Quadratic Equation
1. Standard Form:
\[
ax^2 + bx + c = 0
\]
2. Solutions
The solutions are given by:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
3. Example:
Solve the quadratic equation \( 2x^2 – 4x – 6 = 0 \).
Step 1: Identify \( a = 2 \), \( b = -4 \), and \( c = -6 \).
Step 2: Compute the discriminant \( D \):
\[
D = (-4)^2 – 4(2)(-6) = 16 + 48 = 64
\]
Since \( D > 0 \), the roots are real and unequal.
Step 3: Apply the quadratic formula:
\[
x = \frac{-(-4) \pm \sqrt{64}}{2(2)} = \frac{4 \pm 8}{4}
\]
Step 4: Compute the roots:
\[
x_1 = \frac{4 + 8}{4} = 3, \quad x_2 = \frac{4 – 8}{4} = -1
\]
Solution: The roots are \( x_1 = 3 \) and \( x_2 = -1 \).
Cubic Equation
1. Standard Form:
\[
x^3 + a_1x^2 + a_2x + a_3 = 0
\]
2. Solutions:
Use the discriminant \( D = Q^3 + R^2 \) to classify the roots.
3. Example:
Solve \( x^3 – 6x^2 + 11x – 6 = 0 \).
Step 1: Identify \( a_1 = -6 \), \( a_2 = 11 \), and \( a_3 = -6 \).
Step 2: Compute \( Q \) and \( R \):
\[
Q = \frac{3(11) – (-6)^2}{9} = \frac{33 – 36}{9} = -\frac{1}{3}, \quad R = \frac{9(-6)(11) – 27(-6) – 2(-6)^3}{54} = 0
\]
Since \( R = 0 \), all roots are real and unequal.
Step 3: Factor the cubic equation by trial or synthetic division. We try \( x = 1 \), and it works. So, divide \( x^3 – 6x^2 + 11x – 6 \) by \( (x – 1) \).
Step 4: After division, we get \( (x – 1)(x^2 – 5x + 6) = 0 \).
Step 5: Solve the quadratic \( x^2 – 5x + 6 = 0 \):
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 – 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 – 24}}{2} = \frac{5 \pm 1}{2}
\]
So, \( x_1 = 3 \) and \( x_2 = 2 \).
Solution: The roots are \( x = 1 \), \( x = 2 \), and \( x = 3 \).
Quartic Equation
1. Standard Form:
\[
x^4 + a_1x^3 + a_2x^2 + a_3x + a_4 = 0
\]
2. Solutions:
Solve by factoring into two quadratics or using a related cubic equation.
3. Example:
Solve \( x^4 – 8x^3 + 22x^2 – 24x + 9 = 0 \).
Step 1: Try factoring. We use synthetic division or trial roots. Trying \( x = 1 \), we find it works.
Step 2: Divide \( x^4 – 8x^3 + 22x^2 – 24x + 9 \) by \( (x – 1) \). This gives \( (x – 1)(x^3 – 7x^2 + 15x – 9) = 0 \).
Step 3: Factor the cubic equation \( x^3 – 7x^2 + 15x – 9 \) by trial. We find \( x = 1 \) works again.
Step 4: Divide the cubic by \( (x – 1) \) to get \( (x – 1)(x – 3)(x^2 – 4x + 3) = 0 \).
Step 5: Solve the quadratic \( x^2 – 4x + 3 = 0 \):
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(3)}}{2(1)} = \frac{4 \pm \sqrt{16 – 12}}{2} = \frac{4 \pm 2}{2}
\]
So, \( x_1 = 3 \) and \( x_2 = 1 \).
Solution: The roots are \( x = 1 \) (repeated), \( x = 3 \), and \( x = 3 \).
Formula Explained
Solving Algebraic Equations: Quadratic, Cubic, and Quartic
Algebraic equations can be classified based on their degree, such as quadratic (degree 2), cubic (degree 3), and quartic (degree 4). Here’s how to solve them
1. Quadratic Equations (Degree 2)
A quadratic equation takes the form:
\[ ax^2 + bx + c = 0 \]
Solution Methods:
Quadratic Formula:
The most common way to solve is using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
– Discriminant (\(b^2 – 4ac\)) determines the nature of the roots:
– If \(b^2 – 4ac > 0\), two real and distinct roots.
– If \(b^2 – 4ac = 0\), one real and repeated root.
– If \(b^2 – 4ac < 0\), two complex roots.
Factoring:
If the quadratic can be factored, it’s the easiest method:
\[
ax^2 + bx + c = (px + q)(rx + s) = 0
\]
Solve by setting each factor equal to 0.
Completing the Square:
You can complete the square to transform the equation and solve for \(x\):
\[
x^2 + bx = -(c/a) \quad \Rightarrow \quad \left( x + \frac{b}{2} \right)^2 = d
\]
2. Cubic Equations (Degree 3)
A cubic equation has the form:
\[ ax^3 + bx^2 + cx + d = 0 \]
Solution Methods:
Factorization:
– Try factoring out common terms or using synthetic division.
– If one root \(r\) is known (by trial or factorization), you can use synthetic division or polynomial division to reduce the cubic equation to a quadratic, which can be solved.
Cardano’s Method (General Cubic Formula):
For more complex cubics, Cardano’s method is used, though it’s quite involved. It involves a substitution of variables and then solving a depressed cubic.
3. Quartic Equations (Degree 4)
A quartic equation has the form:
\[ ax^4 + bx^3 + cx^2 + dx + e = 0 \]
Solution Methods:
Factoring:
Similar to cubic equations, if you can find one or more roots, factor out and reduce the equation to a quadratic or cubic.
Ferrari’s Method:
This is a general method to solve quartic equations. It’s more complex and involves converting the quartic into a cubic equation by completing the square and solving step by step.
– Special Case (Bi-quadratic equations):
If the equation is of the form \(ax^4 + bx^2 + c = 0\), it can be solved by substituting \(y = x^2\), solving the resulting quadratic in \(y\), and then solving for \(x\).