 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.

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.

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.