Question

How do you combine and simplify \[\sqrt {\dfrac{{25}}{{128}}} \]?

Answer 1

Hint: Simplify the given algebraic expression by the use of formula \[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\] and \[\sqrt {a \times a} = a\], here the symbol \[\sqrt {} \] is used for square root. Also use the rationalization of a fraction if there is a number in the denominator that cannot be square rooted and in a form \[\sqrt a \].

Complete step by step solution:
The given algebraic expression is \[\sqrt {\dfrac{{25}}{{128}}} \]. It expresses the square root of the fraction that consists of number twenty-five as numerator and one hundred twenty-eight as denominator.
The symbol for square root is \[\sqrt {} \]. It is known that in algebraic expressions, square roots can be divided in numerator and denominator for a fraction. Therefore, we can write the given expression as shown below.
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{{\sqrt {25} }}{{\sqrt {128} }}\]
Now, write the prime factorize form of numbers that are in the square root symbol in numerator and in denominator by the use of formula \[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\] as shown below.
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{{\sqrt {5 \times 5} }}{{\sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} }}\]
Now, perform the square root operation by taking out one of the numbers from the square root for each pair of same numbers by the use of formula \[\sqrt {a \times a} = a\] as shown below.
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{5}{{2 \times 2 \times 2\sqrt 2 }}\]
Simplify the expression further as shown below.
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{5}{{8\sqrt 2 }}\]
Now, rationalize the fraction as any irrational number is not allowed in the denominator. Therefore multiply the result \[\dfrac{5}{{8\sqrt 2 }}\] by the unit in form of fraction as \[\dfrac{{\sqrt 2 }}{{\sqrt 2 }}\] and simplify as shown below.
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{5}{{8\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\]
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{{5\sqrt 2 }}{{8\sqrt {2 \times 2} }}\]
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{{5\sqrt 2 }}{{8 \times 2}}\]
\[ \Rightarrow \sqrt {\dfrac{{25}}{{128}}} = \dfrac{{5\sqrt 2 }}{{16}}\]
Thus, the combined and simplified expression for \[\sqrt {\dfrac{{25}}{{128}}} \] is \[\dfrac{{5\sqrt 2 }}{{16}}\].

Note:
Some basic algebraic expressions that are helpful to solve and simplify these types of expressions are \[\sqrt {ab} = \sqrt a \sqrt b \], \[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\], \[\sqrt {a \times a} = a\] and \[\dfrac{1}{{\sqrt a }} = \dfrac{{\sqrt a }}{a}\]. Where numbers \[a\] and \[b\] are positive integers.