How do you answer this?

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

Reply 1

JN17

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.

Reply 2

3121

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

Reply 3

RDKGames

Study Forum Helper

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.

(edited 7 years ago)

Reply 4

RDKGames

Study Forum Helper

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?

(edited 7 years ago)

Reply 5

RDKGames

Study Forum Helper

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.

Reply 6

3121

Original post by RDKGames

After trying it on paper I also don't see my point :tongue:

think I'll just steer clear of the maths forums for a while

Reply 7

Lucofthewoods

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