Question

Find the Value of α and b if :
$\frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{2 - \sqrt{5} } = a + b \sqrt{5}$
​

Answer 1

Given,

$\frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{ 2- \sqrt{5} } = a + b \sqrt{5}$

To Find,

• The value of α and b

Solution,

$::\implies \frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{2 - \sqrt{5} } = a + b \sqrt{5} \\ \\ ::\implies \frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{2 - \sqrt{5} } \times \frac{2 + \sqrt{5} }{2 + \sqrt{5} } = a + b \sqrt{5} \\ \\::\implies \frac{(7 + 3 \sqrt{5} )(2 - \sqrt{5}) }{ {(2)}^{2} - {( \sqrt{5} )}^{2} } - \frac{(7 - 3 \sqrt{5} )(2 + \sqrt{5}) }{ {(2)}^{2} - {( \sqrt{5} )}^{2} } = a + b \sqrt{5} \\ \\::\implies \frac{7(2 - \sqrt{5}) + 3 \sqrt{5}(2 - \sqrt{5}) }{ 4 - 5} - \frac{7(2 + \sqrt{5}) - 3 \sqrt{5}(2 + \sqrt{5}) }{ 4 - 5} = a + b \sqrt{5} \\ \\ ::\implies \frac{14 - 7 \sqrt{5} + 6 \sqrt{5} - 15 }{ - 1} - \frac{14 + 7 \sqrt{5} - 6 \sqrt{5} - 15}{4 - 5} = a + b \sqrt{5} \\ \\ ::\implies \frac{ - 1 - \sqrt{5} }{ - 1} - \frac{ - 1 + \sqrt{5} }{ - 1} = a + b \sqrt{5} \\ \\::\implies \frac{- 1 \times (1 + \sqrt{5} )}{ - 1} - (- 1( - 1 + \sqrt{5} ) = a + b \sqrt{5} \\ \\ ::\implies 1 + \sqrt{5} - 1 + \sqrt{5} = a + b \sqrt{5} \\ \\ ::\implies 0 + 2 \sqrt{5} = a + b \sqrt{5}$

Required Answer,

• α = 0
• b = 2

Answer 2

$\huge \bf \color{blue}Given$

$\begin{gathered} \frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{ 2- \sqrt{5} } = a + b \sqrt{5} \\ \\ \end{gathered} \\ </p><p>$

$\huge \bf \color{green}{To \: \: Find}$

The value of α and b

$\huge \bf \color{red}{Solution}$

$\begin{gathered}::\implies \frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{2 - \sqrt{5} } = a + b \sqrt{5} \\ \\ ::\implies \frac{7 + 3 \sqrt{5} }{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{2 - \sqrt{5} } \times \frac{2 + \sqrt{5} }{2 + \sqrt{5} } = a + b \sqrt{5} \\ \\::\implies \frac{(7 + 3 \sqrt{5} )(2 - \sqrt{5}) }{ {(2)}^{2} - {( \sqrt{5} )}^{2} } - \frac{(7 - 3 \sqrt{5} )(2 + \sqrt{5}) }{ {(2)}^{2} - {( \sqrt{5} )}^{2} } = a + b \sqrt{5} \\ \\::\implies \frac{7(2 - \sqrt{5}) + 3 \sqrt{5}(2 - \sqrt{5}) }{ 4 - 5} - \frac{7(2 + \sqrt{5}) - 3 \sqrt{5}(2 + \sqrt{5}) }{ 4 - 5} = a + b \sqrt{5} \\ \\ ::\implies \frac{14 - 7 \sqrt{5} + 6 \sqrt{5} - 15 }{ - 1} - \frac{14 + 7 \sqrt{5} - 6 \sqrt{5} - 15}{4 - 5} = a + b \sqrt{5} \\ \\ ::\implies \frac{ - 1 - \sqrt{5} }{ - 1} - \frac{ - 1 + \sqrt{5} }{ - 1} = a + b \sqrt{5} \\ \\::\implies \frac{- 1 \times (1 + \sqrt{5} )}{ - 1} - (- 1( - 1 + \sqrt{5} ) = a + b \sqrt{5} \\ \\ ::\implies 1 + \sqrt{5} - 1 + \sqrt{5} = a + b \sqrt{5} \\ \\ ::\implies 0 + 2 \sqrt{5} = a + b \sqrt{5} \end{gathered}$

$\huge \bf \color{purple}{Required \: \: Answer,}$

α = 0

α = 0b = 2

$\huge \bf \color{orange}{hope \: it's \: helpful \: to \: you} \\ \\ \color {pink}@TheGoodGirl$