To solve the equation [tex]\(\frac{\sqrt{7}-1}{\sqrt{7}+1} - \frac{\sqrt{7}+1}{\sqrt{7}-1} = a + b\sqrt{7}\)[/tex], we follow these steps:
1. Rationalize both fractions:
For the first fraction [tex]\(\frac{\sqrt{7}-1}{\sqrt{7}+1}\)[/tex]:
[tex]\[
\frac{\sqrt{7}-1}{\sqrt{7}+1} \times \frac{\sqrt{7}-1}{\sqrt{7}-1} = \frac{(\sqrt{7}-1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)}
\][/tex]
Simplify the numerator:
[tex]\[
(\sqrt{7}-1)^2 = 7 - 2\sqrt{7} + 1 = 8 - 2\sqrt{7}
\][/tex]
Simplify the denominator:
[tex]\[
(\sqrt{7}+1)(\sqrt{7}-1) = 7 - 1 = 6
\][/tex]
So,
[tex]\[
\frac{\sqrt{7}-1}{\sqrt{7}+1} = \frac{8 - 2\sqrt{7}}{6} = \frac{4}{3} - \frac{\sqrt{7}}{3}
\][/tex]
For the second fraction [tex]\(\frac{\sqrt{7}+1}{\sqrt{7}-1}\)[/tex]:
[tex]\[
\frac{\sqrt{7}+1}{\sqrt{7}-1} \times \frac{\sqrt{7}+1}{\sqrt{7}+1} = \frac{(\sqrt{7}+1)^2}{(\sqrt{7}-1)(\sqrt{7}+1)}
\][/tex]
Simplify the numerator:
[tex]\[
(\sqrt{7}+1)^2 = 7 + 2\sqrt{7} + 1 = 8 + 2\sqrt{7}
\][/tex]
Simplify the denominator:
[tex]\[
(\sqrt{7}-1)(\sqrt{7}+1) = 7 - 1 = 6
\][/tex]
So,
[tex]\[
\frac{\sqrt{7}+1}{\sqrt{7}-1} = \frac{8 + 2\sqrt{7}}{6} = \frac{4}{3} + \frac{\sqrt{7}}{3}
\][/tex]
2. Subtract the two fractions:
[tex]\[
\left( \frac{4}{3} - \frac{\sqrt{7}}{3} \right) - \left( \frac{4}{3} + \frac{\sqrt{7}}{3} \right)
\][/tex]
Combine them into one expression:
[tex]\[
\frac{4}{3} - \frac{\sqrt{7}}{3} - \frac{4}{3} - \frac{\sqrt{7}}{3}
\][/tex]
Simplify the terms:
[tex]\[
\left( \frac{4}{3} - \frac{4}{3} \right) - \left( \frac{\sqrt{7}}{3} + \frac{\sqrt{7}}{3} \right) = 0 - \frac{2\sqrt{7}}{3}
\][/tex]
3. Express in the form [tex]\(a + b\sqrt{7}\)[/tex]:
[tex]\[
0 - \frac{2\sqrt{7}}{3}
\][/tex]
Therefore,
[tex]\[
a = 0, \quad b = -\frac{2}{3}
\][/tex]
So the values are [tex]\(a = 0\)[/tex] and [tex]\(b = -\frac{2}{3}\)[/tex].
For the additional value asked for [tex]\(a\)[/tex] and 15:
[tex]\[
a = 0 \text{ and substituting 15, } a + 15 = 0 + 15 = 15
\][/tex]
Thus, the values required are [tex]\(a = 0\)[/tex] and [tex]\(15\)[/tex].
