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pratu043 Group Title

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

  • one year ago
  • one year ago

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  1. pratu043 Group Title
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    |dw:1344951900463:dw|

    • one year ago
  2. mathslover Group Title
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    multiply that by its conjugate @pratu043 what do you get?

    • one year ago
  3. pratu043 Group Title
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    By conjugate you mean: |dw:1344952032255:dw|

    • one year ago
  4. mathslover Group Title
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    \[\large{(\sqrt{2}+\sqrt{7}-\sqrt{10})(\sqrt{2}+\sqrt{7}+\sqrt{10})=?}\]

    • one year ago
  5. mathslover Group Title
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    |dw:1344952080072:dw|

    • one year ago
  6. mathslover Group Title
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    @pratu043 what do you get by multiplying that.. ?

    • one year ago
  7. pratu043 Group Title
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    How do you find the conjugate?

    • one year ago
  8. pratu043 Group Title
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    I only know how to find for binomial surds.

    • one year ago
  9. mathslover Group Title
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    @pratu043 let the expression given be like this: \[\large{(\sqrt{2}+\sqrt{7})-(\sqrt{10})}\]

    • one year ago
  10. mathslover Group Title
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    I meant to say that: a+b-c = (a+b)-(c) right?

    • one year ago
  11. pratu043 Group Title
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    yes

    • one year ago
  12. mathslover Group Title
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    |dw:1344952309566:dw|

    • one year ago
  13. pratu043 Group Title
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    |dw:1344952351465:dw| What do you do with - root10?

    • one year ago
  14. pratu043 Group Title
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    |dw:1344952415376:dw| sorry

    • one year ago
  15. mathslover Group Title
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    ok leave that for once... what will be the conjugate of a+b ?

    • one year ago
  16. pratu043 Group Title
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    a-b

    • one year ago
  17. mathslover Group Title
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    right so let a = c+d and b = b now can u tell me what will be the new conjugate? : (c+d)+(b) conjugate = ?

    • one year ago
  18. pratu043 Group Title
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    (c+d) - (b)

    • one year ago
  19. mathslover Group Title
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    right so what will be the conjugate of : \[\large{(\sqrt{2}+\sqrt{7})-(\sqrt{10})}\]

    • one year ago
  20. pratu043 Group Title
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    |dw:1344952630671:dw| ok, I got the conjugate part.

    • one year ago
  21. mathslover Group Title
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    right now : multiply this conjugate with original expression

    • one year ago
  22. pratu043 Group Title
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    |dw:1344952686106:dw| right?

    • one year ago
  23. mathslover Group Title
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    good so we have: \[\large{2\sqrt{14}-1}\] now:(√2 + √7 - √10)[(√2 + √7 + √10) (2√14 + 1)] do this

    • one year ago
  24. pratu043 Group Title
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    |dw:1344952926110:dw|

    • one year ago
  25. pratu043 Group Title
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    But why should we do this?

    • one year ago
  26. mathslover Group Title
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    it is a step towards the answer

    • one year ago
  27. mathslover Group Title
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    also try to check that out pratu that u did: can that be more simplified?

    • one year ago
  28. mathslover Group Title
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    @pratu043 you there?

    • one year ago
  29. pratu043 Group Title
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    sorry yes i am there.

    • one year ago
  30. pratu043 Group Title
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    do you mean that we should try and convert the irrationals in that expression to mixed surds?

    • one year ago
  31. mathslover Group Title
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    yes

    • one year ago
  32. pratu043 Group Title
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    I think I can handle it on my own now, thanks!!

    • one year ago
  33. mathslover Group Title
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    (√2 + √7 - √10)[(√2 + √7 + √10) (2√14 + 1)] \[\large{(\sqrt{2}+\sqrt{7}-\sqrt{10})(\sqrt{2}+\sqrt{7}+\sqrt{10})(2\sqrt{14}+1)}\] \[\large{(9+2\sqrt{14}-10)(2\sqrt{14}+1)=(2\sqrt{14}-1)(2\sqrt{14}+1)=56-1=55}\]

    • one year ago
  34. mathslover Group Title
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    nice to hear best of luck

    • one year ago
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