AS Maths stuck rationalising these fractions

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swaus
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a)

3 + 4 √5 / 2-√5

3 + 4 √5 / 2-√5 * 2-√5 / 2-√5

= (3+4√5)(2-√5) / (2-√5)(2-√5)

= -20+6+8√5-3√5 / 4-4√5+5

=-14+5√5 / 9-4√5

Tried rationalising again, but this just fed me through a loop. Also have another similar question like this, but maybe I'm just using the wrong method?
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RDKGames
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(Original post by swaus)
a)

3 + 4 √5 / 2

3 + 4 √5 / 2 * 2-√5 / 2-√5

= (3+4√5)(2-√5) / (2-√5)(2-√5)

= -20+6+8√5-3√5 / 4-4√5+5

=-14+5√5 / 9-4√5

Tried rationalising again, but this just fed me through a loop. Also have another similar question like this, but maybe I'm just using the wrong method?
What are you actually trying to do??

You cannot 'rationalise' an expression when it's not.

Since \dfrac{3+4\sqrt{5}}{2} is irrational, it stays that way.
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swaus
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(Original post by RDKGames)
What are you actually trying to do??

You cannot 'rationalise' an expression when it's not.

Since \dfrac{3+4\sqrt{5}}{2} is irrational, it stays that way.
My mistake, copied out the denominator wrong for the first two it's fixed now -- rather than over two, it should be:

3+4√5 / 2-√5
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mrbubbles123
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(Original post by swaus)
My mistake, copied out the denominator wrong for the first two it's fixed now -- rather than over two, it should be:

3+4√5 / 2-√5
to rationalise the denominator, you multiply top and bottom by the conjugate i.e. 2 + root 5, not 2 - root 5. You want a difference of two squares in the bottom to give you the rational denominator.
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RDKGames
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(Original post by swaus)
My mistake, copied out the denominator wrong for the first two it's fixed now -- rather than over two, it should be:

3+4√5 / 2-√5
You need to notice that (2-\sqrt{5})(2+\sqrt{5}) gives something rational.

So you need to multiply by \dfrac{2+\sqrt{5}}{2+\sqrt{5}} instead to rationalise the denominator.
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swaus
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Oh, right, thanks I see now! Thanks for the help all.
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