
How do you simplify \[\sqrt {125{x^2}} \]?
Answer
558.3k+ views
Hint: In this question we should simplify the expression given, first we have to simplify the term that is under the square root, and the constant under the square root should be prime factorised, here it is 125, and we should know that the square-root of \[x\]-squared is simply \[x\], and then further simplification we will get the required answer.
Complete step-by-step answer:
Given expression is \[\sqrt {125{x^2}} \],
The given term is a multiple of the constant 125 and square of the variable \[x\] which is equal to \[{x^2}\].
A square root can also be seen as a power of \[\dfrac{1}{2}\], so the given expression can be written as,\[{\left( {125{x^2}} \right)^{\dfrac{1}{2}}}\] , and we know that the power is distributed across the factors, which means that the square-root can be separated across the factors and this can be written as,
\[ \Rightarrow {\left( {125} \right)^{\dfrac{1}{2}}}{\left( {{x^2}} \right)^{\dfrac{1}{2}}}\],
Now we will prime factorise the constant term which can be written as \[125 = 5 \times 5 \times 5\], which again can be rewritten as,\[125 = {5^2} \times 5\],
So substituting the values in the expression we get,
\[ \Rightarrow \sqrt {125{x^2}} = {\left( {{5^2} \times 5} \right)^{\dfrac{1}{2}}} \cdot {\left( {{x^2}} \right)^{\dfrac{1}{2}}}\],
Now again distributing the powers we get,
\[ \Rightarrow \sqrt {125{x^2}} = {\left( {{5^2}} \right)^{\dfrac{1}{2}}} \times {5^{\dfrac{1}{2}}} \times {\left( {{x^2}} \right)^{\dfrac{1}{2}}}\],
Now taking out the square root we get,
\[ \Rightarrow \sqrt {125{x^2}} = \left( 5 \right) \times {5^{\dfrac{1}{2}}} \times \left( x \right)\],
Now simplifying we get,
\[ \Rightarrow \sqrt {125{x^2}} = \left( {5\sqrt 5 } \right) \cdot \left( x \right)\],
Now multiplying we get,
\[ \Rightarrow \sqrt {125{x^2}} = 5\sqrt 5 x\],
So, the simplified form is \[5\sqrt 5 x\].
\[\therefore \]The simplified form of the given expression \[\sqrt {125{x^2}} \] is equal to \[5\sqrt 5 x\].
Note:
The formula for indices is given by \[\sqrt[n]{x} = {x^{\dfrac{1}{n}}}\], the \[{n^{th}}\] root can be expressed in both the ways, of n value is not mentioned then we always assume that the value of n is 2. It always represents square root values.
Complete step-by-step answer:
Given expression is \[\sqrt {125{x^2}} \],
The given term is a multiple of the constant 125 and square of the variable \[x\] which is equal to \[{x^2}\].
A square root can also be seen as a power of \[\dfrac{1}{2}\], so the given expression can be written as,\[{\left( {125{x^2}} \right)^{\dfrac{1}{2}}}\] , and we know that the power is distributed across the factors, which means that the square-root can be separated across the factors and this can be written as,
\[ \Rightarrow {\left( {125} \right)^{\dfrac{1}{2}}}{\left( {{x^2}} \right)^{\dfrac{1}{2}}}\],
Now we will prime factorise the constant term which can be written as \[125 = 5 \times 5 \times 5\], which again can be rewritten as,\[125 = {5^2} \times 5\],
So substituting the values in the expression we get,
\[ \Rightarrow \sqrt {125{x^2}} = {\left( {{5^2} \times 5} \right)^{\dfrac{1}{2}}} \cdot {\left( {{x^2}} \right)^{\dfrac{1}{2}}}\],
Now again distributing the powers we get,
\[ \Rightarrow \sqrt {125{x^2}} = {\left( {{5^2}} \right)^{\dfrac{1}{2}}} \times {5^{\dfrac{1}{2}}} \times {\left( {{x^2}} \right)^{\dfrac{1}{2}}}\],
Now taking out the square root we get,
\[ \Rightarrow \sqrt {125{x^2}} = \left( 5 \right) \times {5^{\dfrac{1}{2}}} \times \left( x \right)\],
Now simplifying we get,
\[ \Rightarrow \sqrt {125{x^2}} = \left( {5\sqrt 5 } \right) \cdot \left( x \right)\],
Now multiplying we get,
\[ \Rightarrow \sqrt {125{x^2}} = 5\sqrt 5 x\],
So, the simplified form is \[5\sqrt 5 x\].
\[\therefore \]The simplified form of the given expression \[\sqrt {125{x^2}} \] is equal to \[5\sqrt 5 x\].
Note:
The formula for indices is given by \[\sqrt[n]{x} = {x^{\dfrac{1}{n}}}\], the \[{n^{th}}\] root can be expressed in both the ways, of n value is not mentioned then we always assume that the value of n is 2. It always represents square root values.
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