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mayankdevnani Group Title

Please Help me !! Its simple question though but i don't know how to solve it ??

  • one month ago
  • one month ago

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  1. mayankdevnani Group Title
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    How to write \(\large \bf 6x^2+5xy-4y^2=0\) in another form or equation?

    • one month ago
  2. Mokeira Group Title
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    basically, just factor out

    • one month ago
  3. mayankdevnani Group Title
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    Like i have to convert in this :- \[\large \bf (2x-y)(3x+4y)\] i can't judge out that how to obtain this equation ?

    • one month ago
  4. mayankdevnani Group Title
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    @cwrw238 help me!!

    • one month ago
  5. mayankdevnani Group Title
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    @SolomonZelman

    • one month ago
  6. SolomonZelman Group Title
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    well, you just did it, no ?

    • one month ago
  7. cwrw238 Group Title
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    well this is the way i do these - you need to look at factors of the first coefficient and the last that is 6 and -4 6 could be 6 * 1 or 2 * 3 and -4 could be -4*1 or 4 * -1 , or -2 * 2 now we have to find a way to combine these factor so as to get the middle coefficient +5

    • one month ago
  8. mayankdevnani Group Title
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    solomon, i do it many times but i don't get this equation

    • one month ago
  9. cwrw238 Group Title
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    i draw up columns; |dw:1408387576129:dw|

    • one month ago
  10. mayankdevnani Group Title
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    i don't understand your method... @cwrw238

    • one month ago
  11. cwrw238 Group Title
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    so 4 links with 2 and -1 with +3 4 and 2 on the outside and -1 and 3 on inside giving (2x - y)( 3x + 4y) - this works for me

    • one month ago
  12. mayankdevnani Group Title
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    totally confused... ??? @cwrw238

    • one month ago
  13. cwrw238 Group Title
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    yea -- it takes a bit of getting used to

    • one month ago
  14. amistre64 Group Title
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    6x^2 +5xy -4y^2=0 assume it factors rationally, multiply the first and last coeffs, and set up the solution to divide off the first term afterwards 6*4 = 24, factors of 24 that combine to 5 c since the last term is negative, we have different signs, since the middle term is positive, the largest factor is postivie (x+L/6) (x-s/6) 8 and 3 combine to form 5 sooo (x+8/6) (x-3/6) reduce the fractions and if there is any fraction left over, stick the bottom back in front (x+4/3) (x-1/2) (3x+4) (2x-1) thats how i tend to do it

    • one month ago
  15. amistre64 Group Title
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    i forgot the y :)

    • one month ago
  16. cwrw238 Group Title
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    actually therea an error in my last line in drawing it should read 4 * 2 and -1 * 3

    • one month ago
  17. mayankdevnani Group Title
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    my BAD... i don't understand... @amistre64

    • one month ago
  18. amistre64 Group Title
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    its a method is all ...

    • one month ago
  19. mayankdevnani Group Title
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    can you teach me step by step ??

    • one month ago
  20. cwrw238 Group Title
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    its tricky i know - it needs a lot of practice

    • one month ago
  21. amistre64 Group Title
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    it would look just like what i just typed, but sure

    • one month ago
  22. mayankdevnani Group Title
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    but i don't know this method yet,this is first chance to see this long method...@cwrw238 hehehe

    • one month ago
  23. amistre64 Group Title
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    ax^2 + bx + c setup the factor, assuming it has ratioanl roots (x /a) (x /a) and mutliply the first and last coefficients

    • one month ago
  24. mayankdevnani Group Title
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    why its (x/a)(x/a) ?? @amistre64

    • one month ago
  25. amistre64 Group Title
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    ac = k, such that some pair of the factors of k combine to equal b

    • one month ago
  26. amistre64 Group Title
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    its not x/a there is a space left to determine the operations

    • one month ago
  27. cwrw238 Group Title
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    |dw:1408388342676:dw|

    • one month ago
  28. amistre64 Group Title
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    x ___ ( )/a

    • one month ago
  29. amistre64 Group Title
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    im not sure why it works out like that, but i recall when i went thr the class that the hundreds of problems we did it just sort of patterned out this way

    • one month ago
  30. mayankdevnani Group Title
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    like :- \[\large \bf \frac{x-2}{a}\]

    • one month ago
  31. mayankdevnani Group Title
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    this is what you mean to say

    • one month ago
  32. amistre64 Group Title
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    not quite: at first we are just organizing the information, filling it in as we go \[(x \pm\frac{ r_1}{a})~~(x \pm\frac{ r_2}{a})\]

    • one month ago
  33. mayankdevnani Group Title
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    ohkk

    • one month ago
  34. amistre64 Group Title
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    if the last term is positive, then ++ or -- create a positive term if the last term is negative, then +- or -+ create a negative term

    • one month ago
  35. mayankdevnani Group Title
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    yeah..

