Question

Express \[\left( {{5^{\dfrac{1}{2}}} + {9^{\dfrac{1}{8}}}} \right) \div \left( {{5^{\dfrac{1}{2}}} - {9^{\dfrac{1}{8}}}} \right)\] as an equivalent fraction with a rational denominator.

Answer 1

Hint: Write the number in fraction and rationalize it. If the number is not giving a rational denominator, then rationalize it again to get the desired result.

Complete step-by-step answer:
According to question, the given number is \[\left( {{5^{\dfrac{1}{2}}} + {9^{\dfrac{1}{8}}}} \right) \div \left( {{5^{\dfrac{1}{2}}} - {9^{\dfrac{1}{8}}}} \right)\]. Let this number is \[N\]. So we have:
\[ \Rightarrow N = \left( {{5^{\dfrac{1}{2}}} + {9^{\dfrac{1}{8}}}} \right) \div \left( {{5^{\dfrac{1}{2}}} - {9^{\dfrac{1}{8}}}} \right) = \dfrac{{{5^{\dfrac{1}{2}}} + {9^{\dfrac{1}{8}}}}}{{{5^{\dfrac{1}{2}}} - {9^{\dfrac{1}{8}}}}} .....(i)\].
If we put \[9\] as \[{3^2}\], our number will be:
\[ \Rightarrow N = \dfrac{{{5^{\dfrac{1}{2}}} + {3^{\dfrac{1}{4}}}}}{{{5^{\dfrac{1}{2}}} - {3^{\dfrac{1}{4}}}}}\]
Rationalizing the number by multiplying and dividing by the conjugate of denominator, we’ll get:
\[ \Rightarrow N = \dfrac{{{5^{\dfrac{1}{2}}} + {3^{\dfrac{1}{4}}}}}{{{5^{\dfrac{1}{2}}} - {3^{\dfrac{1}{4}}}}} \times \dfrac{{{5^{\dfrac{1}{2}}} + {3^{\dfrac{1}{4}}}}}{{{5^{\dfrac{1}{2}}} + {3^{\dfrac{1}{4}}}}}\]
Using formulas \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] and \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\] we’ll get:
\[ \Rightarrow N = \dfrac{{5 + {3^{\dfrac{1}{2}}} + {{2.5}^{\dfrac{1}{2}}}{{.3}^{\dfrac{1}{4}}}}}{{5 - {3^{\dfrac{1}{2}}}}}\]
Rationalizing it one more time, we’ll get:
\[ \Rightarrow N = \dfrac{{5 + {3^{\dfrac{1}{2}}} + {{2.5}^{\dfrac{1}{2}}}{{.3}^{\dfrac{1}{4}}}}}{{5 - {3^{\dfrac{1}{2}}}}} \times \dfrac{{5 + {3^{\dfrac{1}{2}}}}}{{5 + {3^{\dfrac{1}{2}}}}}\]
Using \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\] in denominator and multiplying step by step in numerator, we’ll get:
\[ \Rightarrow N = \dfrac{{25 + {{5.3}^{\dfrac{1}{2}}} + {{5.3}^{\dfrac{1}{2}}} + 3 + {{2.5}^{\dfrac{3}{2}}}{{.3}^{\dfrac{1}{4}}} + {{2.5}^{\dfrac{1}{2}}}{{.3}^{\dfrac{3}{4}}}}}{{{5^2} - 3}}\]
Simplifying it further, we’ll get:
\[
   \Rightarrow N = \dfrac{{28 + {{2.5.3}^{\dfrac{1}{2}}} + {{2.5}^{\dfrac{3}{2}}}{{.3}^{\dfrac{1}{4}}} + {{2.5}^{\dfrac{1}{2}}}{{.3}^{\dfrac{3}{4}}}}}{{22}} \\
   \Rightarrow N = \dfrac{{2\left( {14 + {{5.3}^{\dfrac{1}{2}}} + {5^{\dfrac{3}{2}}}{{.3}^{\dfrac{1}{4}}} + {5^{\dfrac{1}{2}}}{{.3}^{\dfrac{3}{4}}}} \right)}}{{22}} \\
   \Rightarrow N = \dfrac{{14 + {{5.3}^{\dfrac{1}{2}}} + {5^{\dfrac{3}{2}}}{{.3}^{\dfrac{1}{4}}} + {5^{\dfrac{1}{2}}}{{.3}^{\dfrac{3}{4}}}}}{{11}} \\
\]
Putting back the value of \[N\] from equation \[(i)\], we’ll get:
\[ \Rightarrow \left( {{5^{\dfrac{1}{2}}} + {9^{\dfrac{1}{8}}}} \right) \div \left( {{5^{\dfrac{1}{2}}} - {9^{\dfrac{1}{8}}}} \right) = \dfrac{{14 + {{5.3}^{\dfrac{1}{2}}} + {5^{\dfrac{3}{2}}}{{.3}^{\dfrac{1}{4}}} + {5^{\dfrac{1}{2}}}{{.3}^{\dfrac{3}{4}}}}}{{11}}\]
The equivalent fraction of the number with a rational denominator is \[\dfrac{{14 + {{5.3}^{\dfrac{1}{2}}} + {5^{\dfrac{3}{2}}}{{.3}^{\dfrac{1}{4}}} + {5^{\dfrac{1}{2}}}{{.3}^{\dfrac{3}{4}}}}}{{11}}\].

Note: In order to rationalize the denominator, we have to multiply the numerator and denominator by a radical that will get rid of the radical in the denominator. Make sure that all radicals are simplified. Simplify the fraction if needed.