
How do you simplify \[\sqrt {125{x^2}} \]?
Answer
543.9k+ 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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Write a letter to the principal requesting him to grant class 10 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Discuss the main reasons for poverty in India

10 examples of evaporation in daily life with explanations

