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anonymous

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    x = 5?

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    if u mean 5x^2 u get x = 5 as amistre said

  4. anonymous
    • 5 years ago
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    \[5x^2 - 5 = 120 \]\[\implies 5x^2 = 120 + 5\] \[\implies x^2 = 25\] \[\implies x = \pm 5\]

  5. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Well, write the next one and see how much you can do on your own.

  7. anonymous
    • 5 years ago
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    i dont know what to do !! :( im trying though 6r^2+5v=10

  8. anonymous
    • 5 years ago
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    Well what are you supposed to be solving for?

  9. anonymous
    • 5 years ago
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    i am supposed to solve the equation with the quadratic formula

  10. anonymous
    • 5 years ago
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    I don't understand though you have 2 variables there, v and r.

  11. anonymous
    • 5 years ago
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    it must have been a misprint. we can just use 1 variable. v? so whats my first step?

  12. anonymous
    • 5 years ago
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    Well, what is the form you need for the quadratic formula?

  13. anonymous
    • 5 years ago
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    the general form?

  14. anonymous
    • 5 years ago
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    yeah

  15. anonymous
    • 5 years ago
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    ax2 + bx + c = 0

  16. anonymous
    • 5 years ago
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    Right. So make the equation you have look like that.

  17. anonymous
    • 5 years ago
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    (ax2+bx+c=0) 6x2=5=0?

  18. anonymous
    • 5 years ago
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    no thats not it !! uhm im sorry i am wasting your time.

  19. anonymous
    • 5 years ago
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    Wait, what's the equation you started with?

  20. anonymous
    • 5 years ago
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    (6v)^2+5v=10

  21. anonymous
    • 5 years ago
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    Are there really parentheses there?

  22. anonymous
    • 5 years ago
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    no , i just dont know how to square it on here

  23. anonymous
    • 5 years ago
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    oh, ok. So if you subtract a 10 from both sides wont the left side equal 0?

  24. anonymous
    • 5 years ago
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    so it would be something like ... 6v^2+5v-10=0 ?

  25. anonymous
    • 5 years ago
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    It would be exactly like that. So what is a, b, and c?

  26. anonymous
    • 5 years ago
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    a = 6v or 6 ? b = 5 c= -10

  27. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    =( im going to fail this algebra 2 class !!

  31. anonymous
    • 5 years ago
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    No. You're doing very very well. Have you been taught the quadratic equation?

  32. anonymous
    • 5 years ago
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    yea we have but i was really sick and my teacher has not helped me really .

  33. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    My algebra2 teacher taught it to us using a song.

  35. anonymous
    • 5 years ago
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    x= -5+ the square root 5^2-4(6)(-10) - _____________________________________ 2(6)

  36. anonymous
    • 5 years ago
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    lol sorry for the huge spaces. i tried though . and how does the song go ?

  37. anonymous
    • 5 years ago
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    Yes.

  38. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Yes of course you can use it ;p

  42. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    I'm on pretty often yeah =)

  46. anonymous
    • 5 years ago
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    Or you can always find me a gmail if I'm not online.

  47. anonymous
    • 5 years ago
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    err at gmail rather.

  48. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    my email is selenadowns@yahoo.com or i have a facebook lol just saying

  50. anonymous
    • 5 years ago
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    i meant selenadowns73@yahoo.com

  51. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    huh ? sweetie you lost me lol set what &'d what equal to eachother ?

  55. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    So if y = x+5 and y = \(x^2 - 3x + 10\) then you have \(x+5 = y = x^2 - 3x + 10\)

  57. anonymous
    • 5 years ago
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    So you can get rid of the y.

  58. anonymous
    • 5 years ago
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    and you're left with \(x+5 = x^2 - 3x + 10\)

  59. anonymous
    • 5 years ago
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    -x^2-3x+10=x+5

  60. anonymous
    • 5 years ago
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    o :/ what am i doing wrong ?

  61. anonymous
    • 5 years ago
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    You did it right. I forgot the negative sign on the \(x^2\) term

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

  64. anonymous
    • 5 years ago
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    Yep. Either of those two will work (and give the same result)

  65. anonymous
    • 5 years ago
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    (x+5)(x-1) x=-5 x=1 y=-5+5 y=1+5 y+0 y=6 (-5,0) (1,6)

  66. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    It's still good practice!

  68. anonymous
    • 5 years ago
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    i got all excited because i thought i comprehended

  69. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Heh.. you're doing really well actually!

  71. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    I already did.

  73. anonymous
    • 5 years ago
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    i will check my phone then . well bye =) thank you SOOOOOOOOOOOOOOO MUCH

  74. anonymous
    • 5 years ago
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    hello :)

  75. anonymous
    • 5 years ago
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    lol, hi.

  76. anonymous
    • 5 years ago
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    are you online?

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