Question

who can Slove this question

Answer 1

Heya !!

Identity used :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$



$\frac{4 \sqrt{3} - 6 \sqrt{5} }{7 \sqrt{5} + 3 \sqrt{3} } = a + b \sqrt{15}$

L.H.S.

On rationalizing the denominator we get,

$= \frac{4 \sqrt{3} - 6 \sqrt{5} }{7 \sqrt{5} + 3 \sqrt{3} } \times \frac{7 \sqrt{5} - 3 \sqrt{3} }{7 \sqrt{5} - 3 \sqrt{3} } \\ \\ = \frac{4 \sqrt{3}(7 \sqrt{5} - 3 \sqrt{3} ) - 6 \sqrt{5} (7 \sqrt{5} - 3 \sqrt{3}) }{ {(7 \sqrt{5}) }^{2} - {(3 \sqrt{3}) }^{2} } \\ \\ = \frac{28 \sqrt{15} - 36 - 210 + 18 \sqrt{15} }{245 - 27} \\ \\ = \frac{ - 246 + 46 \sqrt{15} }{218} \\ \\ = \frac{ - 123 + 23 \sqrt{15} }{109}$

On comparing both the sides we get

$\frac{ - 123 + 23 \sqrt{15} }{109} = a + b \sqrt{15} \\ \\ \frac{ - 123}{109} + \frac{23 }{109} \sqrt{15} = a + b \sqrt{15} \\ \\ a = \frac{ - 123}{109} \: : \: b = \frac{23}{109}$

Hope this helps ☺