To solve for [tex]\( p \)[/tex] and [tex]\( q \)[/tex] such that
[tex]\[
\frac{3 + 2 \sqrt{5}}{3 - 2 \sqrt{5}} = p + q \sqrt{5},
\][/tex]
we can approach the problem by rationalizing the denominator. The simplest way to do this involves multiplying both the numerator and the denominator by the conjugate of the denominator.
First, let's identify the conjugate of the denominator [tex]\( 3 - 2 \sqrt{5} \)[/tex], which is [tex]\( 3 + 2 \sqrt{5} \)[/tex].
Hence, the expression becomes:
[tex]\[
\frac{3 + 2 \sqrt{5}}{3 - 2 \sqrt{5}} \cdot \frac{3 + 2 \sqrt{5}}{3 + 2 \sqrt{5}} = \frac{(3 + 2 \sqrt{5})^2}{(3 - 2 \sqrt{5})(3 + 2 \sqrt{5})}.
\][/tex]
Let's first simplify the denominator:
[tex]\[
(3 - 2 \sqrt{5})(3 + 2 \sqrt{5}) = 3^2 - (2 \sqrt{5})^2 = 9 - 4 \cdot 5 = 9 - 20 = -11.
\][/tex]
Now, let's simplify the numerator by expanding [tex]\( (3 + 2 \sqrt{5})^2 \)[/tex]:
[tex]\[
(3 + 2 \sqrt{5})^2 = 3^2 + 2 \cdot 3 \cdot 2 \sqrt{5} + (2 \sqrt{5})^2.
\][/tex]
This can be broken down as:
[tex]\[
3^2 = 9,
\][/tex]
[tex]\[
2 \cdot 3 \cdot 2 \sqrt{5} = 12 \sqrt{5},
\][/tex]
[tex]\[
(2 \sqrt{5})^2 = 4 \cdot 5 = 20.
\][/tex]
So, the numerator becomes:
[tex]\[
9 + 12 \sqrt{5} + 20 = 29 + 12 \sqrt{5}.
\][/tex]
Therefore, the expression simplifies to:
[tex]\[
\frac{29 + 12 \sqrt{5}}{-11}.
\][/tex]
We can separate this into real and irrational parts:
[tex]\[
= \frac{29}{-11} + \frac{12 \sqrt{5}}{-11} = -\frac{29}{11} - \frac{12}{11} \sqrt{5}.
\][/tex]
From this, we can identify [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
[tex]\[
p = -\frac{29}{11} \approx -2.636,
\][/tex]
[tex]\[
q = -\frac{12}{11} \approx -1.091.
\][/tex]
Finally, the values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are:
[tex]\[
p \approx -2.636,
\][/tex]
[tex]\[
q \approx -1.091.
\][/tex]
