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Size:  425.7 KBThe question at the bottom (number 4). I multiplied everything by (1-root2)(-root5-1) but this is wrong and I'm unsure why so please help
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    Not sure if there's a quicker way, but if you multiply top&bottom by (root5 + 1)(1 - root2), the denominator can be done by the difference of two squares, then just expand everything on top and simplify as much as you can.
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    You could just expand the top and bottom then cancel out. They haven't asked for answer in the form of a(root b) so you don't need to cancel out the bottom like I can see they have for question 3
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    (Original post by zayn008)
    You could just expand the top and bottom then cancel out. They haven't asked for answer in the form of a(root b) so you don't need to cancel out the bottom like I can see they have for question 3
    Expand the top and bottom then cancel out?

    You can't cancel anything in fractions until you have something, if not everything, factorised. Trying expanding \frac{a(a+b)}{b(a+b)} FIRST and tell me if you get anywhere when it comes to simplifying it.
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    (Original post by Lucofthewoods)
    The question at the bottom (number 4). I multiplied everything by (1-root2)(-root5-1) but this is wrong and I'm unsure why so please help
    Almost correct. The mistake you made here is that you've multiplied by (-\sqrt5-1) when you should've multiplied by (\sqrt5+1) instead.

    When you are rationalising the denominator, the trick you are using when it comes to choosing what to multiply is the difference of two squares, whereby: a^2-b^2=(a+b)(a-b) where as you can see the left side is the difference of two squared numbes (ie squares), and you take one away from the other (ie the difference)

    So when you have (\sqrt5-1) you can say that a=\sqrt5 and b=1. Applying the equation gives:
    a^2-b^2=(\sqrt5+1)(\sqrt5-1) therefore you're getting rid off any roots.

    Can you see why your chosen (-\sqrt5-1) doesn't work when you apply it here?
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    (Original post by zayn008)
    My point was he didn't need to multiply the top by the adjusted bottom, factorising is needed to cancel out
    Yes but there are no same factors on the numerator and denominator to cancel out beforehand. The denominator is irrational therefore it is not a 'simplified' form, which is what he needs to do; rationalise it, and the process of rationalising the denominator requires him to multiply the top and and bottom by the same factors. He doesn't even need to expand the numerator as long as he rationalises the denominator.

    I don't see your point.
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    (Original post by RDKGames)
    Yes but there are no same factors on the numerator and denominator to cancel out beforehand. The denominator is irrational therefore it is not a 'simplified' form, which is what he needs to do; rationalise it, and the process of rationalising the denominator requires him to multiply the top and and bottom by the same factors. He doesn't even need to expand the numerator as long as he rationalises the denominator.

    I don't see your point.
    After trying it on paper I also don't see my point think I'll just steer clear of the maths forums for a while
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    (Original post by RDKGames)
    Almost correct. The mistake you made here is that you've multiplied by (-\sqrt5-1) when you should've multiplied by (\sqrt5+1) instead.

    When you are rationalising the denominator, the trick you are using when it comes to choosing what to multiply is the difference of two squares, whereby: a^2-b^2=(a+b)(a-b) where as you can see the left side is the difference of two squared numbes (ie squares), and you take one away from the other (ie the difference)

    So when you have (\sqrt5-1) you can say that a=\sqrt5 and b=1. Applying the equation gives:
    a^2-b^2=(\sqrt5+1)(\sqrt5-1) therefore you're getting rid off any roots.

    Can you see why your chosen (-\sqrt5-1) doesn't work when you apply it here?
    Thanks!
 
 
 
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