Question

Find the square root of $73-12\sqrt{35}$ .
(a) $\sqrt{25}-\sqrt{18}$
(b) $\sqrt{25}+\sqrt{18}$
(c) $\sqrt{45}-\sqrt{18}$
(d) $\sqrt{45}-\sqrt{28}$

Answer 1

Try to convert $73-12\sqrt{35}$ to a perfect square to easily remove the square root and simplify it. The simplification requires the knowledge of rational and irrational numbers along with the square root operators. Use the results that $45=5\times 3\times 3={{\left( 3\sqrt{5} \right)}^{2}}$ , $28=2\times 2\times 7={{\left( 2\sqrt{7} \right)}^{2}}$, $12\sqrt{35}=2\times 3\times 2\sqrt{7}\times \sqrt{5}$ .

Complete step-by-step answer :
For our convenience, we let $73-12\sqrt{35}$ to be ${{x}^{2}}$ .
So, our questions becomes:
$\sqrt{73-12\sqrt{35}}=\sqrt{{{x}^{2}}}................(i)$
To proceed in the question, try to simplify ${{x}^{2}}$ ;
${{x}^{2}}=73-12\sqrt{35}$
We can break 73 as the sum of 45 and 28. On doing so, we have
${{x}^{2}}=45+28-12\sqrt{35}$
Here, the terms can be written as:
$45=5\times 3\times 3={{\left( 3\sqrt{5} \right)}^{2}}$
$28=2\times 2\times 7={{\left( 2\sqrt{7} \right)}^{2}}$
$12\sqrt{35}=2\times 3\times 2\sqrt{7}\times \sqrt{5}$
So, our equation becomes:
${{x}^{2}}=45+28-12\sqrt{35}$
$\Rightarrow {{x}^{2}}={{\left( 3\sqrt{5} \right)}^{2}}+{{\left( 2\sqrt{7} \right)}^{2}}-2\times 3\sqrt{5}\times 2\sqrt{7}$
Now, we know:
${{\left( a-b \right)}^{2}}={{b}^{2}}+{{a}^{2}}-2ab$
Using the above formula, we get:
${{x}^{2}}={{\left( 3\sqrt{5} \right)}^{2}}+{{\left( 2\sqrt{7} \right)}^{2}}-2\times 3\sqrt{5}\times 2\sqrt{7}$
$\Rightarrow {{x}^{2}}={{\left( 3\sqrt{5}-2\sqrt{7} \right)}^{2}}$
Now, moving back to equation (i);
$\sqrt{73-12\sqrt{35}}=\sqrt{{{x}^{2}}}$
$\Rightarrow \sqrt{73-12\sqrt{35}}=\sqrt{{{\left( 3\sqrt{5}-2\sqrt{7} \right)}^{2}}}$
We know;
$\sqrt{{{k}^{2}}}=k$
Applying we get;
$\Rightarrow \sqrt{73-12\sqrt{35}}=3\sqrt{5}-2\sqrt{7}=\sqrt{45}-\sqrt{28}$
Hence, the square root of $73-12\sqrt{35}$ is $3\sqrt{5}-2\sqrt{7}=\sqrt{45}-\sqrt{28}$ . Therefore, the answer is option (d).

Note : At first glance of the question you might think that you can directly put in the value of $\sqrt{35}$ and get the answer, but that won’t be simplification, nor you will be able to get an accurate answer, as the answer you would get is approximate of the exact one because you cannot put the exact value of $\sqrt{35}$ as it’s a non-terminating decimal number i.e. have infinite numbers occurring after the decimal points.
In questions including roots you can also give a try to rationalisation to get a simplified result or to initiate the simplification process.