evaluate (a) (3-√5)(6+√2)
Answers
Answer:
Identity used :
(x + y)(x - y) = {x}^{2} - {y}^{2}(x+y)(x−y)=x
2
−y
2
\begin{gathered} \frac{2 \sqrt{6} - \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6} } \\ \end{gathered}
3
5
−2
6
2
6
−
5
On rationalizing the denominator we get,
\begin{gathered} = \frac{2 \sqrt{6} - \sqrt{5} }{3 \sqrt{5} - 2 \sqrt{6} } \times \frac{3 \sqrt{5} + 2 \sqrt{6} }{3 \sqrt{5} + 2 \sqrt{6} } \\ \\ = \frac{2 \sqrt{6}(3 \sqrt{5} + 2 \sqrt{6} ) - \sqrt{5}(3 \sqrt{5} + 2 \sqrt{6} )}{ {(3 \sqrt{5} )}^{2} - {(2 \sqrt{6}) }^{2} } \\ \\ = \frac{6 \sqrt{30} + 24 - 15 - 2 \sqrt{30} }{45 - 24} \\ \\ = \frac{9 + 4 \sqrt{30} }{21} \end{gathered}
=
3
5
−2
6
2
6
−
5
×
3
5
+2
6
3
5
+2
6
=
(3
5
)
2
−(2
6
)
2
2
6
(3
5
+2
6
)−
5
(3
5
+2
6
)
=
45−24
6
30
+24−15−2
30
=
21
9+4
30
Hope this helps ☺