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Find the square root of \[7-4\sqrt{3}.\]
\[\left( a \right)2+\sqrt{3}\]
\[\left( b \right)5-\sqrt{3}\]
\[\left( c \right)2-\sqrt{3}\]
\[\left( d \right)5+\sqrt{3}\]
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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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
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