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quadratic equation

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Spiritual_Null answered on June 2, 2021 Popularity 10/10 Helpfulness 2/10

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Quadratic Equation are equations where the highest power of the variable (usually x) is 2, hence the word Quadratic Equations

The standard form of Quadratic Equations is as follows...

ax^2 + bx + c = 0 where a ≠ 0

This is because, even with an equation such as x^2 = 5, you can always* change it to the standard form.

x^2 = 5
x^2 - 5 = 0
x^2 - 5 + 0x = 0         ### 0 * x is 0 anyways so we can add that
x^2 + 0x - 5 = 0 
1x^2 + 0x - 5 = 0        ### 1 * x^2 is x^2 so we can add that (no change)

There! We got 1x^2 + 0x - 5 = 0!

You can cross check it with the standard form and see its the same!

Did you wonder why "a" can't be zero in the standard form? Well look here.

5x^2 + 10x + 20 = 0                 # PERFECTLY NORMAL QUADRATIC EQUATION 

0x^2 + 10x + 20 = 0                 # changing "a" to 0

10x + 20 = 0

At first glance, one may not notice the issue but let me pin point the definition of Quadratic Equation...

"Quadratic Equation are equations where the highest power of the variable is 2"

The definition states the highest power of x should be 2 but in 10x + 20 = 0 the highest power is 1 (where is no power, there is 1) so this equation became a linear equation! so to prevent that is why "a" or the coefficient of "x^2" can't equal to 0 in a Quadratic Equation

thank u! <3
NOW GO LEARN

-----------------
* = take with a pinch of salt

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Contributed on May 02 2024

Spiritual_Null

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quadratic formula

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116

 (-b±√(b²-4ac))/(2a)

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Contributed on Jun 02 2021

Obedient Oryx

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quadratic formula

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x = [-b ± √(b² - 4ac)] / (2a)

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Contributed on Oct 18 2021

LoneWolf

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quadratic equation

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(-b-sqrt(b**2-4*a*c))/(2*a)
(-b+sqrt(b**2-4*a*c))/(2*a)
If that doesn't work,
((0-b)-sqrt(b**2-4*a*c))/(2*a)
((0-b)+sqrt(b**2-4*a*c))/(2*a)

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Contributed on Dec 31 2021

Thankful Toucan

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quadratic equations

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Formula:
(-b±√(b²-4ac))/(2a)

general form:
ax^2 + bx + c = 0
x^2 + bx + c = 0

some of the other hidden forms:
x^2 = bx - c (quadriatic after transposition)
c/bx+d = x (where you end up with a quadriatic after cross multiplication)

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Contributed on Feb 11 2023

Al Ayman Sarker

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quadratic equation

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Δ=b2−4ac 
(-b±√Δ)/2a

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Contributed on Jan 30 2022

BGOPC

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Solving quadratic equations

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import math

def solve_quadratic(a, b, c):
    # Calculate the discriminant
    discriminant = b**2 - 4*a*c
    
    # Check if the equation has complex roots
    if discriminant < 0:
        return "Complex Roots"
    
    # Calculate the roots
    root1 = (-b + math.sqrt(discriminant)) / (2*a)
    root2 = (-b - math.sqrt(discriminant)) / (2*a)
    
    # Return the roots
    return root1, root2

# Example usage
a = 1
b = -5
c = 6
roots = solve_quadratic(a, b, c)
print("Roots:", roots)

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Contributed on Sep 11 2023

YOU.com

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quadratic equation solver

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import cmath

def solve_quadratic(a, b, c):
    # Calculate the discriminant
    discriminant = (b**2) - (4*a*c)

    # Find two solutions
    solution1 = (-b - cmath.sqrt(discriminant)) / (2*a)
    solution2 = (-b + cmath.sqrt(discriminant)) / (2*a)

    return solution1, solution2

# Example usage
a = 1
b = -5
c = 6
solution1, solution2 = solve_quadratic(a, b, c)

print("Solutions: {0} and {1}".format(solution1, solution2))

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Contributed on Sep 12 2023

YOU.com

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quadratic equation formula

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import math

def solve_quadratic_equation(a, b, c):
    discriminant = b**2 - 4*a*c
    
    if discriminant > 0:
        # Two distinct real roots
        root1 = (-b + math.sqrt(discriminant)) / (2*a)
        root2 = (-b - math.sqrt(discriminant)) / (2*a)
        return root1, root2
    elif discriminant == 0:
        # One real root
        root = -b / (2*a)
        return root
    else:
        # Complex roots
        real_part = -b / (2*a)
        imaginary_part = math.sqrt(abs(discriminant)) / (2*a)
        root1 = complex(real_part, imaginary_part)
        root2 = complex(real_part, -imaginary_part)
        return root1, root2

# Example usage with coefficients a, b, and c
a = 1
b = -3
c = 2

roots = solve_quadratic_equation(a, b, c)
print("Roots:", roots)

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Contributed on Sep 12 2023

YOU.com

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quadratic formula

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from cmath import sqrt as c_sqrt
from math import sqrt as m_sqrt
import tkinter as tk


# creates and runs an instance of the window
window = tk.Tk()

#makes the size of the widow
window.geometry("600x400")

#sets a title for the screen
window.title("intercode")

# creating a new label and adding a text to the window
new_label=tk.Label(window,text='enter a,b,c quadratic formula\n')
new_label.pack()

#creates a entrybox to add words to window
entry_label1=tk.Entry(window)
entry_label1.pack()

entry_label2=tk.Entry(window)
entry_label2.pack()

entry_label3=tk.Entry(window)
entry_label3.pack()
# defines results as string var
results= tk.StringVar(window)



#creating a results label
result_lable=tk.Label(window,textvariable=results)
result_lable.pack()



def callbackfunct():
    a =entry_label1.get()
    b =entry_label2.get()
    c =entry_label3.get()
    try:
        a= float(a)
        b= float(b)
        c= float(c)

    except ValueError:
        results.set("\nEnter number.")


    if (b**2-4*a*c) < 0:
        minus = (-b-c_sqrt(b**2-4*a*c))/(2*a)
        plus = (-b+c_sqrt(b**2-4*a*c))/(2*a)
    else:
        minus = (-b-m_sqrt(b**2-4*a*c))/(2*a)
        plus = (-b+m_sqrt(b**2-4*a*c))/(2*a)
        
    # deletes the text in the text box
    entry_label1.delete(0,tk.END)
    entry_label2.delete(0,tk.END)
    entry_label3.delete(0,tk.END)

    var = (f"'B+'   {plus} , 'B-'   {minus}")
    results.set(var)
# creating a button with a command
buttonez = tk.Button(text="Submit", command=callbackfunct)
buttonez.pack()


window.mainloop()

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Contributed on Nov 03 2022

An7wan

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