Quadratic Equations

A quadratic equation is a polynomial equation of the second degree. This means it contains at least one term that is squared. The general form of a quadratic equation is:
ax² + bx + c = 0

Here, 'x' is the variable we are trying to solve for. The coefficients 'a', 'b', and 'c' are constants, with the crucial condition that a ≠ 0. If 'a' were 0, the x² term would disappear, and it would become a linear equation (bx + c = 0), not a quadratic one.

The solutions to a quadratic equation are also known as its roots. A quadratic equation can have at most two roots. These roots can be real or complex, and they can be distinct or repeated.

Methods to Solve Quadratic Equations

There are several methods to find the roots of a quadratic equation. The most common ones are:

1. Factoring
2. Completing the Square
3. Using the Quadratic Formula

1. Factoring Method

This method involves rewriting the quadratic expression as a product of two linear factors. If we can express ax² + bx + c as (px + q)(rx + s), then setting this product to zero gives us (px + q)(rx + s) = 0. This implies either px + q = 0 or rx + s = 0, which are simple linear equations that can be easily solved for x.

Steps for Factoring:

• Ensure the equation is in the standard form ax² + bx + c = 0.
• Find two numbers that multiply to give 'ac' (the product of the coefficient of x² and the constant term) and add up to 'b' (the coefficient of x).
• Rewrite the middle term (bx) using these two numbers. For example, if the numbers are m and n, rewrite bx as mx + nx.
• Group the terms and factor out the common factors from each group.
• Set each factor to zero and solve for x.

Example: Solve x² + 5x + 6 = 0 by factoring.

Here, a=1, b=5, c=6. We need two numbers that multiply to ac = 1 * 6 = 6 and add up to b = 5. These numbers are 2 and 3.

Rewrite the equation: x² + 2x + 3x + 6 = 0

Group the terms: (x² + 2x) + (3x + 6) = 0

Factor out common terms: x(x + 2) + 3(x + 2) = 0

Factor out the common binomial (x + 2): (x + 2)(x + 3) = 0

Set each factor to zero:

• x + 2 = 0 => x = -2
• x + 3 = 0 => x = -3

The roots are -2 and -3.

2. Completing the Square Method

This method involves manipulating the equation to create a perfect square trinomial on one side. A perfect square trinomial is of the form (x + k)² or (x - k)².

Steps for Completing the Square:

• Start with the standard form: ax² + bx + c = 0.
• If a ≠ 1, divide the entire equation by 'a' to make the coefficient of x² equal to 1: x² + (b/a)x + (c/a) = 0.
• Move the constant term (c/a) to the right side of the equation: x² + (b/a)x = -c/a.
• Take half of the coefficient of the x term (which is b/a), square it ((b/2a)² = b²/4a²), and add it to both sides of the equation. This completes the square on the left side.
• The left side now becomes a perfect square: (x + b/2a)² = -c/a + b²/4a².
• Simplify the right side: (x + b/2a)² = (b² - 4ac) / 4a².
• Take the square root of both sides: x + b/2a = ±√((b² - 4ac) / 4a²).
• Simplify the square root on the right side: x + b/2a = ±(√(b² - 4ac)) / 2a.
• Isolate x: x = -b/2a ± (√(b² - 4ac)) / 2a.
• Combine the terms on the right side: x = (-b ± √(b² - 4ac)) / 2a.

Notice that the final step yields the quadratic formula. This method is fundamental for deriving the formula itself.

Example: Solve x² + 6x + 5 = 0 by completing the square.

1. Move the constant term: x² + 6x = -5.

2. Coefficient of x is 6. Half of it is 3. Square it: 3² = 9. Add 9 to both sides: x² + 6x + 9 = -5 + 9.

3. The left side is now a perfect square: (x + 3)² = 4.

4. Take the square root of both sides: x + 3 = ±√4.

5. Simplify: x + 3 = ±2.

6. Solve for x:

• x + 3 = 2 => x = 2 - 3 = -1
• x + 3 = -2 => x = -2 - 3 = -5

The roots are -1 and -5.

3. Quadratic Formula Method

This is the most direct and universally applicable method for solving any quadratic equation. It is derived from the completing the square method and provides the roots directly in terms of the coefficients a, b, and c.

The quadratic formula is:
x = [-b ± √(b² - 4ac)] / 2a

This formula gives the two roots:
x₁ = [-b + √(b² - 4ac)] / 2a
x₂ = [-b - √(b² - 4ac)] / 2a

Example: Solve 2x² - 5x + 3 = 0 using the quadratic formula.

Here, a = 2, b = -5, c = 3.

