Question

Ples find the a and b the first term is a and the second term is b

Answer 1

Hello friends!!

Here is your answer :

$a + b \sqrt{5} = \frac{1 + 2 \sqrt{5} }{1 - 2 \sqrt{5} } + \frac{1 - 2 \sqrt{5} }{1 + 2 \sqrt{5} }$


First we have to rationalise the denominator..

$a + b \sqrt{5} = \frac{1 + 2 \sqrt{5} }{1 - 2 \sqrt{5} } \times \frac{1 + 2 \sqrt{5} }{1 + 2 \sqrt{5} } + \frac{1 - 2 \sqrt{5} }{1 + 2 \sqrt{5} } \times \frac{1 - 2 \sqrt{5} }{1 - 2 \sqrt{5} }$


$a + b \sqrt{5} = \frac{(1 + 2 \sqrt{5}) (1 + 2 \sqrt{5}) }{(1 - 2 \sqrt{5})(1 + 2 \sqrt{5} ) } + \frac{(1 - 2 \sqrt{5})(1 - 2 \sqrt{5} )}{(1 + 2 \sqrt{5} )(1 - 2 \sqrt{5)} }$



Using identity :

( a + b)² = a² + b² + 2ab

( a + b) ( a - b) = a² - b²


$a + b \sqrt{5} = \frac{ {(1)}^{2} + {(2 \sqrt{5}) }^{2} + 2(1)(2 \sqrt{5}) }{ {(1)}^{2} - {(2 \sqrt{5} )}^{2} } + \frac{ {(1)}^{2} + {(2 \sqrt{5} )}^{2} - 2(1)(2 \sqrt{5}) }{ {(1)}^{2} - {(2 \sqrt{5} )}^{2} }$



$a + b \sqrt{5} = \frac{1 + 20+ 4 \sqrt{5} }{1 - 20} + \frac{1 + 20 - 4 \sqrt{5} }{1 - 20}$



$a + b \sqrt{5} = \frac{21 + 4 \sqrt{5} }{ - 19} + \frac{21 - 4 \sqrt{5} }{ - 19}$



$a + b \sqrt{5} = \frac{21 + 4 \sqrt{5} + 21 - 4 \sqrt{5} }{ - 19}$



$a + b \sqrt{5} = \frac{21 + 21}{ - 19}$


$a + b \sqrt{5} = \frac{ - 42}{19}$



Comparing it, We get the value of a and b.



$a = \frac{ - 42}{19}$


$b = 0$



Hope it helps you..