Question

Simply each of the following by rationalising the denominator.
(i) $ \dfrac{6-4\sqrt{2}}{6+4\sqrt{2}} $
(ii) $ \dfrac{7-\sqrt{5}}{7+\sqrt{5}} $
(ii) $ \dfrac{1}{3\sqrt{2}-2\sqrt{3}} $

Answer 1

Hint: Use the fact that when we multiply an irrational number with its conjugate, it results into a rational number. Hence, to rationalise the denominators, we can multiply and divide the number by its conjugate.

Complete step-by-step answer:
Let us first understand what are rational numbers and what are irrational numbers.
Rational numbers are numbers that can be expressed in the form $ \dfrac{p}{q} $ , where p and q are both integers but $ q\ne 0 $ .
For example, the number 1.5 is a rational number because it can be written as $ \dfrac{3}{2} $ and this satisfies the condition of a rational number.
Irrational numbers are those numbers which cannot be expressed in the form of a rational number.
The number $ \pi $ (pi) and $ \sqrt{2} $ are examples of irrational numbers.
Consider an irrational number $ \sqrt{a}+\sqrt{b} $ , where a and b are integers.
Then the irrational number $ \sqrt{a}-\sqrt{b} $ is considered its conjugate number.
Now, multiply the two numbers.
  $ \Rightarrow \left( \sqrt{a}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)={{\left( \sqrt{a} \right)}^{2}}-\sqrt{a}\sqrt{b}+\sqrt{a}\sqrt{b}-{{\left( \sqrt{b} \right)}^{2}} $
 $ \Rightarrow \left( \sqrt{a}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)=a-b $
Here, $ a-b $ is a rational number because a and b are integers.
This means when we multiply an irrational number with its conjugate, it results into a rational number.
Hence, to rationalise the denominators, we shall multiply and divide the number by its conjugate.

(i) The conjugate of $ 6+4\sqrt{2} $ is $ 6-4\sqrt{2} $ .
Therefore, multiply and divide the number by $ 6-4\sqrt{2} $ .
 $ \Rightarrow \dfrac{6-4\sqrt{2}}{6+4\sqrt{2}}=\dfrac{6-4\sqrt{2}}{6+4\sqrt{2}}\times \dfrac{6-4\sqrt{2}}{6-4\sqrt{2}}=\dfrac{36-24\sqrt{2}-24\sqrt{2}+16(2)}{36-16(2)} $
 $ \Rightarrow \dfrac{6-4\sqrt{2}}{6+4\sqrt{2}}=\dfrac{36+32-48\sqrt{2}}{36-32}=\dfrac{68-48\sqrt{2}}{4}=17-12\sqrt{2} $ .
Therefore, the number $ \dfrac{6-4\sqrt{2}}{6+4\sqrt{2}} $ is simplified to $ 17-12\sqrt{2} $ .
So, the correct answer is “ $ 17-12\sqrt{2} $ ”.

(ii) The conjugate of $ 7+\sqrt{5} $ is $ 7-\sqrt{5} $ .
Therefore, multiply and divide the number by $ 7-\sqrt{5} $ .
 $ \Rightarrow \dfrac{7-\sqrt{5}}{7+\sqrt{5}}=\dfrac{7-\sqrt{5}}{7+\sqrt{5}}\times \dfrac{7-\sqrt{5}}{7-\sqrt{5}}=\dfrac{49-7\sqrt{5}-7\sqrt{5}+5}{49-5} $
 $ \Rightarrow \dfrac{7-\sqrt{5}}{7+\sqrt{5}}=\dfrac{54-14\sqrt{5}}{44}=\dfrac{26-7\sqrt{5}}{22} $
Therefore, the number $ \dfrac{7-\sqrt{5}}{7+\sqrt{5}} $ is simplified to $ \dfrac{26-7\sqrt{5}}{22} $ .
So, the correct answer is “ $ \dfrac{26-7\sqrt{5}}{22} $ ”.

(iii) The conjugate of $ 3\sqrt{2}-2\sqrt{3} $ is $ 3\sqrt{2}+2\sqrt{3} $ .
Therefore, multiply and divide the number by $ 3\sqrt{2}+2\sqrt{3} $ .
 $ \Rightarrow \dfrac{1}{3\sqrt{2}-2\sqrt{3}}=\dfrac{1}{3\sqrt{2}-2\sqrt{3}}\times \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}+2\sqrt{3}}{9(2)-4(3)} $
 $ \Rightarrow \dfrac{1}{3\sqrt{2}-2\sqrt{3}}=\dfrac{3\sqrt{2}+2\sqrt{3}}{18-12}=\dfrac{3\sqrt{2}+2\sqrt{3}}{6} $
Therefore, the number $ \dfrac{1}{3\sqrt{2}-2\sqrt{3}} $ is simplified to $ \dfrac{3\sqrt{2}+2\sqrt{3}}{6} $ .
So, the correct answer is “ $ \dfrac{3\sqrt{2}+2\sqrt{3}}{6} $ ”.

Note: Note that addition and subtraction of two irrational numbers results into an irrational number.
When we add or subtract irrational numbers from a rational number, what we get is also an irrational number.
However, when we multiply or divide two irrational numbers, the result ‘may’ be a rational number.
The multiplication and division of an irrational number by a rational number always gives an irrational number.