Quick Math Solve X
Enter expression in this form
Example (linear): 2x+5
Example (quadratic): 2x^2+5x +1
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.
Quadratic Equations
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 \( x = 3 \).