How can I solve ax^2+bx+c<0 in python? - Python
8/19/2021, 2:15:37 PM
How can I solve ax^2+bx+c<0 in python?

Ответы (31)

8/19/2021, 2:19:09 PM

By using complex math module And you can put this question on google

8/19/2021, 2:19:47 PM

I googled it but I could not find solution

8/19/2021, 2:20:08 PM

what is complex math module?

8/19/2021, 2:20:27 PM

A math module

8/19/2021, 2:20:52 PM

Is there an example?

8/19/2021, 2:21:01 PM

Just put this question and open first or second website You can see examples on google

8/19/2021, 2:21:40 PM

thank you for your help

formulae for the roots of a quadratic equations are well defined. It's high school mathematics

8/19/2021, 2:22:37 PM

yes I wanted to solve it without delta command and automatically.

what do you mean automatically? python isn't magic. It'll just use that formula, of course with some optimizations but 🤷‍♂️

8/19/2021, 2:24:39 PM

😁😁automatically means usng built in procedures or commans

google quadratic solvers and you'd find one 🤔, here's something that will interest you:

8/19/2021, 2:32:57 PM

🙏🙏🙏🌺🌺🌺 thanks a lot infact it is a Quadratic matrix form y^TQy but as I do not know how to solve these types of inequalities I am probing for second order polynomial

oh i just realized that wasn't what you wanted

also sympy has inequality solvers

8/19/2021, 2:36:00 PM

yes it wasnt the main problem is a quadratic form with two variables as these variables are free I am trying to find out this please consider the arbitrary column vector y=np.array([[?],[?]]) , and this matrix P=np.array([[12137.5, 26.381], [26.381,11438.7]]) (I began with vector y=np.array([[0.1],[0.2]]) as an initial guess) I need to unederstand for which vectors "y" This inequality is valid np.transpose(y)*P*y -8.918847<0.000731 What I have down was that I began with an initial guess y=np.array([[0.1],[0.2]] and randomly I changed y but I could not find suitable y

8/19/2021, 2:37:08 PM

as I know it just solve one variable inequalities infact my problem is in terms of two variables

i see now what your problem is

8/19/2021, 2:37:59 PM

Im confused how can I solved it 😔😔

wait a second this is just ytranspose*P*y where is the 2nd variable?

8/19/2021, 2:42:25 PM

Aha sorry for not explaining y is a 2*1 vector with unknown or free components (e.g. y_1 and y_2)

8/19/2021, 2:48:18 PM

If u don't Want to spent time trying to understand how libraries works, just do a for loop in range (0, 1000000000) and inside of it generate a random y, and see when your inequality is ok. Its like a Montecarlo optimization, is not ok in term of computation, but iy could works.

8/19/2021, 2:50:11 PM

thank u so much I do not know how can I relate randome y to index of the loop

hmm, i just reduced that and it solves down to (y_1)²*(12137.5 + 26.381) + (y_2)²*(26.381 + 11438.7) < (8.918847 +0.00731) this is just an equation of a circle 🤔 ax^2 + by^2 < c the inequality represents all points inside circle

come to @pythonofftopic, as this is not python related

8/19/2021, 2:56:14 PM

there would be a term based on y_1*y_2 : 12137.5y_1^2+2(26.381)y_1y_2+11438.7y_2^2

hmm..come to @pythonofftopic. I'll just how you the reduction i did. You can tell me where i am wrong, if i am

8/19/2021, 2:58:13 PM

well you have dropped the second term. There can I state the problem?

8/19/2021, 3:01:55 PM

ok thnx a lot for your help I joined

Ahmad Al-Ateek
8/19/2021, 4:10:24 PM

it is a hard problem

8/19/2021, 4:18:21 PM

yes it is

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