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How do you find the rationalising factor of root2 + root7 - root10?

Mathematics
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|dw:1344951900463:dw|
multiply that by its conjugate @pratu043 what do you get?
By conjugate you mean: |dw:1344952032255:dw|

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Other answers:

\[\large{(\sqrt{2}+\sqrt{7}-\sqrt{10})(\sqrt{2}+\sqrt{7}+\sqrt{10})=?}\]
|dw:1344952080072:dw|
@pratu043 what do you get by multiplying that.. ?
How do you find the conjugate?
I only know how to find for binomial surds.
@pratu043 let the expression given be like this: \[\large{(\sqrt{2}+\sqrt{7})-(\sqrt{10})}\]
I meant to say that: a+b-c = (a+b)-(c) right?
yes
|dw:1344952309566:dw|
|dw:1344952351465:dw| What do you do with - root10?
|dw:1344952415376:dw| sorry
ok leave that for once... what will be the conjugate of a+b ?
a-b
right so let a = c+d and b = b now can u tell me what will be the new conjugate? : (c+d)+(b) conjugate = ?
(c+d) - (b)
right so what will be the conjugate of : \[\large{(\sqrt{2}+\sqrt{7})-(\sqrt{10})}\]
|dw:1344952630671:dw| ok, I got the conjugate part.
right now : multiply this conjugate with original expression
|dw:1344952686106:dw| right?
good so we have: \[\large{2\sqrt{14}-1}\] now:(√2 + √7 - √10)[(√2 + √7 + √10) (2√14 + 1)] do this
|dw:1344952926110:dw|
But why should we do this?
it is a step towards the answer
also try to check that out pratu that u did: can that be more simplified?
@pratu043 you there?
sorry yes i am there.
do you mean that we should try and convert the irrationals in that expression to mixed surds?
yes
I think I can handle it on my own now, thanks!!
(√2 + √7 - √10)[(√2 + √7 + √10) (2√14 + 1)] \[\large{(\sqrt{2}+\sqrt{7}-\sqrt{10})(\sqrt{2}+\sqrt{7}+\sqrt{10})(2\sqrt{14}+1)}\] \[\large{(9+2\sqrt{14}-10)(2\sqrt{14}+1)=(2\sqrt{14}-1)(2\sqrt{14}+1)=56-1=55}\]
nice to hear best of luck

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