7.3 The Quadratic Formula, Completing the Square, & Graphs of Quadratics

7.3 The Quadratic Formula, Completing the Square, & Graphs of Quadratics

While factoring is efficient for simple integer relationships, many real-world quadratic equations cannot be factored using rational numbers. On TABE 13&14 Mathematics Levels D and A, you must be equipped with universal methods—the Quadratic Formula and Completing the Square—to solve any quadratic equation, evaluate the Discriminant, and analyze the geometric anatomy of parabolic graphs.

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1. The Quadratic Formula

The Quadratic Formula is a universal algebraic formula derived by completing the square on the general quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$):

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

The 4-Step Execution Protocol

1. Standard Form: Rearrange the equation so one side equals zero: $ax^2 + bx + c = 0$.
2. Identify Coefficients: Extract $a$, $b$, and $c$ with their precise signs.
3. Compute the Radicand (Discriminant): Evaluate $b^2 - 4ac$.
4. Simplify the Fraction: Evaluate the $\pm$ numerator and divide by $2a$.

Worked Example 1: Rational Solutions

Solve $2x^2 + 5x - 3 = 0$.

• Identify coefficients: $a = 2, \, b = 5, \, c = -3$.
• Calculate discriminant: $b^2 - 4ac = (5)^2 - 4(2)(-3) = 25 + 24 = 49$.
• Apply formula: $$x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4}$$
• Branch into two solutions: $$x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}, \qquad x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$$

Solution Set: $x = \left\{-3, \frac{1}{2}\right\}$

Worked Example 2: Irrational Radical Solutions

Solve $x^2 - 6x + 2 = 0$.

• Identify coefficients: $a = 1, \, b = -6, \, c = 2$.
• Calculate discriminant: $b^2 - 4ac = (-6)^2 - 4(1)(2) = 36 - 8 = 28$.
• Apply formula and simplify radical: $$x = \frac{-(-6) \pm \sqrt{28}}{2(1)} = \frac{6 \pm \sqrt{4 \cdot 7}}{2} = \frac{6 \pm 2\sqrt{7}}{2} = \mathbf{3 \pm \sqrt{7}}$$

Exact Solutions: $x = 3 + \sqrt{7}$ and $x = 3 - \sqrt{7}$ (approx. $5.65$ and $0.35$).

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2. The Discriminant ($D = b^2 - 4ac$) & Solution Diagnosis

The algebraic expression located beneath the radical symbol, $D = b^2 - 4ac$, is called the discriminant. It diagnoses the number and nature of solutions before completing calculations:

Discriminant D = b^2 - 4ac:
├── D > 0 (Positive) ──> Two Distinct Real Solutions (2 x-intercepts)
│     ├── Perfect Square ──> 2 Rational Solutions (Factorable)
│     └── Non-Square    ──> 2 Irrational Solutions (Radicals)
├── D = 0 (Zero)     ──> Exactly One Repeated Real Solution (1 x-intercept / Tangent Vertex)
└── D < 0 (Negative) ──> No Real Solutions / Two Complex Solutions (0 x-intercepts)

Worked Example 3: Determine the nature of solutions for $3x^2 - 4x + 5 = 0$.

• $a = 3, \, b = -4, \, c = 5$
• $D = (-4)^2 - 4(3)(5) = 16 - 60 = -44$
• Conclusion: Because $D < 0$, the equation has no real solutions ($2$ complex conjugate solutions) and its graph never intersects the $x$-axis.

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3. Completing the Square & Vertex Form

Completing the square is an algebraic technique that transforms any monic binomial $x^2 + bx$ into a perfect square trinomial by adding the constant term $\left(\frac{b}{2}\right)^2$:

$$x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$$

Solving Equations by Completing the Square

Worked Example 4: Solve $x^2 + 8x - 9 = 0$.

1. Isolate variable terms: $x^2 + 8x = 9$
2. Add $\left(\frac{b}{2}\right)^2$ to both sides: $\left(\frac{8}{2}\right)^2 = 4^2 = 16$ $$x^2 + 8x + 16 = 9 + 16 \implies (x + 4)^2 = 25$$
3. Take square root of both sides: $x + 4 = \pm \sqrt{25} \implies x + 4 = \pm 5$
4. Solve for $x$: $x = -4 \pm 5 \implies x = 1 \quad \text{or} \quad x = -9$

Converting to Vertex Form: $y = a(x - h)^2 + k$

Vertex form immediately identifies the coordinates of the parabola's turning point (vertex) at $(h, k)$ and its axis of symmetry at $x = h$.

