Solve y^2+a*y+b=0 for y, where a=? and b=?
basically you treat x^{2.125} as a constant and solve for y using the quadratic formula or some other means. Then substitute x^{2.125} back into to your solved for roots.
I get
$$y=0.005 \left(-\sqrt{40000 x^{2.125}+111161}-329\right)$$
and
$$y=0.005 \left(\sqrt{40000 x^{2.125}+111161}-329\right)$$
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=....)
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.
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.