Let's solve the given expression step-by-step for \(\frac{\sqrt{5}-2}{\sqrt{5}+2} - \frac{\sqrt{5}+2}{\sqrt{5}-2}\) and find it in the form \(a + b\sqrt{5}\).
First, let's rationalize the denominators of both fractions.
### Rationalizing \(\frac{\sqrt{5}-2}{\sqrt{5}+2}\):
Multiply the numerator and denominator by the conjugate of the denominator, \(\sqrt{5} - 2\):
[tex]\[
\frac{\sqrt{5}-2}{\sqrt{5}+2} \cdot \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{(\sqrt{5}-2)^2}{(\sqrt{5})^2 - (2)^2}
\][/tex]
Simplify the denominator:
[tex]\[
(\sqrt{5})^2 - (2)^2 = 5 - 4 = 1
\][/tex]
Now expand the numerator using the binomial expansion:
[tex]\[
(\sqrt{5}-2)^2 = (\sqrt{5})^2 - 2 \cdot \sqrt{5} \cdot 2 + 2^2 = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5}
\][/tex]
So,
[tex]\[
\frac{\sqrt{5}-2}{\sqrt{5}+2} = 9 - 4\sqrt{5}
\][/tex]
### Rationalizing \(\frac{\sqrt{5}+2}{\sqrt{5}-2}\):
Multiply the numerator and denominator by the conjugate of the denominator, \(\sqrt{5} + 2\):
[tex]\[
\frac{\sqrt{5}+2}{\sqrt{5}-2} \cdot \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{(\sqrt{5}+2)^2}{(\sqrt{5})^2 - (2)^2}
\][/tex]
Simplify the denominator:
[tex]\[
(\sqrt{5})^2 - (2)^2 = 5 - 4 = 1
\][/tex]
Now expand the numerator using the binomial expansion:
[tex]\[
(\sqrt{5}+2)^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot 2 + 2^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}
\][/tex]
So,
[tex]\[
\frac{\sqrt{5}+2}{\sqrt{5}-2} = 9 + 4\sqrt{5}
\][/tex]
### Subtracting the rationalized expressions:
Now combine the results:
[tex]\[
9 - 4\sqrt{5} - (9 + 4\sqrt{5}) = 9 - 4\sqrt{5} - 9 - 4\sqrt{5} = -8\sqrt{5}
\][/tex]
Thus, we can compare this result with \(a + b\sqrt{5}\):
[tex]\[
-8\sqrt{5} = 0 + (-8)\sqrt{5}
\][/tex]
Therefore, \(a = 0\) and \(b = -8\).
So, the values are:
[tex]\[
a = 0 \quad \text{and} \quad b = -8
\][/tex]
