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2 years ago

7+3√5/3+√5 - 7-3√5/3-√5 = a + b√5

Mathematics

2 answers:

Olenka [21]2 years ago

8 0

$\begin{gathered} \\\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\ \end{gathered}$

$\begin{gathered} \\ \frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \: \sqrt{5} = a + \sqrt{5} \: b\\ \end{gathered}$

we can also write it as ;

$\begin{gathered} \\ \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\ \end{gathered}$

★ Henceforth, the value of a and b are :

→ a = 0

→ b = 1

daser333 [38]2 years ago

6 0

<h3>Given:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

The value of a and b

<h3>Solution:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

$\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: \cancel{4 } \sqrt{5} \: \: }{ \: \: \: \: \cancel{4 }\: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies \: \sqrt{5} = a + \sqrt{5} \: b\\$

we can also write it as ;

$\\ \sf \implies \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\$

★<u> </u><u>Henceforth, the value of a and b are</u> :

→ a = 0

→ b = 1

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