How do you simplify \[\sqrt {\dfrac{{49}}{{16}}} \]?

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 we can add the powers of that element i.e.
$\Rightarrow$\[{p^m} \times {p^n} = {p^{m + n}}\]

Complete step-by-step answer:
We have to simplify the value of \[\sqrt {\dfrac{{49}}{{16}}} \] … (1)
We will write the prime factorization of the numbers 49 and 16
$ \Rightarrow$ \[49 = 7 \times 7\]
$ \Rightarrow$ \[16 = 2 \times 2 \times 2 \times 2\]
Since we know the law of exponents so we can collect the powers of same base
$ \Rightarrow$ \[49 = {7^2}\]
$ \Rightarrow$ \[16 = {(2 \times 2)^2}\]i.e. \[16 = {4^2}\]
Now we substitute the value of 16 and 49 back in equation (1)
\[ \Rightarrow \sqrt {\dfrac{{49}}{{16}}} = \sqrt {\dfrac{{{7^2}}}{{{4^2}}}} \]
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
\[ \Rightarrow \sqrt {\dfrac{{49}}{{16}}} = \sqrt {{{\left( {\dfrac{7}{4}} \right)}^2}} \]
We can cancel the square root by square power in right hand side of the equation
\[ \Rightarrow \sqrt {\dfrac{{49}}{{16}}} = \dfrac{7}{4}\]
\[\therefore \] The value of \[\sqrt {\dfrac{{49}}{{16}}} \] is \[\dfrac{7}{4}\].
 We can further divide 7 by 4 and write answer as 1.75

The correct answer is 1.75

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.