Question

how to cancel 2 root 5 square minus 3 root 3 square​

Answer 1

Answer:

Given :

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

Let's simplify L.H.S by rationalising it's denominator.

\frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } \\ \\ \frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } \times \frac{2 \sqrt{5} + 3 \sqrt{3} }{2 \sqrt{5} + 3 \sqrt{3} } \\ \\ \frac{ \sqrt{5}(2 \sqrt{5} + 3 \sqrt{3} ) + \sqrt{3} (2 \sqrt{5} + 3 \sqrt{3} )}{(2 \sqrt{5}) {}^{2} - (3 \sqrt{3}) {}^{2} } \\ \\ \frac{10 + 3 \sqrt{15} + 2 \sqrt{15} + 9 }{20 - 27} \\ \\ \frac{19 + 5 \sqrt{15} }{ - 7}

Now,

\frac{ - 19}{7} + \frac{5 }{7} \sqrt{15}

On comparing the result of L.H.S from R.H.S , we get ;

\bold{a = \frac{ - 19}{7} \: \: and \: \: b = \frac{5}{7} }

Answer 2

Step-by-step explanation:

hope it helps you dear...