Substitute these values into the formula:

x = [-(-5) ± √((-5)² - 4 * 2 * 3)] / (2 * 2)

x = [5 ± √(25 - 24)] / 4

x = [5 ± √1] / 4

x = (5 ± 1) / 4

So, the two roots are:

• x₁ = (5 + 1) / 4 = 6 / 4 = 3/2
• x₂ = (5 - 1) / 4 = 4 / 4 = 1

The roots are 3/2 and 1.

The Discriminant (D)

The term inside the square root in the quadratic formula, b² - 4ac, is called the discriminant, often denoted by 'D'. The value of the discriminant tells us about the nature of the roots without actually calculating them.

The nature of the roots based on the discriminant:

Discriminant (D = b2 - 4ac)	Nature of Roots
D > 0	Two distinct real roots.
D = 0	Two equal real roots (or one repeated real root).
D < 0	Two complex conjugate roots (no real roots).

Example: Determine the nature of the roots for 3x² - 4x + 5 = 0.

Here, a = 3, b = -4, c = 5.

Calculate the discriminant:

D = b² - 4ac

D = (-4)² - 4 * 3 * 5

D = 16 - 60

D = -44

Since D < 0, the equation has two complex conjugate roots.

Relationship Between Roots and Coefficients

For a quadratic equation ax² + bx + c = 0, if α (alpha) and β (beta) are the roots, then there are important relationships between these roots and the coefficients:

• Sum of the roots (α + β): α + β = -b/a
• Product of the roots (α * β): α * β = c/a

These relationships are extremely useful for solving problems where you might not need to find the individual roots but rather a combination of them.

Example: If the roots of the equation x² - 7x + 10 = 0 are α and β, find the value of α + β and α * β.

Here, a = 1, b = -7, c = 10.

Sum of roots: α + β = -b/a = -(-7)/1 = 7.

Product of roots: α * β = c/a = 10/1 = 10.

We can verify this by finding the roots: x² - 7x + 10 = 0 => (x-2)(x-5) = 0. Roots are 2 and 5. Sum = 2+5=7, Product = 2*5=10.

Constructing a Quadratic Equation from its Roots:

If you know the roots of a quadratic equation, say α and β, you can form the equation using the following structure:
x² - (Sum of roots)x + (Product of roots) = 0
Or, substituting the formulas:
x² - (α + β)x + αβ = 0

Example: Form a quadratic equation whose roots are 4 and -3.

Sum of roots = 4 + (-3) = 1.

Product of roots = 4 * (-3) = -12.

Using the formula: x² - (1)x + (-12) = 0

The equation is: x² - x - 12 = 0.

To check, solve x² - x - 12 = 0: (x-4)(x+3) = 0. Roots are 4 and -3. Correct.

Special Cases and Identities

Sometimes quadratic equations involve specific forms or require the use of algebraic identities.

• Equations reducible to quadratic form: These might involve higher powers of x (like x⁴) or fractions. They can often be solved by substitution. For example, an equation like ax⁴ + bx² + c = 0 can be solved by substituting y = x², turning it into ay² + by + c = 0. Solve for y, and then find x by taking the square root of y.
• Identities: Knowledge of algebraic identities can simplify problems involving roots. For example, if you need to find α² + β², you can use the identity (α + β)² = α² + 2αβ + β². Rearranging gives α² + β² = (α + β)² - 2αβ. Since you know α + β and αβ, you can calculate this value.

Example: If α and β are the roots of x² - 5x + 6 = 0, find the value of α² + β².

From the equation, a=1, b=-5, c=6.

Sum of roots: α + β = -(-5)/1 = 5.

Product of roots: α * β = 6/1 = 6.

We need α² + β². Using the identity:

α² + β² = (α + β)² - 2αβ

Substitute the values:

α² + β² = (5)² - 2(6)

α² + β² = 25 - 12

α² + β² = 13.

Graphical Interpretation

The graph of a quadratic function y = ax² + bx + c is a parabola. The roots of the quadratic equation ax² + bx + c = 0 correspond to the x-intercepts of this parabola (where y = 0).

• If the parabola intersects the x-axis at two distinct points, there are two distinct real roots (D > 0).
• If the parabola touches the x-axis at exactly one point (the vertex), there is one repeated real root (D = 0).
• If the parabola does not intersect the x-axis at all, there are no real roots (D < 0), meaning the roots are complex.

The direction of the parabola depends on the sign of 'a':

• If a > 0, the parabola opens upwards (U-shaped).
• If a < 0, the parabola opens downwards (∩-shaped).

The vertex of the parabola is located at x = -b/2a. This point represents the minimum value of the function if a > 0, or the maximum value if a < 0.