## poopsiedoodle 3 years ago Can someone help explain how to do quadratic equations with the quadratic formula? I'll give you a problem. See comments.

1. poopsiedoodle

$\huge 0 = -4(-4t^2 -8t -1)$

2. hba

Well the first step is to compare your equation with the standard form and get a,b and c $\huge\ ax^2+bx+c=0$

3. poopsiedoodle

So, how would I do that with this one? Multiply the parentheses stuff by -4?

4. hba

Seeing your question i would suggest you to first divide your whole equation by -4 and then compare

5. poopsiedoodle

$\Huge 0=-4t^2 -8t-1$

6. hba

What is a,b and c in this ?

7. poopsiedoodle

a=-4 b=-8 c=-1

8. hba

Good,Now you can use the quad formula. $\huge\ x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$

9. hba

The other think which you can do with your equation is that you can take -1 common and then compare so everything becomes positive.you can also do it straight away :)

10. poopsiedoodle

so, t = -8*√64-4*-4 ------------------ 2*-4 ?

11. poopsiedoodle

which would then be t=-8*(8+4) ----------- -8

12. poopsiedoodle

am I getting it right?

13. hba

It should be +- not *

14. poopsiedoodle

I don't get it... wouldn't you be multiplying -8 and $\pm √64-4*-4$?

15. hba

t=-8(+-)(8+4) ----------- -8

16. poopsiedoodle

Now I'm rather confused.

17. hba

Lemme show you.

18. hba

I appreciate you working with me though.

19. poopsiedoodle

I guess I could do that $$\Huge\mathsf{:P}$$

20. hba

a=-4 b=-8 c=-1 $\huge\ t=\frac{ -(-8) \pm \sqrt{(-8)^2-4(-4)(-1)} }{ 2(-4) }$

21. poopsiedoodle

Ok, but what does the $\pm$ do? what's its significance?

22. hba

Well when we simplify it we would be taking the adding part on one side and subracting part on the other side.

23. poopsiedoodle

so it'd be 8+√blahblahblah --------------- 2-(-4)?

24. hba

Either, 8+√blahblahblah --------------- 2-(-4)? Or, 8-√blahblahblah --------------- 2-(-4)?

25. poopsiedoodle

so the $\pm$ just breaks up between the denominator and the numerator?

26. hba

Exactly.

27. poopsiedoodle

Ah. I guess I'll try another problem here and ask you if it's right or something.

28. hba

Sure :)

29. poopsiedoodle

$\huge 0=-2x ^{2} +14x +20$ $\large x=\frac{ -14\pm \sqrt{14^2 - 4(-2+20)} }{ -4 }$ $\large x= \frac{ -14+14-72 }{ -4 }$

30. poopsiedoodle

and the answer is $\huge \frac{ -7\pm \sqrt{89} }{ 2 }$. Would I be right?

31. hba

Its is -4(a)(c) not -4(a+c)

32. poopsiedoodle

oh. Well that ruins that then.

33. hba

Yes it does :) Do it on a paper and tell me what you got from that equation :)

34. poopsiedoodle

$\Large x=\frac{ -14\pm \sqrt{14^2 - 4(-2*20)} }{ -4 }$ $\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }$ $\Large x=\frac{ -14\pm 14-160} { -4}$ $\Large x=\frac{ -14\pm -146} {-4}$ set x to 0 and divide all by -2 to get $\Large \frac{ 7\pm 73} { 2}$

35. poopsiedoodle

and now I'm confused.

36. poopsiedoodle

oh wait, I forgot to keep the x's in there. Crap.

37. poopsiedoodle

Think you could help me out here? My mind is about to implode.

38. hba

x=7+73/2,x=7-73/2

39. hba

Btw, (-) *(-)=+

40. hba

So just because of signs it goes wrong.

41. hba

$\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }$ Continue from here.

42. poopsiedoodle

But I left the xs out of my work. ._.

43. hba

44. hba

Continue working from here.

45. poopsiedoodle

$\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }$ $\Large x=\frac{ -14\pm 14+160} {-4}$ $\Large x=\frac{ -14\pm 174} {-4}$ $\Large x=\frac {\frac{ -14\pm 174} {-4}} {-2}$ $\Large x=\frac {7\pm -87} {2}$

46. poopsiedoodle

And now what do I do?

47. hba

I don't know what you are upto ,where did that divide by -2 come in between ?

48. poopsiedoodle

Well, I'm not sure. I just thought I had to do that. I'm rather confused ._.

49. hba

Hmm,If OS would be working on chrome i would have worked one problem for you. Can we do this later ?

50. poopsiedoodle

I'm using chrome right now o_o But, I guess that'd be OK :P

51. hba

Thanks.

52. poopsiedoodle

Thanks to you too :3