7.2 Quadratic Equations, Factoring & Quadratic Formula

7.2 Quadratic Equations, Factoring & Quadratic Formula

Quadratic equations represent a major segment of the advanced algebra questions on the AFQT Mathematics Knowledge subtest. A quadratic equation is a single-variable polynomial equation of degree 2. Mastering multiple methods to solve quadratics—by factoring, extracting square roots, and applying the quadratic formula—ensures you can select the fastest and most reliable approach under exam timed conditions.

1. Standard Form of a Quadratic Equation

Every quadratic equation can be written in standard form:

$$ax^2 + bx + c = 0$$

where $a$, $b$, and $c$ are real constants, and $a \neq 0$.

• $ax^2$: The quadratic term ($a$ is the quadratic coefficient).
• $bx$: The linear term ($b$ is the linear coefficient).
• $c$: The constant term.

The solutions to a quadratic equation are also called the roots, zeros, or x-intercepts (where the graph of the parabola $y = ax^2 + bx + c$ crosses the x-axis).

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2. Solving Quadratics by Factoring & Zero-Product Property

Factoring is usually the fastest method to solve a quadratic equation on the AFQT when the coefficients are simple integers.

The Zero-Product Property

If the product of two real expressions $A$ and $B$ is zero:

$$A \cdot B = 0 \implies A = 0 \quad \text{or} \quad B = 0$$

Step-by-Step Factoring Method

1. Set the equation equal to zero: Move all terms to one side so the equation takes the form $ax^2 + bx + c = 0$.
2. Factor out any GCF if applicable.
3. Factor the trinomial into two linear binomial factors: $(px + q)(rx + s) = 0$.
4. Apply the Zero-Product Property: Set each linear factor equal to zero.
5. Solve each linear equation for $x$.

Worked Example 1: Basic Factoring

Solve $x^2 - 7x + 12 = 0$.

1. The equation is already in standard form with $a = 1, b = -7, c = 12$.
2. Find two factors of $+12$ that add up to $-7$. The numbers are $-3$ and $-4$ (since $(-3)(-4) = 12$ and $-3 + (-4) = -7$).
3. Factor the trinomial: $$(x - 3)(x - 4) = 0$$
4. Set each factor to zero: $$x - 3 = 0 \implies x = 3$$ $$x - 4 = 0 \implies x = 4$$
5. The solutions are $x = 3$ and $x = 4$.

Worked Example 2: Non-zero Right-Hand Side

Solve $2x^2 + 5x = 12$.

1. Rearrange to standard form: $$2x^2 + 5x - 12 = 0$$
2. Factor by grouping or inspection:
   • We need factors of $a \cdot c = 2 \cdot (-12) = -24$ that sum to $b = 5$.
   • The factors are $+8$ and $-3$ (since $8 \cdot (-3) = -24$ and $8 + (-3) = 5$).
   • Rewrite middle term: $2x^2 + 8x - 3x - 12 = 0$.
   • Group terms: $2x(x + 4) - 3(x + 4) = 0$.
   • Factored form: $(2x - 3)(x + 4) = 0$.
3. Set factors to zero: $$2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$$ $$x + 4 = 0 \implies x = -4$$
4. The solutions are $x = \frac{3}{2}$ and $x = -4$.

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3. Solving by Taking Square Roots

When the linear term is absent ($b = 0$), the equation simplifies to $ax^2 + c = 0 \implies x^2 = k$. You can solve directly by isolating $x^2$ and taking the square root of both sides.

Critical Rule: Remember to include both positive and negative roots: $\sqrt{x^2} = |x| \implies x = \pm\sqrt{k}$.

Worked Example 3: Extraction of Roots

Solve $3x^2 - 48 = 0$.

1. Isolate $x^2$: $$3x^2 = 48 \implies x^2 = 16$$
2. Take the square root of both sides: $$x = \pm\sqrt{16} \implies x = \pm 4$$
3. The roots are $x = 4$ and $x = -4$.

Worked Example 4: Binomial Squared

Solve $(x - 5)^2 = 36$.

1. Take square roots of both sides: $$x - 5 = \pm\sqrt{36} = \pm 6$$
2. Split into two equations: $$x - 5 = 6 \implies x = 11$$ $$x - 5 = -6 \implies x = -1$$
3. The solutions are $x = 11$ and $x = -1$.

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4. The Quadratic Formula

When a quadratic equation cannot be easily factored using integers, the Quadratic Formula provides a guaranteed solution for any quadratic $ax^2 + bx + c = 0$:

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

Breakdown of Components

• $-b$: The opposite of the linear coefficient.
• $b^2 - 4ac$: The discriminant under the square root.
• $2a$: The denominator (entire numerator must be divided by $2a$).

Worked Example 5: Using the Quadratic Formula

Solve $x^2 - 6x + 4 = 0$.

1. Identify coefficients: $a = 1, b = -6, c = 4$.
2. Substitute into the formula: $$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)}$$
3. Simplify step-by-step: $$x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2}$$
4. Simplify the radical ($\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$): $$x = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$$
5. The solutions are $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}$.

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5. The Discriminant ($b^2 - 4ac$)

The portion of the quadratic formula under the square root, $D = b^2 - 4ac$, is called the discriminant. It reveals the number and nature of the roots without requiring full resolution of the equation.

Worked Example 6: Discriminant Analysis

Determine the number of real roots for $3x^2 - 4x + 5 = 0$.

1. Identify $a = 3, b = -4, c = 5$.
2. Compute $D = b^2 - 4ac$: $$D = (-4)^2 - 4(3)(5) = 16 - 60 = -44$$
3. Since $D = -44 < 0$, the equation has 0 real roots (two complex conjugate roots).

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6. Word Problems Involving Quadratic Equations

On the AFQT, quadratic word problems frequently involve geometry (area of rectangles), consecutive integers, or motion.

Worked Example 7: Geometric Word Problem

The length of a rectangular tactical field is $4$ meters longer than its width. If the area of the field is $60$ square meters, find its dimensions.

1. Define variables: Let width $= w$. Then length $= w + 4$.
2. Set up equation: Area $=$ length $\cdot$ width $\implies w(w + 4) = 60$.
3. Expand and put in standard form: $$w^2 + 4w - 60 = 0$$
4. Factor the quadratic:
   • Find two numbers multiplying to $-60$ and adding to $+4$: $+10$ and $-6$.
   $$(w + 10)(w - 6) = 0$$
5. Solve for $w$: $$w = -10 \quad \text{or} \quad w = 6$$
6. Reject non-physical solutions: A distance/width cannot be negative, so $w = 6$ meters.
7. Calculate length: Length $= w + 4 = 6 + 4 = 10$ meters.
   • Check: $6 \cdot 10 = 60\text{ m}^2$. Correct!

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7. AFQT Strategies, Tips & Common Errors

• Don't Forget the $\pm$ Sign: When taking square roots of both sides of $x^2 = 25$, $x = \pm 5$. Leaving off the negative solution is a frequent error.
• Never Divide by the Variable: Given $x^2 = 5x$, do NOT divide both sides by $x$ to get $x = 5$, because you lose the root $x = 0$! Instead, subtract $5x$ to get $x^2 - 5x = 0 \implies x(x - 5) = 0$, giving solutions $x = 0$ and $x = 5$.
• Back-Solving on Multiple-Choice Questions: If an AFQT question asks for the solution to a quadratic equation, plug the given answer choices back into the original equation to find which number makes the statement true!