Question

plz help to finde this answer<br />$\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } - a + b \sqrt{15}$<br />


find a and b

Answer 1

$\sf{\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } = a + b \sqrt{15} }$

rationalise the denominator

$\sf\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15}$

$\sf\frac{{ (\sqrt{5 } - \sqrt{3})}^{2} }{{( \sqrt{5})}^{2} - {(\sqrt{3}) }^{2}} = a + b \sqrt{15}$

$\sf\frac{{ {(\sqrt{5 } )}^{2} + (\sqrt{3})}^{2} - 2( \sqrt{5} )( \sqrt{3} )}{{5} - 3} = a + b \sqrt{15}$

$\sf\frac{5 + 3 - 2\sqrt{15}}{5 - 3} = a + b \sqrt{15}$

$\sf\frac{8 - 2\sqrt{15}}{2} = a + b \sqrt{15}$

$\sf{ \frac{2(4 + \sqrt{15} )}{2} = a + b \sqrt{15} }$

$\sf{4 + \sqrt{15} = a + b \sqrt{15} }$

on comparing both sides

$\sf{a = 4} \\ \sf{b = 1}$

Answer 2

Answer:

Step-by-step explanation:

$\sf{\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } = a + b \sqrt{15} }$

rationalise the denominator

$\sf\frac{ \sqrt{5 } - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15}$

$\sf\frac{{ (\sqrt{5 } - \sqrt{3})}^{2} }{{( \sqrt{5})}^{2} - {(\sqrt{3}) }^{2}} = a + b \sqrt{15}$

$\sf\frac{{ {(\sqrt{5 } )}^{2} + (\sqrt{3})}^{2} - 2( \sqrt{5} )( \sqrt{3} )}{{5} - 3} = a + b \sqrt{15}$

[tex] \sf\frac{5 + 3 - 2\sqrt{15}}{5 - 3} = a + b \sqrt{15}

\sf\frac{8 - 2\sqrt{15}}{2} = a + b \sqrt{15}

\sf{ \frac{2(4 + \sqrt{15} )}{2} = a + b \sqrt{15} }

\sf{4 + \sqrt{15} = a + b \sqrt{15} }

on comparing both sides

\sf{a = 4} \\ \sf{b = 1}