Solve quadratic equations. - GED Math

SECTION 1

Historical Context & Motivation

People have been solving quadratic equations for thousands of years, long before anyone called them "quadratic." Ancient civilizations needed to calculate areas of land, design buildings, and predict the paths of projectiles. Whenever the quantity you are looking for is squared—multiplied by itself—you are dealing with a quadratic equation. Understanding where these equations come from helps you see why they appear so often on the GED and in everyday life.

~2000 BCE • Babylonian Clay Tablets

Babylonian scribes solved area problems that we would now write as quadratic equations, using a method similar to completing the square—all without modern notation.

~300 BCE • Greek Geometric Algebra

Euclid's Elements showed how to solve quadratic problems by constructing geometric shapes, connecting algebra to visual reasoning.

~820 CE • Al-Khwarizmi's Algebra

The Persian mathematician al-Khwarizmi published systematic rules for solving quadratics. The word "algebra" itself comes from the title of his book.

1637 • Descartes & Modern Notation

René Descartes introduced the notation we use today—letters like x for unknowns and superscripts for exponents—making quadratic equations easier to write and solve.

Today, quadratic equations show up in contexts ranging from calculating the trajectory of a basketball to figuring out profit in a business plan. On the GED, roughly 55 percent of the test focuses on algebraic problem solving, and quadratics are a core part of that. The good news: there are clear, step-by-step methods you can learn to solve every quadratic equation you will encounter.

SECTION 2

Core Principles & Definitions

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b, and c are numbers and a is not zero. The highest power of the variable x is 2, which is why the graph of a quadratic makes a U-shaped curve called a parabola. Before you can solve a quadratic, you need to understand a few foundational ideas.

Standard Form

Always rearrange the equation so it reads ax² + bx + c = 0. Everything on one side, zero on the other. This is your starting point for every method.

Two Solutions (Usually)

Because the variable is squared, a quadratic equation can have two solutions, one solution, or no real solutions. The GED focuses on equations with one or two real solutions.

Three Solving Methods

You can solve by factoring, by using the quadratic formula, or (less commonly on the GED) by completing the square. Choosing the right method saves time.

Check Your Work

Plug each solution back into the original equation. If both sides are equal, your answer is correct. This verification step catches sign errors and arithmetic mistakes.

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KEY TAKEAWAY

Think of a quadratic equation like asking, "At what two times does a ball thrown in the air reach a height of 20 feet?" The ball passes through that height once on the way up and once on the way down—that is why quadratics usually have two solutions.

SECTION 3

Visual Explanation — The Parabola and Its Solutions

The graph of a quadratic equation is a parabola. The solutions (also called roots or x-intercepts) are the x-values where the parabola crosses the horizontal axis. The diagram below shows the parabola for y = x² − 2x − 3, which factors into (x − 3)(x + 1) = 0. Notice the curve crosses the x-axis at x = −1 and x = 3.

The parabola for y = x² − 2x − 3 crosses the x-axis at the two solutions: x = −1 and x = 3. The vertex at (1, −4) is the lowest point of the curve.

When you solve a quadratic equation, you are finding those x-intercept values—the places where the parabola touches or crosses the x-axis. If the parabola just touches the axis at one point, the equation has one solution. If the parabola floats above or below the axis without crossing, there are no real solutions. On the GED, you will almost always work with equations that have one or two real solutions.

SECTION 4

Mathematical Framework — Three Solving Methods

There are three main methods for solving quadratic equations. Each one works, but some are faster depending on the equation. The GED formula sheet provides the quadratic formula, so you never need to memorize it—but you do need to know how to use it. Let's look at each method.

Method 1: Factoring

Factoring works when you can rewrite the quadratic as a product of two binomials. You rely on the zero product property: if A × B = 0, then either A = 0 or B = 0. So once you factor, set each factor equal to zero and solve.

FACTORING PATTERN

ax² + bx + c = 0 → (x + p)(x + q) = 0 → x = −p or x = −q

Find two numbers p and q that multiply to c and add to b (when a = 1). Set each factor to zero to find the solutions.

Method 2: The Quadratic Formula

The quadratic formula works for every quadratic equation, whether or not it factors neatly. It is provided on the GED formula sheet, so your job is to identify a, b, and c from standard form, plug them in, and simplify carefully.

QUADRATIC FORMULA

x = (−b ± √(b² − 4ac)) / (2a)

a = coefficient of x², b = coefficient of x, c = constant term. The ± symbol means you compute two values: one with + and one with −.

The Discriminant — How Many Solutions?

DISCRIMINANT

D = b² − 4ac

If D > 0 → two real solutions. If D = 0 → one real solution. If D < 0 → no real solutions. The discriminant is the expression under the square root in the quadratic formula.

Method 3: Square Root Method

When the equation has no bx term (b = 0) or can be written as (x + k)² = d, you can solve by taking the square root of both sides. Remember to include both the positive and negative square root.

SQUARE ROOT METHOD

x² = d → x = ±√d

This method is the fastest when there is no middle term (bx). For example, x² = 25 gives x = 5 or x = −5.