Worked Example 5: Convert $y = x^2 - 6x + 13$ to vertex form.

$$\begin{aligned} y &= (x^2 - 6x) + 13 \\ y &= \left(x^2 - 6x + 9\right) + 13 - 9 \quad \text{(Add and subtract } (-6/2)^2 = 9\text{)} \\ y &= (x - 3)^2 + 4 \end{aligned}$$

Parabola Characteristics: Vertex is at $(3, 4)$; Axis of Symmetry is $x = 3$; minimum value is $y = 4$.

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4. Anatomy of a Parabola Graph

The graph of any quadratic function $f(x) = ax^2 + bx + c$ is a smooth, U-shaped curve called a parabola.

Parabola Anatomy (a > 0, Opens Upward):
           │              Axis of Symmetry (x = -b / 2a)
           │                     │
      \    │    /                │
       \   │   /                 │
  ──────\──┼──/─────── x-intercepts (Roots where y = 0)
         \ │ /                   │
          \│/                    │
        ───●─── y-intercept (0, c)│
           │                     │
           └──────────────●──────┘
                     Vertex (h, k) [Global Minimum]

Core Geometric Features of $f(x) = ax^2 + bx + c$

1. Orientation (Concavity):
   • If $a > 0$, the parabola opens upward ($\bigcup$), and the vertex is a minimum point.
   • If $a < 0$, the parabola opens downward ($\bigcap$), and the vertex is a maximum point.
2. Axis of Symmetry: The vertical line of bilateral symmetry passing through the vertex: $$x = -\frac{b}{2a}$$
3. Vertex Coordinates: The turning point $(h, k)$: $$h = -\frac{b}{2a}, \qquad k = f\left(-\frac{b}{2a}\right)$$
4. $y$-Intercept: Found by evaluating $f(0)$: located at $(0, c)$.
5. $x$-Intercepts (Roots / Zeros): Found by setting $f(x) = 0$ and solving via factoring or the Quadratic Formula.

Worked Example 6: Find all key features of $f(x) = -2x^2 + 12x - 10$.

• Orientation: $a = -2 < 0 \implies$ opens downward (vertex is a maximum).
• Axis of Symmetry: $x = -\frac{12}{2(-2)} = -\frac{12}{-4} = \mathbf{3}$.
• Vertex: $f(3) = -2(3)^2 + 12(3) - 10 = -18 + 36 - 10 = 8 \implies \mathbf{(3, 8)}$.
• $y$-Intercept: $f(0) = -10 \implies \mathbf{(0, -10)}$.
• $x$-Intercepts: $-2(x^2 - 6x + 5) = 0 \implies -2(x - 1)(x - 5) = 0 \implies \mathbf{(1, 0) \text{ and } (5, 0)}$.

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5. Real-World Applications: Projectile Motion & Trajectory Modeling

In classical mechanics, the vertical position of an object propelled into the air under the influence of Earth's gravity is modeled by the standard quadratic equation:

$$h(t) = -16t^2 + v_0 t + h_0 \quad (\text{feet}) \qquad \text{or} \qquad h(t) = -4.9t^2 + v_0 t + h_0 \quad (\text{meters})$$

• $h(t) = \text{height at time } t$
• $t = \text{time in seconds}$
• $v_0 = \text{initial vertical velocity}$
• $h_0 = \text{initial launch height}$
• $-16 = \frac{1}{2}g = \frac{1}{2}(-32\text{ ft/s}^2)$

Applied Physics Problem: Flare Trajectory

A marine emergency flare is fired upward from the deck of a boat $32\text{ feet}$ above sea level with an initial vertical velocity of $80\text{ ft/s}$. Its height equation is:

$$h(t) = -16t^2 + 80t + 32$$

1. Time to Reach Maximum Height: $$t_{\text{max}} = -\frac{b}{2a} = -\frac{80}{2(-16)} = \frac{80}{32} = \mathbf{2.5\text{ seconds}}$$
2. Maximum Height Attained: $$h(2.5) = -16(2.5)^2 + 80(2.5) + 32 = -16(6.25) + 200 + 32 = -100 + 200 + 32 = \mathbf{132\text{ feet}}$$
3. Total Time of Flight (When Flare Hits the Water $h = 0$): $$-16t^2 + 80t + 32 = 0 \implies -16(t^2 - 5t - 2) = 0 \implies t^2 - 5t - 2 = 0$$ $$t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-2)}}{2(1)} = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 \pm \sqrt{33}}{2}$$ Since time must be positive ($t > 0$): $$t = \frac{5 + \sqrt{33}}{2} \approx \frac{5 + 5.74}{2} = \frac{10.74}{2} \approx \mathbf{5.37\text{ seconds}}$$