Quadratic Formula Explained

Quadratic equations are second-degree polynomials of the form \( ax^2 + bx + c = 0 \), where \( a \neq 0 \). These equations model parabolas when graphed and are fundamental in algebra, appearing in physics (e.g., projectile motion), economics (e.g., profit maximization), and engineering. The quadratic formula offers a universal method to find the roots (values of \( x \) that satisfy the equation), even when factoring or completing the square is challenging. This guide explores the formula, its derivation, diverse examples including complex roots, graphical representations, and practical applications to deepen your understanding.

The Quadratic Formula

The quadratic formula provides the roots of \( ax^2 + bx + c = 0 \):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where:

  • \( a \): Coefficient of \( x^2 \) (must be non-zero)
  • \( b \): Coefficient of \( x \)
  • \( c \): Constant term
  • \( b^2 - 4ac \): Discriminant, determining the nature of roots:
    • \( > 0 \): Two distinct real roots
    • \( = 0 \): One real root (repeated)
    • \( < 0 \): Two complex conjugate roots

The \( \pm \) symbol indicates two solutions, corresponding to the parabola’s intersection with the x-axis (if real). This formula is derived from completing the square and applies to any quadratic equation, making it a powerful tool in algebraic problem-solving.

Examples: Various Quadratic Equations, Complex Roots, Word Problems, Discriminant Analysis, Optimization

Let’s solve different quadratic equations using the quadratic formula with detailed steps.

Example 1: Basic Quadratic

Solve \( x^2 - 5x + 6 = 0 \) (\( a = 1 \), \( b = -5 \), \( c = 6 \)).

\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} \] \[ x = \frac{5 \pm \sqrt{25 - 24}}{2} \] \[ x = \frac{5 \pm \sqrt{1}}{2} \] \[ x = \frac{5 \pm 1}{2} \]

Solutions:

\[ x = \frac{5 + 1}{2} = \frac{6}{2} = 3 \] \[ x = \frac{5 - 1}{2} = \frac{4}{2} = 2 \]

Roots are \( x = 3 \) and \( x = 2 \).

Example 2: Quadratic with Decimals

Solve \( 0.2x^2 + 1.5x - 3 = 0 \) (\( a = 0.2 \), \( b = 1.5 \), \( c = -3 \)).

\[ x = \frac{-1.5 \pm \sqrt{(1.5)^2 - 4(0.2)(-3)}}{2(0.2)} \] \[ x = \frac{-1.5 \pm \sqrt{2.25 - (-2.4)}}{0.4} \] \[ x = \frac{-1.5 \pm \sqrt{2.25 + 2.4}}{0.4} \] \[ x = \frac{-1.5 \pm \sqrt{4.65}}{0.4} \] \[ \sqrt{4.65} \approx 2.156 \] \[ x = \frac{-1.5 \pm 2.156}{0.4} \]

Solutions:

\[ x = \frac{-1.5 + 2.156}{0.4} \approx \frac{0.656}{0.4} \approx 1.64 \] \[ x = \frac{-1.5 - 2.156}{0.4} \approx \frac{-3.656}{0.4} \approx -9.14 \]

Roots are approximately \( x \approx 1.64 \) and \( x \approx -9.14 \).

Example 3: Complex Roots

Solve \( x^2 + 2x + 5 = 0 \) (\( a = 1 \), \( b = 2 \), \( c = 5 \)).

\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(5)}}{2(1)} \] \[ x = \frac{-2 \pm \sqrt{4 - 20}}{2} \] \[ x = \frac{-2 \pm \sqrt{-16}}{2} \] \[ \sqrt{-16} = 4i \] \[ x = \frac{-2 \pm 4i}{2} \]

Solutions:

\[ x = \frac{-2 + 4i}{2} = -1 + 2i \] \[ x = \frac{-2 - 4i}{2} = -1 - 2i \]

Roots are \( x = -1 + 2i \) and \( x = -1 - 2i \).

Example 4: Word Problem

A ball is thrown upward with initial velocity 20 m/s from 2 m height. Height \( h = -5t^2 + 20t + 2 \) (where \( t \) is time in seconds). When does it hit the ground?

\[ -5t^2 + 20t + 2 = 0 \] \[ \text{Multiply by -1: } 5t^2 - 20t - 2 = 0 \] \[ x = t, a = 5, b = -20, c = -2 \] \[ t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(5)(-2)}}{2(5)} \] \[ t = \frac{20 \pm \sqrt{400 - (-40)}}{10} \] \[ t = \frac{20 \pm \sqrt{440}}{10} \] \[ \sqrt{440} \approx 20.98 \] \[ t = \frac{20 \pm 20.98}{10} \]

Solutions:

\[ t = \frac{20 + 20.98}{10} \approx \frac{40.98}{10} \approx 4.10 \] \[ t = \frac{20 - 20.98}{10} \approx \frac{-0.98}{10} \approx -0.10 \]

Positive root \( t \approx 4.10 \) seconds is the solution.