SECTION 5

Choosing the Right Method

Knowing which method to use is just as important as knowing how to use it. On the timed GED, picking the most efficient approach can save you valuable minutes. The decision tree below walks you through the process, and the comparison table summarizes when each method shines.

Start at the top and follow the arrows. If the bx term is missing, use the square root method. If the equation factors easily, use factoring. Otherwise, use the quadratic formula.

Method	Best When...	Example
Square Root	No bx term; equation is x² = number	x² − 16 = 0 → x = ±4
Factoring	Coefficients are small integers; factors are easy to spot	x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0
Quadratic Formula	Factoring is not obvious; messy coefficients; or you want a guaranteed approach	2x² − 3x − 7 = 0 → use the formula

Comparison of quadratic solving methods

SECTION 6

Worked Examples

Example A — Solving by Factoring

Solve x² + 2x − 15 = 0 by Factoring

1
Step 1 — Confirm Standard Form

The equation is already in the form ax² + bx + c = 0. Here, a = 1, b = 2, and c = −15.
2
Step 2 — Find Two Numbers That Multiply to c and Add to b

We need two numbers that multiply to −15 and add to 2. Try factor pairs of −15: (1, −15), (−1, 15), (3, −5), (−3, 5). The pair −3 and 5 works because −3 × 5 = −15 and −3 + 5 = 2.

Numbers: −3 and 5
3
Step 3 — Write in Factored Form

x² + 2x − 15 = (x − 3)(x + 5) = 0
4
Step 4 — Apply the Zero Product Property

Set each factor equal to zero: x − 3 = 0 gives x = 3. x + 5 = 0 gives x = −5.

x = 3 or x = −5
5
Step 5 — Check Both Solutions

For x = 3: (3)² + 2(3) − 15 = 9 + 6 − 15 = 0 ✓. For x = −5: (−5)² + 2(−5) − 15 = 25 − 10 − 15 = 0 ✓. Both solutions check out.

Example B — Solving with the Quadratic Formula

Solve 2x² − 5x − 3 = 0 Using the Quadratic Formula

1
Step 1 — Identify a, b, and c

From 2x² − 5x − 3 = 0: a = 2, b = −5, c = −3. Be careful with signs—b is negative.
2
Step 2 — Calculate the Discriminant

D = b² − 4ac = (−5)² − 4(2)(−3) = 25 + 24 = 49. Since D > 0, there are two real solutions.

D = 49
3
Step 3 — Plug into the Formula

x = (−b ± √D) / (2a) = (−(−5) ± √49) / (2 × 2) = (5 ± 7) / 4
4
Step 4 — Compute Both Solutions

Using +: x = (5 + 7) / 4 = 12/4 = 3. Using −: x = (5 − 7) / 4 = −2/4 = −1/2.

x = 3 or x = −0.5
5
Step 5 — Verify

For x = 3: 2(9) − 5(3) − 3 = 18 − 15 − 3 = 0 ✓. For x = −0.5: 2(0.25) − 5(−0.5) − 3 = 0.5 + 2.5 − 3 = 0 ✓.

SECTION 7

Common Mistakes & How to Avoid Them

Quadratic equations are not inherently difficult, but small errors in arithmetic or sign-handling can derail an entire problem. The table below summarizes the most frequent mistakes GED test-takers make and how to prevent them.

Common Mistake	Why It Happens	How to Fix It
Forgetting the ± sign	Students take only the positive square root	Always write ± immediately when taking a square root
Wrong sign for b	The formula says −b, so if b is already negative, −(−b) = +b	Write down a, b, c with their signs first, then substitute
Not setting equation = 0	Trying to factor while terms are on both sides	Move everything to one side before solving
Dividing by x instead of factoring	This loses the solution x = 0	Factor out x instead: x(x + 3) = 0 gives x = 0 or x = −3
Arithmetic errors in b² − 4ac	Squaring negative numbers or multiplying signs incorrectly	Compute b² first, then 4ac separately, then subtract

Five most common quadratic mistakes on the GED

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GED TEST TIP

On the GED, you can use the TI-30XS calculator for Part 2 to double-check your discriminant calculation. Compute b² first, store it, then compute 4ac, and subtract. This two-step approach catches sign errors. Remember: Part 1 has no calculator, so for those five questions, practice doing these calculations by hand with small, clean numbers.

SECTION 8

Connections to Advanced Topics

The skills you build solving quadratic equations form the foundation for more advanced algebra and beyond. While the GED tests the basics, understanding where these concepts lead can motivate your study and prepare you for college-level math.

GED Level	College Level
Solve ax² + bx + c = 0 for x	Solve polynomial equations of degree 3, 4, or higher
Graph a parabola and find x-intercepts	Analyze families of curves, transformations, and conic sections
Use the discriminant to count solutions	Work with complex numbers when D < 0
Apply quadratics to area and projectile problems	Use calculus to optimize area, profit, and motion

How GED quadratic skills connect to college math

If you score 165 or higher on the GED Mathematical Reasoning test, you earn the "College Ready" designation, which can help you place directly into credit-bearing college math courses. Mastering quadratics is one of the most important steps toward that goal because these equations appear in nearly every branch of higher mathematics and science.

