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imqwerty

  • one year ago

real math ( ͡~ ͜ʖ ͡°)

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  1. imqwerty
    • one year ago
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  2. anonymous
    • one year ago
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    D

  3. imqwerty
    • one year ago
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    correct @mukushla :)

  4. anonymous
    • one year ago
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    nice question

  5. anonymous
    • one year ago
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    we need to find out that \(b^2=ac\)

  6. imqwerty
    • one year ago
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    yes correct

  7. ganeshie8
    • one year ago
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    *

  8. Zale101
    • one year ago
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    Multiply by the conjugate to both the numerator and the denominator. Foil the numerator and the denominator and you'll get the answer.

  9. imqwerty
    • one year ago
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    well done @Zale

  10. imqwerty
    • one year ago
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    i forgot to write 1 step in the solution so heres the new solution

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  11. Sachintha
    • one year ago
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    @perl Can you explain it?

  12. perl
    • one year ago
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    yes

  13. Sachintha
    • one year ago
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    What I didn't get is how \[b^2=ac\]

  14. perl
    • one year ago
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    We want to force the entire expression to be rational, given a,b,c are integers. You have two possibilities, b^2 -ac = 0 or sqrt(2) b^2 -ac cannot equal to sqrt(2) , because of the denominator. So you are left with only b^2 -ac = 0

  15. Sachintha
    • one year ago
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    I still didn't get why \[b^2-ac=0\] Is it taken so that \[\sqrt2\] can be removed?

  16. perl
    • one year ago
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    otherwise you are left with a radical 2 in the numerator, so the entire expression is irrational

  17. perl
    • one year ago
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    yes to remove the square root 2

  18. perl
    • one year ago
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    do you see why b^2 -ac cannot equal to square root 2 ?

  19. perl
    • one year ago
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    edit** \[ \Large{ \frac{a\sqrt{2}+b}{b\sqrt{2} + c }\cdot \frac{b\sqrt{2}-c}{b\sqrt{2} -c } \\~\\= \frac{2ab + b^2\sqrt 2 - ac \sqrt 2 - bc}{2b^2 - c^2 } \\~\\= \frac{2ab + \sqrt 2 (b^2 - ac) - bc}{2b^2 - c^2 } }\]

  20. Sachintha
    • one year ago
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    Ohh I forgot all about sets of real numbers including rational and irrational .. :D But what does number are a,ar and ar^2 mean?

  21. perl
    • one year ago
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    that part is a bit unclear. we can try to figure it out

  22. Sachintha
    • one year ago
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    I think, \[c=ar^2\]

  23. Sachintha
    • one year ago
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    r is not defined here :(

  24. Sachintha
    • one year ago
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    So b/a is taken as r ?

  25. Sachintha
    • one year ago
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    Is it possible to equalize it like that?

  26. Sachintha
    • one year ago
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    Well it does work too. :)

  27. Sachintha
    • one year ago
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    So we equalize it to another value for the ease of calculations? Sorry for being so dumb.

  28. perl
    • one year ago
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    these are good questions :)

  29. perl
    • one year ago
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    since b^2 = ac , divide both sides by a $$ \Large c = \frac{b^2}{ a} = a \cdot \frac{b^2}{ a^2 } = a \cdot \left(\frac b a \right)^2 $$ Let r = (b/a) Now we have a= a ar = a (b/a) = b ar^2= a (b/a)^2 = c

  30. Sachintha
    • one year ago
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    I got it. :D Was confused because of r.

  31. perl
    • one year ago
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    Yes it looked like it came from out of no where :D

  32. perl
    • one year ago
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    In part d) it appears they completed long division.

  33. Sachintha
    • one year ago
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    Did long division for myself. :D (whew) It factors to \[(r^2-r+1)(r^2+r+1)\]

  34. perl
    • one year ago
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    i see, great job

  35. Sachintha
    • one year ago
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    Thank you so much for taking your time and explaining me. :)

  36. perl
    • one year ago
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    why can't b^2 -ac = sqrt(2) ?

  37. Sachintha
    • one year ago
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    Because we still get a irrational number at the denominator. I am correct?

  38. perl
    • one year ago
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    yes can you show why

  39. Sachintha
    • one year ago
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    Because \[b^2=\sqrt2+ac\]

  40. perl
    • one year ago
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    right

  41. Sachintha
    • one year ago
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    Am I wrong?

  42. Sachintha
    • one year ago
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    Maybe my answer is not adequate. :D

  43. perl
    • one year ago
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    if we plug in the denominator we get 2b^2 -c^2 = 2 ( √2 + ac) - c^2 = 2√2 + 2ac - c^2 , which is irrational

  44. perl
    • one year ago
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    because of the 2 sqrt(2)

  45. Sachintha
    • one year ago
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    yep :)

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