Example 5: Discriminant Analysis

Analyze \( 2x^2 + 3x - 5 = 0 \) (\( a = 2 \), \( b = 3 \), \( c = -5 \)).

\[ \text{Discriminant} = b^2 - 4ac \] \[ = (3)^2 - 4(2)(-5) \] \[ = 9 - (-40) \] \[ = 9 + 40 \] \[ = 49 \] \[ \sqrt{49} = 7 \] \[ x = \frac{-3 \pm 7}{2(2)} \]

Solutions:

\[ x = \frac{-3 + 7}{4} = \frac{4}{4} = 1 \] \[ x = \frac{-3 - 7}{4} = \frac{-10}{4} = -2.5 \]

Discriminant > 0 indicates two real roots: \( x = 1 \) and \( x = -2.5 \).

Example 6: Optimization

Maximize area \( A = -x^2 + 10x \) (a downward parabola). Vertex (maximum) occurs at \( x = -\frac{b}{2a} \).

\[ a = -1, b = 10 \] \[ x = -\frac{10}{2(-1)} \] \[ x = \frac{-10}{-2} \] \[ x = 5 \] \[ A = -(5)^2 + 10(5) \] \[ = -25 + 50 \] \[ = 25 \]

Maximum area is 25 at \( x = 5 \).

Derivation (Completing the Square)

Start with \( ax^2 + bx + c = 0 \), divide by \( a \) (assuming \( a \neq 0 \)):

\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

Move \( \frac{c}{a} \) to the right:

\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]

Add \( \left(\frac{b}{2a}\right)^2 \) to both sides to complete the square:

\[ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \] \[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} \] \[ = \frac{b^2 - 4ac}{4a^2} \]

Take the square root:

\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \]

Solve for \( x \):

\[ x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Graphical View

The roots are where the parabola \( y = ax^2 + bx + c \) intersects the x-axis. Let’s visualize some examples.

Graph 1: \( y = x^2 - 5x + 6 \)

Roots at \( x = 2 \) and \( x = 3 \):

Graph 2: \( y = x^2 + 2x + 5 \)

No real roots (complex roots), vertex below x-axis:

Applications

Quadratic equations model various real-world phenomena. Here are detailed examples:

  • Projectile Motion: Height \( h = -16t^2 + 40t \) (feet, seconds). When does it hit the ground?
    \[ -16t^2 + 40t = 0 \] \[ t(-16t + 40) = 0 \] \[ t = 0 \text{ or } -16t + 40 = 0 \] \[ -16t = -40 \] \[ t = 2.5 \]
    Hits ground at \( t = 2.5 \) seconds.
  • Profit Maximization: Profit \( P = -2x^2 + 100x - 500 \). Find maximum profit.
    \[ x = -\frac{100}{2(-2)} \] \[ x = 25 \] \[ P = -2(25)^2 + 100(25) - 500 \] \[ = -1250 + 2500 - 500 \] \[ = 750 \]
    Maximum profit is $750 at \( x = 25 \).
  • Area Optimization: A rectangular field with perimeter 100 m, area \( A = x(50 - x) \). Maximize area.
    \[ A = -x^2 + 50x \] \[ x = -\frac{50}{2(-1)} \] \[ x = 25 \] \[ A = 25(50 - 25) \] \[ = 25 \times 25 \] \[ = 625 \]
    Maximum area is 625 m².
  • Engineering - Beam Deflection: Deflection \( d = 0.01x^2 - 0.5x + 5 \). Find zero deflection.
    \[ 0.01x^2 - 0.5x + 5 = 0 \] \[ x = \frac{-(-0.5) \pm \sqrt{(-0.5)^2 - 4(0.01)(5)}}{2(0.01)} \] \[ x = \frac{0.5 \pm \sqrt{0.25 - 0.2}}{0.02} \] \[ x = \frac{0.5 \pm \sqrt{0.05}}{0.02} \] \[ \sqrt{0.05} \approx 0.223 \] \[ x = \frac{0.5 \pm 0.223}{0.02} \]

    Solutions:

    \[ x = \frac{0.723}{0.02} \approx 36.15 \] \[ x = \frac{0.277}{0.02} \approx 13.85 \]
    Zero deflection at \( x \approx 13.85 \) or \( 36.15 \).

  • Economics - Cost Function: Cost \( C = 0.03x^2 - 2x + 100 \). Find break-even points (C = 0).
    \[ 0.03x^2 - 2x + 100 = 0 \] \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(0.03)(100)}}{2(0.03)} \] \[ x = \frac{2 \pm \sqrt{4 - 12}}{0.06} \] \[ x = \frac{2 \pm \sqrt{-8}}{0.06} \] \[ \sqrt{-8} = 2\sqrt{2}i \approx 2.828i \] \[ x = \frac{2 \pm 2.828i}{0.06} \]

    Complex roots indicate no real break-even points.

  • Agriculture - Crop Yield: Yield \( Y = -0.01x^2 + 4x \) (tons, acres). Maximize yield.
    \[ x = -\frac{4}{2(-0.01)} \] \[ x = 200 \] \[ Y = -0.01(200)^2 + 4(200) \] \[ = -400 + 800 \] \[ = 400 \]
    Maximum yield is 400 tons at 200 acres.