computerv1

A simple program that solves a polynomial second or lower degree equation

Usage

git clone "..." ; cd computerv1
cargo run [your equation]

-> every entity of the equation need to be formatted like n * X^y like

` " 5 * X^0 + 4 * X^1 - 9.3 * X^2 = 1 * X^0 " ` 

Solving a Second Degree Equation

A quadratic equation is given in the form:

ax^2 + bx + c = 0

where:

• a ≠ 0 (if a = 0, it is not a quadratic equation),
• b and c are coefficients.

1. Calculating the Discriminant

The discriminant is calculated as:

Δ = b^2 - 4ac

• If Δ > 0, there are two distinct real solutions.
• If Δ = 0, there is one real solution (a double root).
• If Δ < 0, there are two complex conjugate solutions.

2. Formula for the Solutions

The solutions of the quadratic equation are given by:

x₁, x₂ = (-b ± √Δ) / (2a)

• If Δ > 0: The two solutions are real and distinct:

  x₁ = (-b + √Δ) / (2a) x₂ = (-b - √Δ) / (2a)

• If Δ = 0: The solution is real and repeated:

  x = -b / (2a)

• If Δ < 0: The solutions are complex:

  x₁ = (-b + i√|Δ|) / (2a) x₂ = (-b - i√|Δ|) / (2a)

Special Cases in Solving Equations

Special Case 1: Infinite Solutions

An equation can have infinitely many solutions when the equality holds true for any value of the variable.

Example:

4x + 1 + x = 3 + 5x - 2

Simplify both sides:

4x + x - 5x = 3 - 2 - 1
0 = 0

The resulting equality, 0 = 0, is always true regardless of the value of ( x ).
Thus, the equation has infinitely many solutions.

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Special Case 2: No Solution

An equation may have no solution when the equality is false for any value of the variable.

Example:

3x + 2 = 2x + 5 + x

Simplify both sides:

3x - 2x - x = 5 - 2

0 = 3

The resulting equality, 0 = 3 , is false.
Thus, the equation has no solution.