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OCR C1 Rationalising

Rationalise the Denominator...

3-Root 3

Isnt it 3-Root3 x 3+Root 3? and that would equal 9-3root3 + 3 +3root3? simplified to 12?
A the back of the book it says root 7 + root 3?

How do you work this out?
To rationalise a denominator, you need a fraction ...
Reply 2
Avatar for ieatcheeseyo
ieatcheeseyo
OP
12
Original post by Mr M
To rationalise a denominator, you need a fraction ...


4
3 - root 3
Original post by ieatcheeseyo
4
3 - root 3


Ok the answer to that question is not the answer you say is in the back of the book.
Reply 4
Avatar for TenOfThem
TenOfThem
18
the 12 is wrong

but then again so is the answer you have given
Reply 5
Avatar for ieatcheeseyo
ieatcheeseyo
OP
12
Original post by Mr M
Ok the answer to that question is not the answer you say is in the back of the book.



Original post by TenOfThem
the 12 is wrong

but then again so is the answer you have given


dunno and dont have time to upload pictures onto TSR
basically the answer at the back is wrong AND my method is wrong... Thanks for your help
Reply 6
Avatar for TenOfThem
TenOfThem
18
when you did +srqt3 x -sqrt3 you made an error
Reply 7
Avatar for ieatcheeseyo
ieatcheeseyo
OP
12
Original post by TenOfThem
when you did +srqt3 x -sqrt3 you made an error


alright whats the asnwer for it?
dont see how it will make it root7 plus root 3
Reply 8
Avatar for TenOfThem
TenOfThem
18
it won't since the answer in the book is wrong

but you said 3∗−3=+3\sqrt{3} * -\sqrt{3} = +3
Original post by ieatcheeseyo
alright whats the asnwer for it?
dont see how it will make it root7 plus root 3


We told you the answer in the book was wrong.

I presume you own a Natural Display calculator - type it in and see for yourself.
Reply 10
Avatar for Kravez
Kravez
1
43−3\dfrac{4}{3 - \sqrt3} x 3+33+3\dfrac{3 + \sqrt3}{3 + \sqrt3}

12+439+33−33−3\dfrac {12 + 4\sqrt3}{9 + 3\sqrt3 - 3\sqrt3 - 3}

12+436\dfrac {12 + 4\sqrt3}{6}

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