the adjoining figure shows two circle with the same centre . the radius of the larger scale is 10cm and the radius
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Answer:
Given−
\frac{(2 - \sqrt{5)} }{(2 + 3 \sqrt{5)} } = \sqrt{5} a + b
(2+3
5)
(2−
5)
=
5
a+b
\small\star\underline\bold\red{To\: find-}⋆
Tofind−
Value of a and b.
\small\star\underline\bold\red{Solution-}⋆
Solution−
Rationalising the LHS , we get ,
\frac{(2 - \sqrt{5)} }{(2 + 3 \sqrt{5)} } \times \frac{(2 - 3\sqrt{5)} }{(2 - 3 \sqrt{5)} }
(2+3
5)
(2−
5)
×
(2−3
5)
(2−3
5)
= \frac{(2 - \sqrt{5})(2 - 3 \sqrt{5} )}{ {(2)}^{2} - {(3 \sqrt{5}) }^{2} } =
(2)
2
−(3
5
)
2
(2−
5
)(2−3
5
)
= \frac{4 - 6 \sqrt{5} - 2 \sqrt{5} + 15 }{4 - 45} =
4−45
4−6
5
−2
5
+15
= \frac{19 - 8 \sqrt{5} }{ - 41} =
−41
19−8
5
= \frac{8 \sqrt{5} }{41} - \frac{19}{41} =
41
8
5
−
41
19
Hence,
a = \frac{8}{41}
41
8
b = \frac{-19}{41}
41
−19
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