax+b = 0.$

Dear Student,

Please find below the solution to the asked query:

Correct query : x  = $\frac{\sqrt{\mathrm{a} + 2 \mathrm{b}} + \sqrt{\mathrm{a} - 2 \mathrm{b}}}{\sqrt{\mathrm{a} + 2 \mathrm{b}} - \sqrt{\mathrm{a} - 2 \mathrm{b}}}$

We know from Componendo and Dividendo :  If we have :  $\frac{\mathrm{a}}{\mathrm{b}} = \frac{\mathrm{c}}{\mathrm{d}} \mathrm{Then} \frac{\mathrm{a} + \mathrm{b}}{\mathrm{a} - \mathrm{b}} = \frac{\mathrm{c} +\hspace{0.17em}\mathrm{d}}{\mathrm{c} - \mathrm{d}}$ , So we apply same rule here and get :

 $$\begin{gathered} \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{\mathrm{a} + 2 \mathrm{b}} + \sqrt{\mathrm{a} - 2 \mathrm{b}} + \left(\sqrt{\mathrm{a} + 2 \mathrm{b}} - \sqrt{\mathrm{a} - 2 \mathrm{b}}\right)}{\sqrt{\mathrm{a} + 2 \mathrm{b}} + \sqrt{\mathrm{a} - 2 \mathrm{b}} - \left(\sqrt{\mathrm{a} + 2 \mathrm{b}} - \sqrt{\mathrm{a} - 2 \mathrm{b}}\right)} \\ \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{\mathrm{a} + 2 \mathrm{b}} + \sqrt{\mathrm{a} - 2 \mathrm{b}} + \sqrt{\mathrm{a} + 2 \mathrm{b}} - \sqrt{\mathrm{a} - 2 \mathrm{b}}}{\sqrt{\mathrm{a} + 2 \mathrm{b}} + \sqrt{\mathrm{a} - 2 \mathrm{b}} - \sqrt{\mathrm{a} + 2 \mathrm{b}} + \sqrt{\mathrm{a} - 2 \mathrm{b}}} \\ \Rightarrow \frac{x + 1}{x - 1} = \frac{2 \sqrt{\mathrm{a} + 2 \mathrm{b}} }{2 \sqrt{\mathrm{a} - 2 \mathrm{b}} } \\ \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{\mathrm{a} + 2 \mathrm{b}} }{\sqrt{\mathrm{a} - 2 \mathrm{b}} } \\ \mathrm{Taking} \mathrm{shole} \mathrm{square} \mathrm{on} \mathrm{both} \mathrm{hand} \mathrm{side} \mathrm{we} \mathrm{get} : \\ \Rightarrow {\left(\frac{x + 1}{x - 1}\right)}^2 = {\left(\frac{\sqrt{\mathrm{a} + 2 \mathrm{b}} }{\sqrt{\mathrm{a} - 2 \mathrm{b}} }\right)}^2 \\ \Rightarrow \frac{{\left(x + 1\right)}^2}{{\left(x - 1\right)}^2} = \frac{{\left(\sqrt{\mathrm{a} + 2 \mathrm{b}}\right)}^2 }{{\left(\sqrt{\mathrm{a} - 2 \mathrm{b}}\right)}^2 } ( \mathrm{We} \mathrm{know} : {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}} = \frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{b}}^{\mathrm{m}}} ) \\ \mathrm{We} \mathrm{know} : ( \mathrm{a} \pm \mathrm{b} {)}^2 = {\mathrm{a}}^2 + {\mathrm{b}}^2 \pm 2 \mathrm{a} \mathrm{b} , \mathrm{So} \\ \Rightarrow \frac{x^2 +\hspace{0.17em}1 + 2 x}{x^2 +\hspace{0.17em}1 - 2 x} = \frac{ \mathrm{a} + 2 \mathrm{b} }{\mathrm{a} - 2 \mathrm{b}} \\ \mathrm{We} \mathrm{apply} \mathrm{Componendo} \mathrm{and} \mathrm{Dividendo} \mathrm{again} \mathrm{and} \mathrm{get} : \\ \Rightarrow \frac{x^2 +\hspace{0.17em}1 + 2 x + \left(x^2 +\hspace{0.17em}1 - 2 x\right)}{x^2 +\hspace{0.17em}1 + 2 x - \left(x^2 +\hspace{0.17em}1 - 2 x\right)} = \frac{ \mathrm{a} + 2 \mathrm{b} +\left(\mathrm{a} - 2 \mathrm{b}\right)}{ \mathrm{a} + 2 \mathrm{b} -\left(\mathrm{a} - 2 \mathrm{b}\right)} \\ \Rightarrow \frac{x^2 +\hspace{0.17em}1 + 2 x + x^2 +\hspace{0.17em}1 - 2 x}{x^2 +\hspace{0.17em}1 + 2 x - x^2 -1 + 2 x} = \frac{ \mathrm{a} + 2 \mathrm{b} + \mathrm{a} - 2 \mathrm{b}}{ \mathrm{a} + 2 \mathrm{b} - \mathrm{a} + 2 \mathrm{b}} \\ \Rightarrow \frac{2 x^2 +\hspace{0.17em}2}{4 x} = \frac{ 2 \mathrm{a}}{ 4 \mathrm{b}} \\ \Rightarrow \frac{2\left( x^2 +1\right)\hspace{0.17em}}{4 x} = \frac{ 2 \mathrm{a}}{ 4 \mathrm{b}} \\ \Rightarrow \frac{ x^2 +1}{x} = \frac{ \mathrm{a}}{ \mathrm{b}} \\ \Rightarrow \mathrm{b} x^2 +\mathrm{b} = \mathrm{a} x \\ \mathbf{\Rightarrow }\mathbf{ }\mathbf{b}\mathbf{ }{\mathit{x}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{a}\mathbf{ }\mathit{x}\mathbf{+}\mathbf{b}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Hence}\mathbf{ }\mathbf{proved}\mathbf{ }\mathbf{)} \end{gathered}$$
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