Question

Rationalize the denominator $\dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}}$ and hence find its value correct to 2 decimals. (Given$\sqrt 2 = 1.414,\sqrt 6 = 2.449$).

Answer 1

Hint: We have to rationalize the denominator by multiplying and dividing the given expression with the conjugate of the denominator, and we also make use of the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]for further simplification.

Complete step-by-step answer:
Given expression is,$\dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}}$,
Here given values are$\sqrt 2 = 1.414,\sqrt 6 = 2.449$,
And now we should multiply and divide the expression with the conjugate of the denominator i.e.,
$\Rightarrow$\[\]\[\sqrt 3 + \sqrt 2 + 1\], i.e.,
So, we get,
$\Rightarrow$\[\dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} \times \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{\sqrt 3 + \sqrt 2 + 1}}\],
Now simplifying the denominator we get,
\[ \Rightarrow \dfrac{1}{{\sqrt 3 - \left( {\sqrt 2 + 1} \right)}} \times \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{\sqrt 3 + \left( {\sqrt 2 + 1} \right)}}\],
Now applying \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]in the denominator we gte,
\[ \Rightarrow \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 + 1} \right)}^2}}}\],
Now removing the square root and also using the formula\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we get.
\[ \Rightarrow \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{3 - \left( {{{\left( {\sqrt 2 } \right)}^2} + {1^2} + 2\left( {\sqrt 2 } \right)\left( 1 \right)} \right)}}\],
Now simplifying we get,
\[ \Rightarrow \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{3 - \left( {2 + 1 + 2\sqrt 2 } \right)}}\]
Now releasing the brackets we get,
\[ \Rightarrow \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{3 - 3 - 2\sqrt 2 }}\],
Now simplifying the denominator we get,
\[ \Rightarrow \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{ - 2\sqrt 2 }}\],
Again we have to rationalize the denominator by multiplying and dividing with\[\sqrt 2 \], we get,
\[ \Rightarrow \dfrac{{\sqrt 3 + \sqrt 2 + 1}}{{ - 2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\],
Now multiplying we get,
\[ \Rightarrow \dfrac{{\sqrt 6 + 2 + \sqrt 2 }}{{ - 2{{\left( {\sqrt 2 } \right)}^2}}}\],
Now simplifying the denominator we get,
\[ \Rightarrow \dfrac{{\sqrt 6 + 2 + \sqrt 2 }}{{ - 4}}\],
Now we got the required result i.e.,
\[ \Rightarrow \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = \dfrac{{\sqrt 6 + 2 + \sqrt 2 }}{{ - 4}}\],
Now substituting the given values of the roots in the result obtained we gte,
\[ \Rightarrow \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = \dfrac{{2.449 + 2 + 1.414}}{{ - 4}}\],
Now adding carefully we get,
\[ \Rightarrow \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = \dfrac{{4.449 + 1.414}}{{ - 4}}\],
Now adding the remaining result we get,
\[ \Rightarrow \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = \dfrac{{5.863}}{{ - 4}}\],
Now by dividing we get,
\[ \Rightarrow \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = - 1.46575\],
Now we have to find the value correct to 2 decimal points, first we will know how can we find the value correct to two decimal points.
Rounding off to the nearest hundredths is the same as rounding it to 2 decimal places. To round off a number to 2 decimal places, look at the digit in the thousandths place. If the digit in the thousandths place is greater or equal to 5, the hundredths digit is increased by one unit.
Now we have to round 1.46575 correct to 2 decimals, To round off -1.46575 to the nearest hundredths, the thousandth digit is checked and is equal to 5.Add one unit to the hundredth digit and drop all the digits that come after it.,The answer is therefore, -1.47.

$\therefore \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = - 1.47$ $\therefore \dfrac{1}{{\sqrt 3 - \sqrt 2 - 1}} = - 1.47$.

Note:
When we get an irrational number in the denominator in any question, we can rationalize the irrational number by multiplying and dividing with the conjugate of the denominator will be a rational number.