Find the value of $\sqrt{5}$ correct to two decimal places, then find the value of $\frac{3-\sqrt{5}}{3+\sqrt{5}}$ correct to two decimal places.

Given :

The given value is $\sqrt{5}$

To find :

We have to find the value of $\frac{3-\sqrt{5}}{3+\sqrt{5}}$

Solution :

The value of √5 can be found by long division method.

	2.23
2 2	5.0000 4
42    2	100    84
443       3	1600     1329
	271

The value of √5 correct to two decimal places is 2.23.

Therefore,

$\frac{3-\sqrt{5}}{3+\sqrt{5}} = \frac{3-2.23}{3+2.23}$

                                                       $=\frac{0.77}{5.23}$

                                                        $=\frac{77}{523} = 0.14$

The value of $\frac{3-\sqrt{5}}{3+\sqrt{5}}$ is 0.14

 

$\begin{array}{l}\frac{3-\sqrt{5}}{3+\sqrt{5}}=\frac{3-2.23}{3+2.23}=\frac{0.77}{5.23}=\frac{77}{523}=0.14\end{array}$.