Question

If it is given that the following equation is true: $\dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}}=a+b\sqrt{10}$ , then find the value of $a$ and $b$ .

Answer 1

Hint: First, we will write down the left hand side of the given equation and rationalise the denominator, we will think of a number such that we get a real number in the denominator so we will be multiplying both the parts of the fraction that is the numerator and denominator with $\left( 3\sqrt{2}-\sqrt{5} \right)$, then we will compare the left hand side and the right hand side and we will get same expression in the left hand side with one real number and another one multiplied by $\sqrt{10}$ , thus we will get the value of a and b.

Complete step by step answer:
Now, we are given that: $\dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}}=a+b\sqrt{10}$
So first we will take the left hand side: $\dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}}$ , now we will rationalise the denominator by multiplying the numerator and the denominator by $\left( 3\sqrt{2}-\sqrt{5} \right)$ , therefore we will get:
$\dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}}\times \dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}-\sqrt{5}}\Rightarrow \dfrac{{{\left( 3\sqrt{2}-\sqrt{5} \right)}^{2}}}{\left( 3\sqrt{2}+\sqrt{5} \right)\left( 3\sqrt{2}-\sqrt{5} \right)}$
Now, we will apply the identity: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to the denominator, therefore:
$\Rightarrow \dfrac{{{\left( 3\sqrt{2}-\sqrt{5} \right)}^{2}}}{\left( 3\sqrt{2}+\sqrt{5} \right)\left( 3\sqrt{2}-\sqrt{5} \right)}\Rightarrow \dfrac{{{\left( 3\sqrt{2}-\sqrt{5} \right)}^{2}}}{{{\left( 3\sqrt{2} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}}\Rightarrow \dfrac{{{\left( 3\sqrt{2}-\sqrt{5} \right)}^{2}}}{\left( 18-5 \right)}\Rightarrow \dfrac{{{\left( 3\sqrt{2}-\sqrt{5} \right)}^{2}}}{13}$
Now, we will apply the identity: ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to the numerator, therefore:
$\Rightarrow \dfrac{{{\left( 3\sqrt{2}-\sqrt{5} \right)}^{2}}}{13}\Rightarrow \dfrac{{{\left( 3\sqrt{2} \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}-2\left( 3\sqrt{2} \right)\left( \sqrt{5} \right)}{13}\Rightarrow \dfrac{18+5-6\sqrt{10}}{13}\Rightarrow \dfrac{23-6\sqrt{10}}{13}$
Now, we will rewrite the given fraction: $\Rightarrow \dfrac{23-6\sqrt{10}}{13}=\dfrac{23}{13}-\dfrac{6\sqrt{10}}{13}=\dfrac{23}{13}-\dfrac{6}{13}\sqrt{10}$
Now, we will put this value of left hand side in the original question:
$\dfrac{23}{13}-\dfrac{6}{13}\sqrt{10}=a+b\sqrt{10}$

Now, we will compare the left hand side and the right hand side and then we will get the value of a and b as follows:
$a=\dfrac{23}{13}$ and $b=-\dfrac{6}{13}$

Note: Student must take a number carefully to rationalise the denominator, for example in: $\dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}}$ , if we multiply the numerator and denominator by $3\sqrt{2}+\sqrt{5}$ , then we will get:
$\dfrac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}}\times \dfrac{3\sqrt{2}+\sqrt{5}}{3\sqrt{2}+\sqrt{5}}\Rightarrow \dfrac{{{\left( 3\sqrt{2} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}}{{{\left( 3\sqrt{2}+\sqrt{5} \right)}^{2}}}\Rightarrow \dfrac{18-5}{18+5+6\sqrt{10}}\Rightarrow \dfrac{13}{23+6\sqrt{10}}$
 So, we will see that we have a rationalised numerator not the denominator and it would be complicated to get the value of a and b, therefore we should always choose a correct number to rationalise the denominator.