Question

Find a and b if 2-√5 / 2+3√5 = √5 a+b​

Answer 1

$\small\star\underline\bold\red{Given-}$

$\frac{(2 - \sqrt{5)} }{(2 + 3 \sqrt{5)} } = \sqrt{5} a + b$

$\small\star\underline\bold\red{To\: find-}$

• Value of a and b.

$\small\star\underline\bold\red{Solution-}$

• Rationalising the LHS , we get ,

$\frac{(2 - \sqrt{5)} }{(2 + 3 \sqrt{5)} } \times \frac{(2 - 3\sqrt{5)} }{(2 - 3 \sqrt{5)} }$

$= \frac{(2 - \sqrt{5})(2 - 3 \sqrt{5} )}{ {(2)}^{2} - {(3 \sqrt{5}) }^{2} }$

$= \frac{4 - 6 \sqrt{5} - 2 \sqrt{5} + 15 }{4 - 45}$

$= \frac{19 - 8 \sqrt{5} }{ - 41}$

$= \frac{8 \sqrt{5} }{41} - \frac{19}{41}$

Hence,

• a = $\frac{8}{41}$

• b = $\frac{-19}{41}$

Answer 2

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