Question

If x=4/8+√60 then 1/2[√x+2/√x]=

Answer 1

$\mathbf{If \ x=\frac{4}{8+\sqrt{60}}, then \ \frac{1}{2}\left ( \sqrt{x}+\frac{2}{\sqrt{x}} \right )=\sqrt{5}}$

Step-by-step explanation:

•        Given that

       $\mathbf{x=\frac{4}{8+\sqrt{60}}}$

•      Rationalise the denominator

       $\mathbf{x=\frac{4}{(8+\sqrt{60})}\times \frac{(8-\sqrt{60})}{(8-\sqrt{60})}}$

     

      So

      $\mathbf{x=4\times \frac{(8-\sqrt{60})}{4}=8-\sqrt{60}}$

•      Which can be written as

     $\mathbf{x=8-\sqrt{60}=5+3-2\sqrt{15}}$

     $\mathbf{x=8-\sqrt{60}=(\sqrt{5})^{2}+(\sqrt{3})^{2}-2\sqrt{5}\times\sqrt{3}}$

    therefore

     $\mathbf{x=(\sqrt{5}-\sqrt{3})^{2}}$

     So

     $\mathbf{\sqrt{x}=\sqrt{5}-\sqrt{3}}$                         ...1)

     Then

     $\mathbf{\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{5}-\sqrt{3}}}$

•      Again rationalise the denominator

     $\mathbf{\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}}$

     $\mathbf{\frac{1}{\sqrt{x}}=\frac{\sqrt{5}+\sqrt{3}}{5-3}=\frac{\sqrt{5}+\sqrt{3}}{2}}$             ...2)

    Now from

    $\mathbf{\frac{1}{2}\left ( \sqrt{x}+2\times \frac{1}{\sqrt{x}} \right )}$

    $\mathbf{\frac{1}{2}\left ( \sqrt{5}-\sqrt{3}+2\times \frac{\sqrt{5}+\sqrt{3}}{2} \right )}$

    $\mathbf{\frac{1}{2}\left ( \sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3} \right )}$

   

    $\mathbf{\frac{1}{2}\left ( 2\sqrt{5} \right )}$

    =$\mathbf{\sqrt{5}}$