Find the values of each of the following correct to three places of decimals, it being given that $\sqrt2= 1.4142, \sqrt3= 1.732, \sqrt5 = 2.2360, \sqrt6= 2.4495$ and $\sqrt{10}= 3.162$.$ \frac{3-\sqrt{5}}{3+2 \sqrt{5}} $

Given:

$\sqrt2= 1.4142, \sqrt3= 1.732, \sqrt5 = 2.2360, \sqrt6=  2.4495$ and $\sqrt{10}= 3.162$.

To do: 

We have to find the value of \( \frac{3-\sqrt{5}}{3+2 \sqrt{5}} \) correct to three places of decimals.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

$\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=\frac{(3-\sqrt{5})(3-2 \sqrt{5})}{(3+2 \sqrt{5})(3-2 \sqrt{5})}$

$=\frac{9-6 \sqrt{5}-3 \sqrt{5}+2 \times 5}{(3)^{2}-(2 \sqrt{5})^{2}}$

$=\frac{9+10-9 \sqrt{5}}{9-20}$

$=\frac{19-9 \sqrt{5}}{-11}$

$=\frac{19-9 \times 2.2360}{-11}$

$=\frac{19-20.124}{-11}$

$=\frac{-1.124}{-11}$

$=0.102$

The value of \( \frac{3-\sqrt{5}}{3+2 \sqrt{5}} \) is $0.102$. 

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Updated on: 10-Oct-2022

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