Find the square root of the following term.
\[8+2\sqrt{15}\]
Answer
648k+ views
Hint: To solve this question, we will use the formula or the identity or the square of two numbers given as \[{{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}.\] Here, we have to obtain the square root of \[8+2\sqrt{15}\] so we will try to compute \[8+2\sqrt{15}\] in terms of some square and finally using the fact that \[\sqrt{\left( {{a}^{2}} \right)}=a\] itself, we will get the result.
Complete step by step answer:
We have to find the square root of \[8+2\sqrt{15}.....\left( i \right)\]
Let us consider \[2\sqrt{15}.\] It can be split as
\[2\sqrt{15}=2\sqrt{5\times 3}=2\sqrt{5}\sqrt{3}\]
As, we know that \[\sqrt{ab}=\sqrt{a}\sqrt{b}\] when both a > 0 and b > 0 and here, 5 > 0 and 3 > 0.
\[\Rightarrow \sqrt{15}=\sqrt{5}\sqrt{3}\]
Therefore, we can rewrite the equation as shown below,
\[\Rightarrow 2\sqrt{15}=2\sqrt{5}\sqrt{3}.....\left( ii \right)\]
Using the identity of the square of the sum of two numbers given as \[{{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}.\] and putting \[x=\sqrt{5}\] and \[y=\sqrt{3}\] in the above, we get,
\[{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+2\times \sqrt{5}\sqrt{3}+{{\left( \sqrt{3} \right)}^{2}}\]
As, we know that \[{{\left( \sqrt{a} \right)}^{2}}=a\] itself, we get,
\[\Rightarrow {{\left( \sqrt{5} \right)}^{2}}=5\]
\[\Rightarrow {{\left( \sqrt{3} \right)}^{2}}=3\]
\[\Rightarrow {{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=5+3+2\sqrt{5}\sqrt{3}\]
Now, using equation (ii) in the above, we get,
\[\Rightarrow {{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=5+3+2\sqrt{15}\]
\[\Rightarrow {{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=8+2\sqrt{15}.....\left( iii \right)\]
Finally, comparing the equations (i) and (ii), we see that they both are equal.
\[\Rightarrow 8+2\sqrt{15}={{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}\]
Hence, \[8+2\sqrt{15}\] is nothing but the square of \[\left( \sqrt{5}+\sqrt{3} \right).\] Now, let us compute the square root of \[8+2\sqrt{15}.\]
\[\Rightarrow \sqrt{8+2\sqrt{15}}=\sqrt{{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}}\]
Then, as \[\sqrt{{{\left( a \right)}^{2}}}=a\] itself, applying this in the RHS of above equation, we have,
\[\sqrt{8+2\sqrt{15}}=\sqrt{5}+\sqrt{3}\]
Hence, the square root of \[8+2\sqrt{15}=\sqrt{5}+\sqrt{3}.\]
Note: The fact or statement \[\sqrt{ab}=\sqrt{a}\sqrt{b}\] is not always true. It is true when a > 0, b > 0. Also, when we extend a, b to complex numbers, then this statement is again not true. Whenever we get such questions, the basic approach should be the same, i.e to convert the given number in the form of \[{{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}\] in order to take its square root easily.
Complete step by step answer:
We have to find the square root of \[8+2\sqrt{15}.....\left( i \right)\]
Let us consider \[2\sqrt{15}.\] It can be split as
\[2\sqrt{15}=2\sqrt{5\times 3}=2\sqrt{5}\sqrt{3}\]
As, we know that \[\sqrt{ab}=\sqrt{a}\sqrt{b}\] when both a > 0 and b > 0 and here, 5 > 0 and 3 > 0.
\[\Rightarrow \sqrt{15}=\sqrt{5}\sqrt{3}\]
Therefore, we can rewrite the equation as shown below,
\[\Rightarrow 2\sqrt{15}=2\sqrt{5}\sqrt{3}.....\left( ii \right)\]
Using the identity of the square of the sum of two numbers given as \[{{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}.\] and putting \[x=\sqrt{5}\] and \[y=\sqrt{3}\] in the above, we get,
\[{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+2\times \sqrt{5}\sqrt{3}+{{\left( \sqrt{3} \right)}^{2}}\]
As, we know that \[{{\left( \sqrt{a} \right)}^{2}}=a\] itself, we get,
\[\Rightarrow {{\left( \sqrt{5} \right)}^{2}}=5\]
\[\Rightarrow {{\left( \sqrt{3} \right)}^{2}}=3\]
\[\Rightarrow {{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=5+3+2\sqrt{5}\sqrt{3}\]
Now, using equation (ii) in the above, we get,
\[\Rightarrow {{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=5+3+2\sqrt{15}\]
\[\Rightarrow {{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=8+2\sqrt{15}.....\left( iii \right)\]
Finally, comparing the equations (i) and (ii), we see that they both are equal.
\[\Rightarrow 8+2\sqrt{15}={{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}\]
Hence, \[8+2\sqrt{15}\] is nothing but the square of \[\left( \sqrt{5}+\sqrt{3} \right).\] Now, let us compute the square root of \[8+2\sqrt{15}.\]
\[\Rightarrow \sqrt{8+2\sqrt{15}}=\sqrt{{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}}\]
Then, as \[\sqrt{{{\left( a \right)}^{2}}}=a\] itself, applying this in the RHS of above equation, we have,
\[\sqrt{8+2\sqrt{15}}=\sqrt{5}+\sqrt{3}\]
Hence, the square root of \[8+2\sqrt{15}=\sqrt{5}+\sqrt{3}.\]
Note: The fact or statement \[\sqrt{ab}=\sqrt{a}\sqrt{b}\] is not always true. It is true when a > 0, b > 0. Also, when we extend a, b to complex numbers, then this statement is again not true. Whenever we get such questions, the basic approach should be the same, i.e to convert the given number in the form of \[{{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}\] in order to take its square root easily.
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