How do you simplify \[\sqrt{\dfrac{1}{6}}\]?
Answer
601.2k+ views
Hint: For the given solution we are given that to solve the problem \[\sqrt{\dfrac{1}{6}}\]. For that we have to find the root for the fraction \[\dfrac{1}{6}\] which is not a perfect square. We have to rationalise the given problem to get the solution for the given question.
Complete step by step solution:
\[\sqrt{\dfrac{1}{6}}\]
First of all we have to rewrite \[\sqrt{\dfrac{1}{6}}\]as \[\dfrac{\sqrt{1}}{\sqrt{6}}\], we get
\[\Rightarrow \dfrac{\sqrt{1}}{\sqrt{6}}\]
And we all know that the root 1 is 1, we get
\[\Rightarrow \dfrac{1}{\sqrt{6}}\]
Now we should multiply \[\dfrac{1}{\sqrt{6}}\]by \[\dfrac{\sqrt{6}}{\sqrt{6}}\], we get
\[\Rightarrow \dfrac{1}{\sqrt{6}}\cdot \dfrac{\sqrt{6}}{\sqrt{6}}\]
Now we should combine and simplify the denominator, we get
\[\Rightarrow \dfrac{\sqrt{6}}{\sqrt{6}\cdot \sqrt{6}}\]
Now we should raise \[\sqrt{6}\]to the power of 1 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{1}}\cdot \sqrt{6}}\]
Now we should raise \[\sqrt{6}\]to the power of 1 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{1}}\cdot {{\sqrt{6}}^{1}}}\]
Now we should use the power rule \[{{a}^{m}}{{a}^{n}}={{a}^{m+n}}\] to combine exponents, we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{1+1}}}\]
Now we should add 1 and 1 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{2}}}\]
Now we should rewrite \[{{\sqrt{6}}^{2}}\]as \[6\]and use some rule to rewrite \[\sqrt{6}\]as \[{{6}^{\dfrac{1}{2}}}\] , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\left( {{6}^{\dfrac{1}{2}}} \right)}^{2}}}\]
And now we have to apply the power rule and multiply components, we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{\dfrac{1}{2}\cdot 2}}}\]
By multiplying \[\dfrac{1}{2}\] and 2 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{\dfrac{2}{2}}}}\]
Now we have to cancel the common factor and divide 1 by 1, we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{1}}}\]
The exact form of the solution is
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{{}}}}\]
Therefore, the above solution is the exact form of the required solution of the problem given.
Note: We should be aware of the roots of rational numbers to solve this type of problems. If the given fraction is not a perfect square and the denominator contains a square root then we have to choose the rationalisation method.
Complete step by step solution:
\[\sqrt{\dfrac{1}{6}}\]
First of all we have to rewrite \[\sqrt{\dfrac{1}{6}}\]as \[\dfrac{\sqrt{1}}{\sqrt{6}}\], we get
\[\Rightarrow \dfrac{\sqrt{1}}{\sqrt{6}}\]
And we all know that the root 1 is 1, we get
\[\Rightarrow \dfrac{1}{\sqrt{6}}\]
Now we should multiply \[\dfrac{1}{\sqrt{6}}\]by \[\dfrac{\sqrt{6}}{\sqrt{6}}\], we get
\[\Rightarrow \dfrac{1}{\sqrt{6}}\cdot \dfrac{\sqrt{6}}{\sqrt{6}}\]
Now we should combine and simplify the denominator, we get
\[\Rightarrow \dfrac{\sqrt{6}}{\sqrt{6}\cdot \sqrt{6}}\]
Now we should raise \[\sqrt{6}\]to the power of 1 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{1}}\cdot \sqrt{6}}\]
Now we should raise \[\sqrt{6}\]to the power of 1 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{1}}\cdot {{\sqrt{6}}^{1}}}\]
Now we should use the power rule \[{{a}^{m}}{{a}^{n}}={{a}^{m+n}}\] to combine exponents, we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{1+1}}}\]
Now we should add 1 and 1 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\sqrt{6}}^{2}}}\]
Now we should rewrite \[{{\sqrt{6}}^{2}}\]as \[6\]and use some rule to rewrite \[\sqrt{6}\]as \[{{6}^{\dfrac{1}{2}}}\] , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{\left( {{6}^{\dfrac{1}{2}}} \right)}^{2}}}\]
And now we have to apply the power rule and multiply components, we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{\dfrac{1}{2}\cdot 2}}}\]
By multiplying \[\dfrac{1}{2}\] and 2 , we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{\dfrac{2}{2}}}}\]
Now we have to cancel the common factor and divide 1 by 1, we get
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{1}}}\]
The exact form of the solution is
\[\Rightarrow \dfrac{\sqrt{6}}{{{6}^{{}}}}\]
Therefore, the above solution is the exact form of the required solution of the problem given.
Note: We should be aware of the roots of rational numbers to solve this type of problems. If the given fraction is not a perfect square and the denominator contains a square root then we have to choose the rationalisation method.
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