anonymous 5 years ago 5x^2-5=120 ? please help (5x to the second power minus 5 equals 120)

1. amistre64

x = 5?

2. anonymous

well i think u mean (5x)^2 as apposed to 5x^2 so then u get (5x)^2 =125 when u add 5 to both sides. so then square 5x giving u 25x^2 = 125. divide both sides and u get x = square root of 5

3. anonymous

if u mean 5x^2 u get x = 5 as amistre said

4. anonymous

$5x^2 - 5 = 120$$\implies 5x^2 = 120 + 5$ $\implies x^2 = 25$ $\implies x = \pm 5$

5. anonymous

is that all i do ? it says solve each equation by using square root .man i guess you are right .! i have more too ! please keep helping me

6. anonymous

Well, write the next one and see how much you can do on your own.

7. anonymous

i dont know what to do !! :( im trying though 6r^2+5v=10

8. anonymous

Well what are you supposed to be solving for?

9. anonymous

i am supposed to solve the equation with the quadratic formula

10. anonymous

I don't understand though you have 2 variables there, v and r.

11. anonymous

it must have been a misprint. we can just use 1 variable. v? so whats my first step?

12. anonymous

Well, what is the form you need for the quadratic formula?

13. anonymous

the general form?

14. anonymous

yeah

15. anonymous

ax2 + bx + c = 0

16. anonymous

Right. So make the equation you have look like that.

17. anonymous

(ax2+bx+c=0) 6x2=5=0?

18. anonymous

no thats not it !! uhm im sorry i am wasting your time.

19. anonymous

Wait, what's the equation you started with?

20. anonymous

(6v)^2+5v=10

21. anonymous

Are there really parentheses there?

22. anonymous

no , i just dont know how to square it on here

23. anonymous

oh, ok. So if you subtract a 10 from both sides wont the left side equal 0?

24. anonymous

so it would be something like ... 6v^2+5v-10=0 ?

25. anonymous

It would be exactly like that. So what is a, b, and c?

26. anonymous

a = 6v or 6 ? b = 5 c= -10

27. anonymous

Well, the standard form is ax^2 + bx + c, so the a would just be 6. It's the coefficient in front of the x^2 (or in this case v^2) term.

28. anonymous

ooooo , ok i think im starting to get this a little more. so what do i need to do next in this equation ?

29. anonymous

Well, given the new form we have here.. You just plug in your a, b and c to the quadratic equation to find what x is. $ax^2 + bx + c =0 \implies x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

30. anonymous

=( im going to fail this algebra 2 class !!

31. anonymous

No. You're doing very very well. Have you been taught the quadratic equation?

32. anonymous

yea we have but i was really sick and my teacher has not helped me really .

33. anonymous

Ok, well that is what it is. If you have something in the form $ax^2 + bx + c = 0$ you can find what x equals by plugging in a, b, and c into that equation.

34. anonymous

My algebra2 teacher taught it to us using a song.

35. anonymous

x= -5+ the square root 5^2-4(6)(-10) - _____________________________________ 2(6)

36. anonymous

lol sorry for the huge spaces. i tried though . and how does the song go ?

37. anonymous

Yes.

38. anonymous

It's to the tune of jingle bells. Quadratic equations / Dashing through the snow often make you cry. / in a one horse open sleigh But now there is a way / or' the fields we go to wave your tears bye bye. / laughing all the way Don't need to factor now. / Bells on bob-tails ring Don't need complete the square. / making spirits bright Just one simple forumla this secret I will share. Oh x equals / Jingle bells minus b / Jingle bells plus or minus square root of / Jingle all the way b squared minus 4ac / oh what fun it is to ride all divided by 2a! / in a one horse open sleigh.

39. anonymous

omg !!! ok can i please use this song , i would love to write it down and use this, we have a test thursday and this could help me extremly considering the fact i love to sing !!!

40. anonymous

In anycase, it helps to take the square root part first. The part in the square root is called the discriminant. Then you can take the square root of the discriminant and add and subtract it from the -b to give you your two values for x.

41. anonymous

Yes of course you can use it ;p

42. anonymous

you are a life saver ! &'d you are extremely smart at this lol i know this is besides the point but how old are you genius ?

43. anonymous

Heh.. I'm 33. I found this video of people singing it. Theres a lot of other videos of the quadratic forumula sung to many different songs, but this is the only one I could find for jingle-bells. Anyway, hope it helps. http://www.youtube.com/watch?v=CnJT1ojHT28&feature=related

44. anonymous

lol it will ! are you on here all the time ? i will most likely start getting on here daily with this work. that is if i dont have to work .

45. anonymous

I'm on pretty often yeah =)

46. anonymous

Or you can always find me a gmail if I'm not online.

47. anonymous

err at gmail rather.

48. anonymous

well i really hope each time i get on i can find you...im new to this lol i just made me a membership once my question was on here. i dont know how to really work this sight

49. anonymous

my email is selenadowns@yahoo.com or i have a facebook lol just saying

50. anonymous

51. anonymous

ok well feel free to drop me a line if you have questions, or you can post them here and someone will no doubt answer eventually =)

52. anonymous

one more question before i log off &'d head to work what are the steps to quadratic systems ? the next problem is y=-x^2-3x+10 y=x+5

53. anonymous

Well. If y = x+5 and y = $$x^2 - 3x + 10$$ Then can't you set those two right hand parts equal to eachother?

54. anonymous

huh ? sweetie you lost me lol set what &'d what equal to eachother ?

55. anonymous

When working with a system, you're finding solutions that satisfy both equations. So the x from one equals the x from the other, and same for the y.

56. anonymous

So if y = x+5 and y = $$x^2 - 3x + 10$$ then you have $$x+5 = y = x^2 - 3x + 10$$

57. anonymous

So you can get rid of the y.

58. anonymous

and you're left with $$x+5 = x^2 - 3x + 10$$

59. anonymous

-x^2-3x+10=x+5

60. anonymous

o :/ what am i doing wrong ?

61. anonymous

You did it right. I forgot the negative sign on the $$x^2$$ term

62. anonymous

Then just solve the same way we did the last one. move everything over so you have a 0 on one side and ax^2 + bx + c on the other.

63. anonymous

from what i had... i subtract the -x-5 on both sides? -x^2-4x+5=0 0=x^2+4x-5 ????????????

64. anonymous

Yep. Either of those two will work (and give the same result)

65. anonymous

(x+5)(x-1) x=-5 x=1 y=-5+5 y=1+5 y+0 y=6 (-5,0) (1,6)

66. anonymous

HOLD ON IM VERY MAD!!!! i just had a moment........that problem i just did were my notes that were already done . lol wow im stupid

67. anonymous

It's still good practice!

68. anonymous

i got all excited because i thought i comprehended

69. anonymous

i really hope that in the future i marry someone who has mathmatic potential as yourself because if not i dont know what i will do lol

70. anonymous

Heh.. you're doing really well actually!

71. anonymous

thanks sweetie. well i am off to work. hopefully i get your help soon. email me asap . i get them to my phone . i will really need more help when i get off work

72. anonymous

73. anonymous

i will check my phone then . well bye =) thank you SOOOOOOOOOOOOOOO MUCH

74. anonymous

hello :)

75. anonymous

lol, hi.

76. anonymous

are you online?