# Quick Math Solve X- Quadratic, Cubic, Quartic Equations

## Quick Math Solve X

1
##### Enter expression in this form

Example (linear):  2x+5

Example (cubic):  x^3+2x-5x+1

Use this calculator to resolve quick answer on simple equation, quadratic, cubic and quartic equations.

### Quadratic and Cubic Equations in Algebra

Quadratic and cubic equations are polynomial equations of degree 2 and 3, respectively. Solving these equations involves finding the values of the variable that satisfy the equation.

A quadratic equation has the general form:
$ax^2 + bx + c = 0$
where $$a$$, $$b$$, and $$c$$ are constants, and $$a \neq 0$$.

#### Methods to Solve Quadratic Equations:

1. Factoring: If the quadratic can be factored into two binomials.
2. Completing the Square: Rewriting the quadratic in the form $$(x – p)^2 = q$$.
3. Quadratic Formula: Using the formula $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$.
4. Graphing: Finding the x-intercepts of the quadratic function $$y = ax^2 + bx + c$$.

#### Example:

Solve $$2x^2 – 4x – 6 = 0$$ using the quadratic formula.

1. Identify $$a$$, $$b$$, and $$c$$:
$a = 2, \quad b = -4, \quad c = -6$

2. Plug into the quadratic formula:
$x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}$
$x = \frac{4 \pm \sqrt{16 + 48}}{4}$
$x = \frac{4 \pm \sqrt{64}}{4}$
$x = \frac{4 \pm 8}{4}$

3. Solve for $$x$$:
$x = \frac{4 + 8}{4} = 3 \quad \text{and} \quad x = \frac{4 – 8}{4} = -1$

So, the solutions are $$x = 3$$ and $$x = -1$$.

#### Cubic Equations

A cubic equation has the general form:
$ax^3 + bx^2 + cx + d = 0$
where $$a$$, $$b$$, $$c$$, and $$d$$ are constants, and $$a \neq 0$$.

#### Methods to Solve Cubic Equations:

1. Factoring by grouping: If possible, factor the cubic equation into simpler polynomials.
2. Rational Root Theorem: Test possible rational roots and factor the polynomial.
3. Cardano’s Formula: A formula for solving cubic equations (more complex and less commonly used).
4. Graphing: Finding the x-intercepts of the cubic function $$y = ax^3 + bx^2 + cx + d$$.

#### Example:

Solve $$x^3 – 6x^2 + 11x – 6 = 0$$ by factoring.

1. Factor by grouping:
$x^3 – 6x^2 + 11x – 6 = (x – 1)(x^2 – 5x + 6)$

2. Further factor $$x^2 – 5x + 6$$:
$x^2 – 5x + 6 = (x – 2)(x – 3)$

3. Combine factors:
$x^3 – 6x^2 + 11x – 6 = (x – 1)(x – 2)(x – 3)$

4. Set each factor to zero:
$x – 1 = 0 \quad \Rightarrow \quad x = 1$
$x – 2 = 0 \quad \Rightarrow \quad x = 2$
$x – 3 = 0 \quad \Rightarrow \quad x = 3$

So, the solutions are $$x = 1$$, $$x = 2$$, and $$x = 3$$.