## anonymous 5 years ago can someone explain the quadratic formula to me please

1. amistre64

Yes, the quadratic formula is what you end up with when you "complete the square". "Completing the square" is proof that the quad formula works. Anything specific you wanna know?

2. anonymous

$5x ^{2} +3x+ 7= 0$how would i do it with the equation

3. amistre64

Just curious, do you know what "quadratic" means? at first I thought it had to do with "4" ... but it is just a fancy name for a squared function like x^2.

4. anonymous

yea im learning about it in math but hes confusing me

5. amistre64

Lets start with the basics and then I will get to your equation ok? Tell me what the quad formula is so I know you have a grasp on it.

6. anonymous

how to find continuity of function of 2 vars??

7. amistre64

Hi think, that doesnt ring a bell. sorry... Are the variables independant or does one depend on the other?

8. anonymous

$-b \sqrt{b ^{2}- 4 (bc} \div 2 (a)$

9. anonymous

its like, lim(x,y)->(1,2){x/sqrt[2x+y]}}

10. anonymous

that just confused me

11. anonymous

david, for your question, a quadratic equation is which has x power 2, and what you mentioned now is the 'discriminant'

12. amistre64

@think: that doesnt look familiar to me. David; it looks like you know the formula; but might have troubles with formatting it with the website... But yes, you are basically right. lets define a, b, and c. a = the first term; b= the middle term, and c= the last term. Does that make sense?

13. anonymous

that is a way to find out the roots

14. anonymous

yeah after that im lost and i asked him to help me and he confused me even more

15. anonymous

@amistre, its like limit of (x,y)->(1,2) with the function as x/ sqrt(2x + y)

16. amistre64

lets re-write the top like this: -(middle) +-sqrt(middle^2 - (4)(first)(last)) How does that look to you?

17. anonymous

that actually makes sense

18. amistre64

@think, I would assume that any values of (x,y) that make the bottom a zero have something to do with it. But I am just not sure.

19. amistre64

@David: good then lets call the bottom part of the formula: 2(first) does that help?

20. anonymous

yeah

21. amistre64

So what was your original equation again and lets see if we can work it out now.

22. anonymous

i think twas 5x^2 + 3x + 7 =0

23. anonymous

right?

24. anonymous

yea

25. amistre64

@thnk: what happens when you plug in (1,2) into the equation, does it get a number or an indeterminiate value?

26. anonymous

thinker wat grade r u in

27. amistre64

5x^2 + 3x + 7 first = (+5) ; middle = (+3); and last = (+7) top of quad form: -3 +- sqrt(3^2 -(4)(5)(7)) ------------------------ bottom part: 2(5) What do we get?

28. anonymous

10

29. anonymous

sry..it freezed again....i get the answer as 1/2

30. amistre64

10 is the bottom of it; but what is the top part of the quad form equal; remember it is just a big fraction looking monstorcity :)

31. amistre64

then as (x,y) approaches (1,2) that was the original right? then the equation approaches (1/2) if you plugged it in correctly :)

32. anonymous

-3 +- sqrt 9 -140

33. amistre64

good; does (-3 + sqrt(-131)) / 10 have any meaning to you? In other words can we have any real values for sqrt(-131)? Let me ask it this way.... what number when multiplied by itself will give you a "negative" answer?+

34. anonymous

yeah...so what 's the continuity?

35. anonymous

no it has no real solution i think

36. amistre64

@think; rationalize the denominator and see if that helps :)

37. amistre64

@David; very good, that is the correct answer, there are no "real" values of 'x'. Tell me, what does it mean to be a "real" value?

38. anonymous

when a number has a square root??

39. amistre64

@ david, good enough answer :) Have you dealt with "imaginary" numbers yet?

40. anonymous

no

41. anonymous

@amistre, have u anytime seen "Thomas' Calculus" text?

42. anonymous

oh r u tlking about the imaginery 1 in front of x

43. anonymous

@david, no..imaginary numbers are those which do not exist..like sqrt of a negative number etc

44. anonymous

oh

45. amistre64

@think; doesnt ring a bell :) @david; lol....good attempt at, but no. the "1" infront of an x is actually there, we just dont write it because it is

46. amistre64

1(x) = x the only imaginary number there is, is called "i" and we simply define it as : i = sqrt(-1) it is the only way in which we can solve the squareroots of negative numbers.

47. anonymous

ohh..was just a try..not an "algebra" person

48. anonymous

@amistre, have u anytime seen "Thomas' Calculus" text??

49. amistre64

Thomas Calculus text doesnt sound familiar to me...

50. anonymous

ohh..:(..its a huge book..which i have to study from..

51. amistre64

I have a few good books in the college library (public libraries here are a joke) that I use. Some are more helpful than others at times.

52. amistre64

How did we do with your 1/sqrt(2x+y) question so far?

53. anonymous

this function is continous only when the denominator,i.e x,y$\neq$0

54. amistre64

And do we know if y is a function of x; or if x is a function of y; or if they are independant variables?

55. anonymous

there's nothing mentioned...its just that we ve to find the points of continuity/discontinuity

56. amistre64

So when (2x+y) = 0 there is a Vertical Asymptote right? When y = -2x and when x = -y/2

57. anonymous

vertical asymptote? what is that?

58. amistre64

there are a few types of discontinuities available to us: the "hole" is a line is called a jump disconiuity becasue the line will forever get close to it and never touch it. the "vertical" asymptote is a value for "x" that the graph will forever slide right up next to but never touch so instead of a jump across a hole, we get a neverending curve that approaches "x"

59. anonymous

sry..didn't get that:(

60. amistre64

Holes occur when you can cross out like factors top to bottom. VAs occur when whats left after crossing out makes the bottom equal to "0"

61. anonymous

@amistres64 may u pls help me

62. amistre64

that was odd... :) kabelo, what is your question? and maybe I can help

63. amistre64

@think: think of the graph for f(x) = x^2; and restrict the domain to x<4 and x>4. there is no value for f(x) at x=4; it just makes a hole.

64. anonymous

i know is bascially algebra

65. amistre64

kab: basic algebra is good :) what is your question?

66. anonymous

the veranda is covered with tiles(30cm times 30cm) in 5 black and4 white tiles,how many black tiles are used to cover the veranda floor if that pattern is continued.the area is 4,2m

67. amistre64

Kab: Ill go over to that posting , this ones getting rather long ok?

68. anonymous

can i join?

69. anonymous

yip pls do ,yes u can.

70. anonymous

what was the question there? unable to find it:(

71. anonymous

the veranda is covered with tiles(30cm times 30cm) in 5 black and4 white tiles,how many black tiles are used to cover the veranda floor if that pattern is continued.the area is 4,2m