## anonymous 4 years ago Solve: 2r^2-11x+5=0 Thanks :) p.s. explain?

1. anonymous

Do you know the quadratic formula?

2. Hero

The variables should be the same, right?

3. anonymous

Agreed.

4. anonymous

Good eye.

5. Hero

Good eye? It was blatantly obvious, lol :P

6. anonymous

oh right! My teacher made a typo, lol. they should both be r's. and no, I don't remember anything from algebra 1! Which is bad :/

7. Hero

How do you know they shouldn't both be x's?

8. anonymous

well, letters don't really matter. whichever you replace, they would still be the same problem. just different letters.

9. anonymous

I think he's being humorous, sakigirl.

10. anonymous

can someone just help me? lol. i'm kinda stuck on this question >_<

11. Hero

2r^2-11r+5=0 2r^2-10r-1r+5 = 0 2r(r-5)-1(r-5) = 0 (r-5)(2r-1) = 0 r = 1/2 r = 5

12. anonymous

Hero is right.

13. Mimi_x3

You can also use the quadratic formula, which is faster i guess. $\large x=\frac{-(b)\pm\sqrt{(b)^{2}-4(a)(c)}}{2(a)}$

14. Hero

Nope, I'm faster. The quad formula is whack.

15. anonymous

Hahaha!

16. Hero

Mimi uses it because she can't factor :P

17. Mimi_x3

Nope, well..the using the quadratic you won't need to think much as to factor where you mgiht make stupid mistakes.

18. Hero

Mimi, you forgot to put the line that protects your discriminant :P

19. Mimi_x3

what line?

20. Hero

The square root line.....(DUH) :P

21. Mimi_x3

wtf..there's nothing wrong :/ You're confusing me :/

22. Hero

lol, just teasing. Don't cry on me :P

23. anonymous

HERO. one last question. i beg.

24. Mimi_x3

Oh..gawd, confusing a poor kid that just woke up :'(

25. Hero

Here we go again with the "poor little kid" retreat line you always pull :P

26. Mimi_x3

You dont have the right to alternate my phrases i said "poor kid" NOT "poor little kid" :P

27. Hero

poor kid = poor little kid To me it's the exact same thing. A sympathy plea

28. anonymous

kwell, I'm just gonna ask. 2c^2+c-10=0

29. Hero

30. Mimi_x3

Excuse me, its not a "sympathy plea" i never told you to be pity me. GOsh people these days who dont have common sense.

31. anonymous

32. Mimi_x3

Because Hero is unable to factor it :P

33. Mimi_x3

$\large c=\frac{-(b)\pm\sqrt{(b)^{2}-4(a)(c)}}{2(a)}$ $\large c=\frac{-(1)\pm\sqrt{(1)^{2}-4(2)(-10)}}{2(2)}$

34. Hero

2c^2+c-10=0 2c^2+5c-4c-10 = 0 c(2c+5)-2(2c+5) = 0 (2c+5)(c-2) = 0 c = -5/2, 2

35. Hero

Mimi jinxed me.

36. Mimi_x3

lol, i told you that you will make stupid mistakes :P The factor method is such a FAIL!

37. anonymous

38. Hero

Quad Form is an even bigger fail, especially when you depend on it so much Mimi :P

39. Hero

I know about 5 different factor methods

40. Hero

41. anonymous

how would you use the quad form on questions like 25m^2-64=0??

42. Mimi_x3

What? You dont have the right to say that its a fail since i got it right in the first go, not like you who go it WRONG! EVen during exams it would be a waste of time to do it again :P

43. anonymous

a.k.a binomials.

44. Hero

25m^2 - 64 = 0 25m^2 = 64 m^2 = 64/25 m = sqrt{64/25} m = plus or minus 8/5

45. Mimi_x3

Btw, quadratic formula can only be used for ax^3+bx+c

46. anonymous

Oh right, I totally knew that.

47. Mimi_x3

ax^2*

48. Hero

Yeah, the quad form is LIMITED

49. Hero

While I was working hard to provide your solution sakigirl. Mimi, started making useless comments

50. anonymous

hahah, forgetting the line of the square root. Oh silly her.

51. Mimi_x3

I was trying to help her know a better method to solve it which is quicker. I was being nice

52. Hero

I was being nicer :P

53. Mimi_x3

whaat? forgettng the line of the square root?

54. Hero

lol

55. anonymous

Yeah! and Hero commented on it

56. Hero

lol

57. Mimi_x3

lol, sorry to say i think you should scroll up, i never make those carelless mistakes.

58. Hero

you just made one, then auto corrected it. Putting ax^3 , then ax^2. Such silly mistakes :P

59. Mimi_x3

its called a TYPO, you make stupid mistakes as well argh..i should never comment in questions that you are in anymore.

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