How do you simplify $\dfrac{{14}}{{\sqrt 2 }}$?
Answer
605.1k+ views
Hint: First, multiply numerator and denominator with $\sqrt 2 $. Then, combine and simplify the denominator by using the power rule (I) to combine exponents. Then, use property (II) to rewrite $\sqrt 2 $ as ${2^{\dfrac{1}{2}}}$ and apply the power rule (III) to multiply exponents. Then, cancel the common factor of $14$ and $2$. We will get the simplified version of $\dfrac{{14}}{{\sqrt 2 }}$.
Formula used:
Power rule to combine exponents: ${a^m} \times {a^n} = {a^{m + n}}$
$\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$
${\left( {{a^m}} \right)^n} = {a^{mn}}$
Complete step by step solution:
We have to simplify $\dfrac{{14}}{{\sqrt 2 }}$.
Multiply numerator and denominator with $\sqrt 2 $.
$ \Rightarrow \dfrac{{14}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
Combine and simplify the denominator.
Multiply $\dfrac{{14}}{{\sqrt 2 }}$ and $\dfrac{{\sqrt 2 }}{{\sqrt 2 }}$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}$
Raise $\sqrt 2 $ to the power of $1$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^1} \times \sqrt 2 }}$
Raise $\sqrt 2 $ to the power of $1$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^1} \times {{\sqrt 2 }^1}}}$
Use the power rule ${a^m} \times {a^n} = {a^{m + n}}$ to combine exponents.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^{1 + 1}}}}$
Add $1$ and $1$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^2}}}$
Rewrite ${\sqrt 2 ^2}$ as $2$.
Use property $\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$ to rewrite $\sqrt 3 $ as ${3^{\dfrac{1}{2}}}$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\left( {{2^{\dfrac{1}{2}}}} \right)}^2}}}$
Apply the power rule and multiply exponents, ${\left( {{a^m}} \right)^n} = {a^{mn}}$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{2^{\dfrac{1}{2} \times 2}}}}$
Multiply $\dfrac{1}{2}$ and $2$, we get
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{2^1}}}$
It can be written as
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{2}$
Now, cancel the common factor of $14$ and $2$.
Factor $2$ out of $14\sqrt 2 $.
$ \Rightarrow \dfrac{{2\left( {7\sqrt 2 } \right)}}{2}$
Cancel the common factor.
$ \Rightarrow 7\sqrt 2 $
The result can be shown in multiple forms.
Exact Form: $7\sqrt 2 $
Decimal Form: $9.899494937$
Hence, simplified version of $\dfrac{{14}}{{\sqrt 2 }}$ is $7\sqrt 2 $.
Note: By simplifying a fraction, we mean to express the fraction as a ratio of prime numbers or we can say that both the numerator and denominator should be prime numbers, that is, they should be divisible by only $1$ and itself. For simplifying a fraction, we write it as a product of prime factors, and then divide both of them with the common factors. In this question both the numerator and denominator are already prime numbers and thus the fraction cannot be simplified further.
Formula used:
Power rule to combine exponents: ${a^m} \times {a^n} = {a^{m + n}}$
$\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$
${\left( {{a^m}} \right)^n} = {a^{mn}}$
Complete step by step solution:
We have to simplify $\dfrac{{14}}{{\sqrt 2 }}$.
Multiply numerator and denominator with $\sqrt 2 $.
$ \Rightarrow \dfrac{{14}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
Combine and simplify the denominator.
Multiply $\dfrac{{14}}{{\sqrt 2 }}$ and $\dfrac{{\sqrt 2 }}{{\sqrt 2 }}$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}$
Raise $\sqrt 2 $ to the power of $1$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^1} \times \sqrt 2 }}$
Raise $\sqrt 2 $ to the power of $1$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^1} \times {{\sqrt 2 }^1}}}$
Use the power rule ${a^m} \times {a^n} = {a^{m + n}}$ to combine exponents.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^{1 + 1}}}}$
Add $1$ and $1$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\sqrt 2 }^2}}}$
Rewrite ${\sqrt 2 ^2}$ as $2$.
Use property $\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$ to rewrite $\sqrt 3 $ as ${3^{\dfrac{1}{2}}}$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{{\left( {{2^{\dfrac{1}{2}}}} \right)}^2}}}$
Apply the power rule and multiply exponents, ${\left( {{a^m}} \right)^n} = {a^{mn}}$.
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{2^{\dfrac{1}{2} \times 2}}}}$
Multiply $\dfrac{1}{2}$ and $2$, we get
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{{{2^1}}}$
It can be written as
$ \Rightarrow \dfrac{{14 \times \sqrt 2 }}{2}$
Now, cancel the common factor of $14$ and $2$.
Factor $2$ out of $14\sqrt 2 $.
$ \Rightarrow \dfrac{{2\left( {7\sqrt 2 } \right)}}{2}$
Cancel the common factor.
$ \Rightarrow 7\sqrt 2 $
The result can be shown in multiple forms.
Exact Form: $7\sqrt 2 $
Decimal Form: $9.899494937$
Hence, simplified version of $\dfrac{{14}}{{\sqrt 2 }}$ is $7\sqrt 2 $.
Note: By simplifying a fraction, we mean to express the fraction as a ratio of prime numbers or we can say that both the numerator and denominator should be prime numbers, that is, they should be divisible by only $1$ and itself. For simplifying a fraction, we write it as a product of prime factors, and then divide both of them with the common factors. In this question both the numerator and denominator are already prime numbers and thus the fraction cannot be simplified further.
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