## mayankdevnani one year ago Please Help me !! Its simple question though but i don't know how to solve it ??

1. mayankdevnani

How to write $$\large \bf 6x^2+5xy-4y^2=0$$ in another form or equation?

2. Mokeira

basically, just factor out

3. mayankdevnani

Like i have to convert in this :- $\large \bf (2x-y)(3x+4y)$ i can't judge out that how to obtain this equation ?

4. mayankdevnani

@cwrw238 help me!!

5. mayankdevnani

@SolomonZelman

6. SolomonZelman

well, you just did it, no ?

7. cwrw238

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

8. mayankdevnani

solomon, i do it many times but i don't get this equation

9. cwrw238

i draw up columns; |dw:1408387576129:dw|

10. mayankdevnani

i don't understand your method... @cwrw238

11. cwrw238

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

12. mayankdevnani

totally confused... ??? @cwrw238

13. cwrw238

yea -- it takes a bit of getting used to

14. amistre64

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

15. amistre64

i forgot the y :)

16. cwrw238

actually therea an error in my last line in drawing it should read 4 * 2 and -1 * 3

17. mayankdevnani

my BAD... i don't understand... @amistre64

18. amistre64

its a method is all ...

19. mayankdevnani

can you teach me step by step ??

20. cwrw238

its tricky i know - it needs a lot of practice

21. amistre64

it would look just like what i just typed, but sure

22. mayankdevnani

but i don't know this method yet,this is first chance to see this long method...@cwrw238 hehehe

23. amistre64

ax^2 + bx + c setup the factor, assuming it has ratioanl roots (x /a) (x /a) and mutliply the first and last coefficients

24. mayankdevnani

why its (x/a)(x/a) ?? @amistre64

25. amistre64

ac = k, such that some pair of the factors of k combine to equal b

26. amistre64

its not x/a there is a space left to determine the operations

27. cwrw238

|dw:1408388342676:dw|

28. amistre64

x ___ ( )/a

29. amistre64

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

30. mayankdevnani

like :- $\large \bf \frac{x-2}{a}$

31. mayankdevnani

this is what you mean to say

32. amistre64

not quite: at first we are just organizing the information, filling it in as we go $(x \pm\frac{ r_1}{a})~~(x \pm\frac{ r_2}{a})$

33. mayankdevnani

ohkk

34. amistre64

if the last term is positive, then ++ or -- create a positive term if the last term is negative, then +- or -+ create a negative term

35. mayankdevnani

yeah..

36. amistre64

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 $(x +\frac{\color{red}L}{a})~~(x -\frac{ \color{red}s}{a})$

37. mayankdevnani

correct

38. amistre64

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?

39. mayankdevnani

another way of writing this :- 6*(-4y^2)=-24y^2 the factors being: 8y,-3y will combine to make 5y am i correct ? @amistre64

40. amistre64

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 $(x +\frac{{4y}}{3})~~(x -\frac{{1y}}{2})$

41. mayankdevnani

correct

42. amistre64

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

43. amistre64

$ax^2 + bx + c$ $a(x^2 + \frac{b}{a}x + \frac{c}{a})$ then finding the roots such that the factors of c/a add up to b/a gets us $a(x+r_1)(x+r_2)$

44. myininaya

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

45. amistre64

im using the term roots incorrectly most likely .... the factors of c/a that do this and that

46. myininaya

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?

47. mayankdevnani

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 $\large \bf 6x^2+8xy-3xy-4y^2=0$ $\large \bf 2x(3x+4y)-y(3x+4y)=0$ $\large \bf (2x-y)(3x+4y)=0$ $$\large \bf \color{red}{Answer}$$^^^

48. mayankdevnani

am i correct?

49. mayankdevnani

@myininaya mam

50. myininaya

yes you are done with the factoring

51. mayankdevnani

THANK YOU GUYS for helping me......i appreciate your work.. @cwrw238 , @amistre64 (once again thank you so much) and mostly to you @myininaya mam....

52. cwrw238

yw

53. amistre64

youre welcome