
How do you simplify ${\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}}$?
Answer
556.2k+ views
Hint: In this problem we have given a fractional number having some power and whose power is also a fractional number. Now we asked to simplify the given number. We can simplify this given fractional number having power as a fractional number by using some algebraic operations like rules of exponents.
Formula used:
${x^1} = x$ and ${\left( {{x^a}} \right)^b} = {x^{a \times b}}$
Complete step by step answer:
Given number is ${\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}}$.
Here in the given number the power is common to both the numerator and the denominator. So we can put the power $\dfrac{1}{2}$ to both the numerator and the denominator.
Also we can write the given number as ${\left( {\dfrac{{{1^1}}}{{{4^1}}}} \right)^{\dfrac{1}{2}}}$
Now, let’s put the power to both numerator and the denominator, we get
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{{{\left( {{1^1}} \right)}^{\dfrac{1}{2}}}}}{{{{\left( {{4^1}} \right)}^{\dfrac{1}{2}}}}}$
Using the formula, ${\left( {{x^a}} \right)^b} = {x^{a \times b}}$ , we get
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{{1^{1 \times \dfrac{1}{2}}}}}{{{4^{1 \times \dfrac{1}{2}}}}}$,
On multiplying the powers in the numerator and in the denominator, we get
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{{1^{\dfrac{1}{2}}}}}{{{4^{\dfrac{1}{2}}}}}$,
On simplifying we get,
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{\sqrt 1 }}{{\sqrt 4 }} - - - - - (1)$
Taking square root for $1$ in the numerator gives again $1$ and taking square root for$4$in the denominator gives $2$, simplifies the above term gives
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{1}{2}$
This is the required solution.
Note: The square is the number itself. The square is the same as the power of $2$. The square root is the opposite of the square.
Taking a number to the power of $\dfrac{1}{2}$ undoes taking a number to the power of $2$ or squaring it.
In other words, taking a number to the power of $\dfrac{1}{2}$ is the same thing as taking a square root. This is what we have done in equation (1). The square root is actually a fractional index and is equivalent to raising a number to the number $\dfrac{1}{2}$.
Formula used:
${x^1} = x$ and ${\left( {{x^a}} \right)^b} = {x^{a \times b}}$
Complete step by step answer:
Given number is ${\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}}$.
Here in the given number the power is common to both the numerator and the denominator. So we can put the power $\dfrac{1}{2}$ to both the numerator and the denominator.
Also we can write the given number as ${\left( {\dfrac{{{1^1}}}{{{4^1}}}} \right)^{\dfrac{1}{2}}}$
Now, let’s put the power to both numerator and the denominator, we get
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{{{\left( {{1^1}} \right)}^{\dfrac{1}{2}}}}}{{{{\left( {{4^1}} \right)}^{\dfrac{1}{2}}}}}$
Using the formula, ${\left( {{x^a}} \right)^b} = {x^{a \times b}}$ , we get
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{{1^{1 \times \dfrac{1}{2}}}}}{{{4^{1 \times \dfrac{1}{2}}}}}$,
On multiplying the powers in the numerator and in the denominator, we get
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{{1^{\dfrac{1}{2}}}}}{{{4^{\dfrac{1}{2}}}}}$,
On simplifying we get,
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{{\sqrt 1 }}{{\sqrt 4 }} - - - - - (1)$
Taking square root for $1$ in the numerator gives again $1$ and taking square root for$4$in the denominator gives $2$, simplifies the above term gives
$ \Rightarrow {\left( {\dfrac{1}{4}} \right)^{\dfrac{1}{2}}} = \dfrac{1}{2}$
This is the required solution.
Note: The square is the number itself. The square is the same as the power of $2$. The square root is the opposite of the square.
Taking a number to the power of $\dfrac{1}{2}$ undoes taking a number to the power of $2$ or squaring it.
In other words, taking a number to the power of $\dfrac{1}{2}$ is the same thing as taking a square root. This is what we have done in equation (1). The square root is actually a fractional index and is equivalent to raising a number to the number $\dfrac{1}{2}$.
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