Question

Simplify by rationalising the denominator $\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}$.

Answer 1

Hint: In this question, we need to simplify $\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}$ by rationalising the denominator. Hence we need to make the denominator a rational number. Since the denominator is of the form (a-b), so we will multiply and divide the given expression by (a+b). Then in the denominator, we will use $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to remove square roots and hence form rational number. For numerator we will simplify using the property of arithmetic given by ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

Complete step by step answer:
Here, we need to simplify $\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}$ by rationalising the denominator. As we know that, for the denominator term of the form (a-b), rationalising factor becomes (a+b). So here the rationalising factor becomes $\sqrt{6}+\sqrt{3}$. Now we need to multiply and divide the rationalising factor. Hence we get: $\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}\times \dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}$.
Now we get expression as $\Rightarrow \dfrac{{{\left( \sqrt{6}+\sqrt{3} \right)}^{2}}}{\left( \sqrt{6}-\sqrt{3} \right)\left( \sqrt{6}+\sqrt{3} \right)}$.
As we can see denominator is of the form (a-b)(a+b), so let us apply the property of arithmetic given by (a-b)(a+b) we get: $\Rightarrow \dfrac{{{\left( \sqrt{6}+\sqrt{3} \right)}^{2}}}{{{\left( \sqrt{6} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}$
Since square root and square cancels out, so we are left with $\dfrac{{{\left( \sqrt{6}+\sqrt{3} \right)}^{2}}}{6-3}\Rightarrow \dfrac{{{\left( \sqrt{6}+\sqrt{3} \right)}^{2}}}{3}$.
Now, let us simplify the numerator. As we can see, the numerator is of the form ${{\left( a+b \right)}^{2}}$. So let us apply the property of arithmetic given by ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ we get:
$\Rightarrow \dfrac{{{\left( \sqrt{6} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+2\left( \sqrt{6} \right)\left( \sqrt{3} \right)}{3}$.
As we know $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ and ${{\left( \sqrt{a} \right)}^{2}}=a$, so we get: $\Rightarrow \dfrac{6+3+2\sqrt{18}}{3}=\dfrac{9+2\sqrt{18}}{3}$
Let us simplify $\sqrt{18}$ since 18 can be written as $2\times 3\times 3$. So, $\sqrt{18}=\sqrt{2\times 3\times 3}=3\sqrt{2}$. Hence we get:
$\Rightarrow \dfrac{9+\left( 2 \right)\left( 3 \right)\sqrt{2}}{3}=\dfrac{9+6\sqrt{2}}{3}$.
Taking 3 common from numerator and cancelling it by denominator we get:
$\Rightarrow \dfrac{3\left( 3+2\sqrt{2} \right)}{3}=3+2\sqrt{2}$.

Hence, $3+2\sqrt{2}$ is our simplified expression.

Note: Students should carefully decide the rationalising factor as they can make mistakes in signs. If denominator is of the form (a+b) we take (a-b) as rationalising factor and if denominator is of the form (a-b) we take (a+b) as rationalising factor. While using the property $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ students should always take care of the signs. At the end, the denominator is 1 which is a rational number and hence our answer is correct.