# Find the square root of $7-4\sqrt{3}.$$\left( a \right)2+\sqrt{3}$$\left( b \right)5-\sqrt{3}$$\left( c \right)2-\sqrt{3}$$\left( d \right)5+\sqrt{3}$

Verified
206.4k+ views
Hint: We have $7-4\sqrt{3},$ we are asked to find the square root. First, we will rearrange $7-4\sqrt{3}$ in such a way that it is a square of some term. In order to rearrange, first, we will use $2=\sqrt{4}$ and then split 7 = 3 + 4 using these. Then we will get $7-4\sqrt{3}$ as ${{\left( 2-\sqrt{3} \right)}^{2}}.$ So, taking the square root on both sides, we will get our solution.

So, we are asked to find the square root of $7-4\sqrt{3}.$ To do so, we will try to write $7-4\sqrt{3}$ in such a way that it is a square of some numbers by rearranging the terms. Now,
$2=\sqrt{4}$
So,
$7-4\sqrt{3}=7-2\times \left( 2\sqrt{3} \right)$
$\Rightarrow 7-4\sqrt{3}=7-2\sqrt{4}\times \sqrt{3}$
We can write,
$\sqrt{4}\times \sqrt{3}=\sqrt{12}$
So, we get,
$\Rightarrow 7-2\sqrt{12}$
Now, we will split 7 into 2 terms, i.e. 7 = 3 + 4. Now, putting this in the above term, we will get,
$\Rightarrow 3+4-2\sqrt{12}$
We can write 3 as ${{\left( \sqrt{3} \right)}^{2}}$ and 4 as ${{\left( \sqrt{4} \right)}^{2}}.$
And so, we get,
$\Rightarrow {{\left( \sqrt{3} \right)}^{2}}+{{\left( \sqrt{4} \right)}^{2}}-2\sqrt{12}$
We can write, $\sqrt{12}=\sqrt{3}\times \sqrt{4}.$
$\Rightarrow {{\left( \sqrt{3} \right)}^{2}}+{{\left( \sqrt{4} \right)}^{2}}-2\times \sqrt{3}\times \sqrt{4}$
We know that, ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}.$
So, we get,
$\Rightarrow {{\left( \sqrt{4}-\sqrt{3} \right)}^{2}}$
Hence, we have,
$7-2\sqrt{3}={{\left( \sqrt{4}-\sqrt{3} \right)}^{2}}$
Also, we know that $\sqrt{4}=2,$ so we get,
$\left( 7-2\sqrt{3} \right)={{\left( 2-\sqrt{3} \right)}^{2}}$
Now taking the square root on both the sides, we get,
$\Rightarrow \sqrt{7-2\sqrt{3}}=\sqrt{{{\left( 2-\sqrt{3} \right)}^{2}}}$
On further simplification, we get,
$\sqrt{7-2\sqrt{3}}=2-\sqrt{3}$
So, the correct answer is “Option C”.

Note: Another way to look for the square root is, we have 4 options, so we will square each option and see which will end up as $7-2\sqrt{3}$ and that would be the correct option.
$\left( a \right)2+\sqrt{3}$
Squaring $2+\sqrt{3},$ we get,
${{\left( 2+\sqrt{3} \right)}^{2}}={{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+2\times \sqrt{3}\times 2$
$\Rightarrow {{\left( 2+\sqrt{3} \right)}^{2}}=4+3+4\sqrt{3}$
$\Rightarrow {{\left( 2+\sqrt{3} \right)}^{2}}=7+4\sqrt{3}$
Hence, $7+4\sqrt{3}\ne 7-4\sqrt{3}$
Thus, option (a) is not our required answer.
$\left( b \right)\left( 5-\sqrt{3} \right)$
Squaring $\left( 5-\sqrt{3} \right),$ we get,
$\Rightarrow \left( 5-\sqrt{3} \right)={{5}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-5\times 2\times \sqrt{3}$
We know that, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.$
$\Rightarrow \left( 5-\sqrt{3} \right)=25+3-10\sqrt{3}$
$\Rightarrow {{\left( 5-\sqrt{3} \right)}^{2}}=28-10\sqrt{3}$
Hence, $5-\sqrt{3}\ne 7-4\sqrt{3}$
Thus, option (b) is not our required answer.
$\left( c \right)2-\sqrt{3}$
Squaring $\left( 2-\sqrt{3} \right),$ we get,
${{\left( 2-\sqrt{3} \right)}^{2}}={{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2\times 2\times \sqrt{3}$
$\Rightarrow {{\left( 2-\sqrt{3} \right)}^{2}}=4+3-4\sqrt{3}$
$\Rightarrow {{\left( 2-\sqrt{3} \right)}^{2}}=7-4\sqrt{3}$
Hence, $7-4\sqrt{3}=7-4\sqrt{3}$
Thus, option (c) is our required answer.