    • one month ago
  36. amistre64 Group Title
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    since the last term in negative, lets go with: \[(x \color{red}+\frac{ r_1}{a})~~(x \color{red}-\frac{ r_2}{a})\] now, when we add up to get the middle term, the sign tells us which is larger, in this case +5 means the the positive part is larger than the negative part \[(x +\frac{\color{red}L}{a})~~(x -\frac{ \color{red}s}{a})\]

    • one month ago
  37. mayankdevnani Group Title
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    correct

    • one month ago
  38. amistre64 Group Title
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    now, we can determine L and s in a more usual way: 6 * 4y^2 = 24y^2 the factors being: 8y,3y will combine to make 5y agreed?

    • one month ago
  39. mayankdevnani Group Title
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    another way of writing this :- 6*(-4y^2)=-24y^2 the factors being: 8y,-3y will combine to make 5y am i correct ? @amistre64

    • one month ago
  40. amistre64 Group Title
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    yes \[(x +\frac{\color{red}{8y}}{6})~~(x -\frac{ \color{red}{3y}}{6})\] then we simply reduce the fractions, and if any are left oever, stick the bottom back in front \[(x +\frac{{4y}}{3})~~(x -\frac{{1y}}{2})\]

    • one month ago
  41. mayankdevnani Group Title
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    correct

    • one month ago
  42. amistre64 Group Title
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    ideally, we would have factored out a 6 to start with and ended up with:\[6~~(x +\frac{{4y}}{3})~~(x -\frac{{1y}}{2})\] \[3(x +\frac{{4y}}{3})~~2(x -\frac{{1y}}{2})\] \[(3x +4y)~~(2x -y)\] which is just visually putting the bottom of the fraction back in front where we got it from to start with

    • one month ago
  43. amistre64 Group Title
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    \[ax^2 + bx + c\] \[a(x^2 + \frac{b}{a}x + \frac{c}{a})\] then finding the roots such that the factors of c/a add up to b/a gets us \[a(x+r_1)(x+r_2)\]

    • one month ago
  44. myininaya Group Title
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    To factor ax^2+bx+c you find two factors of a*c that have product ac and sum b. It is the same way if you have something in this form: ax^2+bxy+cy^2 Find two factors of a*c that have product ac and sum b. Say b=m+n and ac=mn Then ax^2+mxy+nxy+cy^2 Now we factor by grouping... So pretend we have x^2+6xy+9y^2 ac=9 and b=6 ac=9=3(3) and b=3+3 so we have x^2+3xy+3xy+9y^2 Then factor by grouping x(x+3y)+3y(x+3y) (x+3y)(x+3y) (x+3y)^2

    • one month ago
  45. amistre64 Group Title
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    im using the term roots incorrectly most likely .... the factors of c/a that do this and that

    • one month ago
  46. myininaya Group Title
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    Another example: 3x^2-7xy -6y^2 Let's see if you can factor this one... what is a,b, and c? then you must find a*c. then you must find two factors of a*c such that when multiplied together you get ac and when you add them you get b. Let's do this slowly.. First question what is a,b, and c?

    • one month ago
  47. mayankdevnani Group Title
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    Sorry :( for not answering your question,but i know how to do it :- Here is method(suggest by you):- ab=-24 and a+b=5 Now,we get a=8 and b=3 So,we get \[\large \bf 6x^2+8xy-3xy-4y^2=0\] \[\large \bf 2x(3x+4y)-y(3x+4y)=0\] \[\large \bf (2x-y)(3x+4y)=0\] \(\large \bf \color{red}{Answer}\)^^^

    • one month ago
  48. mayankdevnani Group Title
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    am i correct?

    • one month ago
  49. mayankdevnani Group Title
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    @myininaya mam

    • one month ago
  50. myininaya Group Title
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    yes you are done with the factoring

    • one month ago
  51. mayankdevnani Group Title
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    THANK YOU GUYS for helping me......i appreciate your work.. @cwrw238 , @amistre64 (once again thank you so much) and mostly to you @myininaya mam....

    • one month ago
  52. cwrw238 Group Title
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    yw

    • one month ago
  53. amistre64 Group Title
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    youre welcome

    • one month ago
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