"Simplify 5√2 / 3-√2. Give your answer in the form b+c√2 / d"

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elliepollylu
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Please help me learn how to answer this!
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the bear
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you multiply top and bottom by 3 + √2
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troubadour.
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(Original post by elliepollylu)
Please help me learn how to answer this!
Are you looking for other people to do all of your homework? :/

As the bear said, multiply top and bottom by 3 + √2 and then simplify.
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S27
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Can you use brackets just so i know exactly what you mean.
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Muttley79
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The reason you multiply by 3 + √2 is so that the denominator becomes the difference of two squares and so the roots disappear. The process is called rationalising the denominator ... ie remove the irrational surds.
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elliepollylu
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(Original post by the bear)
you multiply top and bottom by 3 + √2
Can I ask where you got 3+√2 from?
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elliepollylu
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(Original post by Hydeman)
Are you looking for other people to do all of your homework? :/

As the bear said, multiply top and bottom by 3 + √2 and then simplify.
I'm really not trying to get other people to do my homework for me, I'm just trying to understand how to answer the questions :/
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troubadour.
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(Original post by elliepollylu)
Can I ask where you got 3+√2 from?
See Muttley79's answer.

(Original post by Muttley79)
The reason you multipy by the complex conjugate is so that the denominator becomes the difference of two squares and so the roots disappear. The process is called rationalising the denominator ... ie remove the irrational surds.
It's basically multiplying both by the denominator but with the signs reversed.
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Muttley79
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(Original post by elliepollylu)
Can I ask where you got 3+√2 from?
Did you read my post?

You multiply the numerator and denominator by the complex conjugate.
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elliepollylu
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(Original post by Muttley79)
Did you read my post?

You multiply the numerator and denominator by the complex conjugate.
Thank you very much
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elliepollylu
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(Original post by Hydeman)
See Muttley79's answer.



It's basically multiplying both by the denominator but with the signs reversed.
Thank you
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Muttley79
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(Original post by elliepollylu)
Thank you very much
Post your working if you don't get the answer in the book.

It's a really neat technique much loved by the examiners
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hideNfreak
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I've done it and showed my working out in two steps in the attached file. sorry I've done step 2 wrong let me do it again i've probably confused you more
Attached files
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hideNfreak
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You also need to know your surd rules thoroughly to do this question
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Rabadon
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It's called rationalizing the denominator if you wanted to read up about it
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username638250
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(Original post by Muttley79)
The reason you multipy by the complex conjugate is so that the denominator becomes the difference of two squares and so the roots disappear. The process is called rationalising the denominator ... ie remove the irrational surds.
(Original post by elliepollylu)
Thank you
You're not multiplying by the complex conjugate; there is no 'i' involved here. You're simply rationalising the denominator. The complex conjugation of a number, say, a+bi, is a-bi, so if we had a fraction like (3+5i)/(2+i), then yes, we would multiply top and bottom by 2-i, the complex conjugate of 2+i.
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Muttley79
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(Original post by yl95)
You're not multiplying by the complex conjugate; there is no 'i' involved here. You're simply rationalising the denominator. The complex conjugation of a number, say, a+bi, is a-bi, so if we had a fraction like (3+5i)/(2+i), then yes, we would multiply top and bottom by 2-i, the complex conjugate of 2+i.
Well it is effectively the same idea - just no i component
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username638250
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(Original post by Muttley79)
Well it is effectively the same idea - just no i component
Yes, but the wrong terminology, just in case the student decided to use 'complex conjugate' to describe this.
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hideNfreak
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Here's the corrected step 2
Attached files
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Dingooose
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(Original post by yl95)
Yes, but the wrong terminology, just in case the student decided to use 'complex conjugate' to describe this.
You're right. The correct term in this case is conjugate, not complex conjugate.
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