How do you simplify \[\sqrt {\dfrac{{500}}{{720}}} \]?
Answer
625.5k+ views
Hint: We write the prime factorization of the numbers inside the square root and try to pair and make squares of values such that we are able to cancel the square root by square power.
Prime factorization is a process of writing a number in multiple of its factors where all factors are prime numbers.
Law of exponent states that when the base is same is same in division we can subtract the powers i.e. \[\dfrac{{{p^m}}}{{{p^n}}} = {p^{m - n}}\]
Complete step-by-step answer:
We have to simplify the value of \[\sqrt {\dfrac{{500}}{{720}}} \] … (1)
We will write the prime factorization of the numbers 500 and 720
\[500 = 2 \times 2 \times 5 \times 5 \times 5\]
\[720 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5\]
Since we know the law of exponents so we can collect the powers of same base
\[500 = {2^2} \times {5^3}\]
\[720 = {2^4} \times {3^2} \times {5^1}\]
Now we substitute the value of 500 and 720 back in equation (1)
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{2^2} \times {5^3}}}{{{2^4} \times {3^2} \times {5^1}}}} \]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{2^2}}}{{{2^4}}} \times \dfrac{{{5^3}}}{{{5^1}}} \times \dfrac{1}{{{3^2}}}} \]
We can write \[1 = {3^0}\]to make the base the same for division.
We now use the rule of exponents that when the base is same in division we subtract the powers i.e. \[\dfrac{{{p^m}}}{{{p^n}}} = {p^{m - n}}\]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {{2^{2 - 4}} \times {5^{3 - 1}} \times {3^{0 - 2}}} \]
Calculate the powers
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {{2^{ - 2}} \times {5^2} \times {3^{ - 2}}} \]
Write all powers with negative sign in denominator as we know \[{x^{ - 1}} = \dfrac{1}{x}\]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{5^2}}}{{{2^2} \times {3^2}}}} \]
Since the powers in the denominator are same we can multiply the bases i.e. \[{p^m} \times {q^m} = {(p \times q)^m}\]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{5^2}}}{{{{(2 \times 3)}^2}}}} \]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{5^2}}}{{{6^2}}}} \]
Now we know that we can pair the terms using the rule \[\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}\]in right hand side of the equation
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {{{\left( {\dfrac{5}{6}} \right)}^2}} \]
We can cancel the square root by square power in right hand side of the equation
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \dfrac{5}{6}\]
\[\therefore \]The value of \[\sqrt {\dfrac{{500}}{{720}}} = \]is \[\dfrac{5}{6}\].
We can further divide 5 by 6 and write the answer as 0.83333333.
Note:
Many students make the mistake of calculating the division inside the square root which will be a long decimal value and then use a long division method to calculate its square root. Keep in mind if we know the square values of the first few natural numbers we can directly write the values of square root as natural numbers be it in division and then after removing square root we divide the values.
Prime factorization is a process of writing a number in multiple of its factors where all factors are prime numbers.
Law of exponent states that when the base is same is same in division we can subtract the powers i.e. \[\dfrac{{{p^m}}}{{{p^n}}} = {p^{m - n}}\]
Complete step-by-step answer:
We have to simplify the value of \[\sqrt {\dfrac{{500}}{{720}}} \] … (1)
We will write the prime factorization of the numbers 500 and 720
\[500 = 2 \times 2 \times 5 \times 5 \times 5\]
\[720 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5\]
Since we know the law of exponents so we can collect the powers of same base
\[500 = {2^2} \times {5^3}\]
\[720 = {2^4} \times {3^2} \times {5^1}\]
Now we substitute the value of 500 and 720 back in equation (1)
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{2^2} \times {5^3}}}{{{2^4} \times {3^2} \times {5^1}}}} \]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{2^2}}}{{{2^4}}} \times \dfrac{{{5^3}}}{{{5^1}}} \times \dfrac{1}{{{3^2}}}} \]
We can write \[1 = {3^0}\]to make the base the same for division.
We now use the rule of exponents that when the base is same in division we subtract the powers i.e. \[\dfrac{{{p^m}}}{{{p^n}}} = {p^{m - n}}\]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {{2^{2 - 4}} \times {5^{3 - 1}} \times {3^{0 - 2}}} \]
Calculate the powers
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {{2^{ - 2}} \times {5^2} \times {3^{ - 2}}} \]
Write all powers with negative sign in denominator as we know \[{x^{ - 1}} = \dfrac{1}{x}\]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{5^2}}}{{{2^2} \times {3^2}}}} \]
Since the powers in the denominator are same we can multiply the bases i.e. \[{p^m} \times {q^m} = {(p \times q)^m}\]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{5^2}}}{{{{(2 \times 3)}^2}}}} \]
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {\dfrac{{{5^2}}}{{{6^2}}}} \]
Now we know that we can pair the terms using the rule \[\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}\]in right hand side of the equation
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \sqrt {{{\left( {\dfrac{5}{6}} \right)}^2}} \]
We can cancel the square root by square power in right hand side of the equation
\[ \Rightarrow \sqrt {\dfrac{{500}}{{720}}} = \dfrac{5}{6}\]
\[\therefore \]The value of \[\sqrt {\dfrac{{500}}{{720}}} = \]is \[\dfrac{5}{6}\].
We can further divide 5 by 6 and write the answer as 0.83333333.
Note:
Many students make the mistake of calculating the division inside the square root which will be a long decimal value and then use a long division method to calculate its square root. Keep in mind if we know the square values of the first few natural numbers we can directly write the values of square root as natural numbers be it in division and then after removing square root we divide the values.
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