Results 1 to 9 of 9
Like Tree1Thanks
  • 1 Post By JJacquelin

Math Help - Can I re-arrange the formula to find "y"

  1. #1
    Newbie
    Joined
    Feb 2014
    From
    Asheville, NC
    Posts
    4

    Can I re-arrange the formula to find "y"

    I have this formula that creates a graph:
    y^2+3.29*y-0.073=x^2.125

    I need to re-arrange it so I will be able to find "y" when I know "x", so it looks like:
    y=.......

    It is possible?

    Thank you in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2011
    Posts
    244
    Thanks
    56

    Re: Can I re-arrange the formula to find "y"

    Solve y^2+a*y+b=0 for y, where a=? and b=?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,454
    Thanks
    944

    Re: Can I re-arrange the formula to find "y"

    Quote Originally Posted by eastwoodreaders View Post
    I have this formula that creates a graph:
    y^2+3.29*y-0.073=x^2.125

    I need to re-arrange it so I will be able to find "y" when I know "x", so it looks like:
    y=.......

    It is possible?

    Thank you in advance.
    basically you treat x2.125 as a constant and solve for y using the quadratic formula or some other means. Then substitute x2.125 back into to your solved for roots.

    I get

    $$y=0.005 \left(-\sqrt{40000 x^{2.125}+111161}-329\right)$$

    and

    $$y=0.005 \left(\sqrt{40000 x^{2.125}+111161}-329\right)$$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2014
    From
    Asheville, NC
    Posts
    4

    Re: Can I re-arrange the formula to find "y"

    Thank you for all your answers.

    JJacquelin:
    I guess to solve a*y^2+b*y+c=0 is y=(-bą√(b^2-4ac))/2a

    romsek:
    That does not work
    I need the solution that would match this blue graph: http://freebookexchange.org/temp/Clipboard02.jpg

    Equation Grapher

    Current formula: y^2+3.29*y-0.073=x^2.125

    Your solution looks different: y=0.005(sqrt(40000*x^2.125+111161-329))

    The curve is correct, but it is not in a right place. (I'm really sorry for my total luck of math knowledge. I feel so dumb.)

    I guess I should change my question:
    Wow can I re-write formula (a*y^2+b*y-c=x^d) so if would look like (y=....)
    Last edited by eastwoodreaders; February 18th 2014 at 08:06 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2014
    From
    Asheville, NC
    Posts
    4

    Re: Can I re-arrange the formula to find "y"

    If you can solve this: re-write formula (a*y^2+b*y-c=x^d) so if would look like (y=....)

    It should be most optimal, elegant solution for computer to calculate fast.

    Actually you might have even better solution for me.
    I'm trying to write a formula that would match as close as possible this linear one (y=(x+0.60-2.80+0.49)*100/82.75)
    You can see in red here: http://freebookexchange.org/temp/Clipboard03.jpg
    But it can not go into negative as it does right now after x is less than 1.72

    What I need is for it to make a gentle curve around x=2.5, y=0.95 and eventually eventually end up at x=0, y=0.25
    (at x>2.5 and above it turns into straight line) "X" value will be supplied and can be anything from 0.25 up to 999 and I need to be able to calculate "y" as close as possible to the linear one with "y" ever going into negative.

    Best I could come up with was (y^2+3.29*y-0.073=x^2.125) (Blue curve) by randomly sticking numbers into in on Equation Grapher

    Eventually formula will be plugged into computer so it should be as simple as possible to assure fast calculations.

    As a thank you, I will order for you any book of your choice (or anything else) from Amazon (as long as it is under $25.00) if you provide me with shipping address.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2011
    Posts
    244
    Thanks
    56

    Re: Can I re-arrange the formula to find "y"

    y^2+3.29*y-0.073=x^2.125
    y^2+3.29*y+(-0.073-x^2.125)=0
    a = 3.29
    b = (-0.073-x^2.125)
    Solve y^2+a*y+b=0 for y.
    Then replace a and b by the terms above.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2014
    From
    Asheville, NC
    Posts
    4

    Re: Can I re-arrange the formula to find "y"

    1. a*y^2 + b*y + c = x^2.125
    2. a*y^2 + b*y + (c-x^2.125)=0
    3. y=(-b+√(b^2-4*a*(c-x^2.125)))/2*a

    1. 1*y^2+3.29*y-0.073=x^2.125
    2. 1*y^2+3.29*y+(-0.073-x^2.125)=0
    3. a=1, b = 3.29, c = (-0.073-x^2.125)
    4. y=(-3.29+√(3.29^2-4*(-0.073-x^2.125)))/2 (Which one is faster for computer to calculate? each millisecond counts)
    5. y=(-3.29+√(11.1161+4*x^2.125))/2 (Which one is faster for computer to calculate? each millisecond counts)
    6. y=√(11.1161+4*x^2.125)/2-1.645 (Which one is faster for computer to calculate? each millisecond counts)

    It worked!!!!.... Thank you so much JJacquelin. Would you like a gift from Amazon as promised? Just send me a link to what it is you would like and your shipping address. Also thank you for not giving full answer and making me work a little, I feel little bit better about myself.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,454
    Thanks
    944

    Re: Can I re-arrange the formula to find "y"

    it didn't work because you didn't input the correct formula into your grapher. The 329 is outside the radical.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2011
    Posts
    244
    Thanks
    56

    Re: Can I re-arrange the formula to find "y"

    Don't mention it, eastwoodreaders !
    Thanks from eastwoodreaders
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: September 22nd 2012, 01:56 PM
  2. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  3. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  4. Replies: 3
    Last Post: April 30th 2009, 02:39 PM
  5. Replies: 5
    Last Post: August 12th 2007, 10:46 AM

Search Tags


/mathhelpforum @mathhelpforum