SECTION 9

Practice Problems

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Before You Begin

Problems 1–2 should be attempted without a calculator, just like Part 1 of the GED. For Problems 3–5, feel free to use the TI-30XS or any calculator. Remember: the quadratic formula is on the GED formula sheet.

Question 1: CONCEPTUAL

A physics student is solving the equation x² − 9 = 0 as part of a projectile problem on a homework assignment. She recognizes that this equation has no middle term (no bx term). Which method is the most efficient way to solve this equation? Choose the best answer.

A. Use the quadratic formula with a = 1, b = 0, c = −9
B. Take the square root of both sides after adding 9
C. Factor into (x − 9)(x + 9) = 0
D. Graph the equation and estimate the roots

Show answer (B)

The correct answer is B. When there is no bx term, the square root method is the fastest: add 9 to both sides to get x² = 9, then x = ±3. Choice A (using the quadratic formula) would produce the correct answer but is unnecessarily slow for such a simple equation—it's like using a sledgehammer to hang a picture frame. Choice C is incorrect because x² − 9 factors as (x − 3)(x + 3), a difference of squares, not (x − 9)(x + 9). The product (x − 9)(x + 9) would equal x² − 81, not x² − 9. Choice D is the least efficient approach and is impractical without a graphing tool; estimating roots visually can easily lead to rounding errors.

Question 2: BASIC CALCULATION

Solve by factoring: x² + 7x + 12 = 0 What are the solutions?

A. x = 3 and x = 4
B. x = −3 and x = −4
C. x = −3 and x = 4
D. x = 3 and x = −4

Show answer (B)

The correct answer is B. We need two numbers that multiply to +12 and add to +7. The pair 3 and 4 multiplies to 12 and adds to 7, so x² + 7x + 12 = (x + 3)(x + 4) = 0. Setting each factor to zero: x + 3 = 0 gives x = −3, and x + 4 = 0 gives x = −4. Choice A reverses the signs—3 and 4 are the numbers used in the factors, but the solutions are their negatives (since x + 3 = 0 means x = −3). Choice C (x = −3 and x = 4) would come from factoring (x + 3)(x − 4) = 0, which gives x² − x − 12 = 0, not our equation. Choice D (x = 3 and x = −4) would come from (x − 3)(x + 4) = 0, which gives x² + x − 12 = 0, also not our equation.

Question 3: INTERMEDIATE

Use the quadratic formula to solve 3x² + 2x − 5 = 0. What are the two solutions?

A. x = 1 and x = −5/3
B. x = −1 and x = 5/3
C. x = 5 and x = −1/3
D. x = −5 and x = 1/3

Show answer (A)

The correct answer is A. With a = 3, b = 2, c = −5: the discriminant is b² − 4ac = 4 − 4(3)(−5) = 4 + 60 = 64. So √64 = 8. Plugging in: x = (−2 ± 8) / (2 × 3) = (−2 ± 8) / 6. Using +: x = (−2 + 8)/6 = 6/6 = 1. Using −: x = (−2 − 8)/6 = −10/6 = −5/3. Choice B has the signs reversed (x = −1 and x = 5/3), which happens when students forget the negative sign on b in the −b part of the formula. Choice C (x = 5 and x = −1/3) results from incorrectly computing −b as +2 and also making an error in the division step. Choice D (x = −5 and x = 1/3) results from confusing factors and solutions or mishandling signs throughout.

Question 4: APPLIED

A rectangular garden has a length that is 3 feet more than its width. The area of the garden is 70 square feet. If w represents the width in feet, the equation w² + 3w − 70 = 0 models this situation. What is the width of the garden?

A. 7 feet
B. 10 feet
C. −10 feet
D. 5 feet

Show answer (A)

The correct answer is A. Factor w² + 3w − 70 = 0 by finding two numbers that multiply to −70 and add to 3: those are 10 and −7. So (w + 10)(w − 7) = 0, giving w = −10 or w = 7. Since a width cannot be negative, w = 7 feet. The length is 7 + 3 = 10 feet, and 7 × 10 = 70 ✓. Choice B (10 feet) is the length, not the width. Choice C (−10 feet) is the rejected negative solution—a physical dimension cannot be negative. Choice D (5 feet) would give an area of 5 × 8 = 40 square feet, not 70.

Question 5: CRITICAL THINKING

3 pts

A ball is launched upward from a 6-foot-high platform. Its height in feet after t seconds is modeled by h = −16t² + 32t + 6. At what time does the ball hit the ground? Set h = 0 and solve using the quadratic formula. Round your answer to the nearest hundredth of a second. (Show your work.)

Show answer

Rubric (3 points): Point 1 — Set up: Student correctly sets −16t² + 32t + 6 = 0 and identifies a = −16, b = 32, c = 6 (or equivalently multiplies through by −1 to get 16t² − 32t − 6 = 0, or divides by −2 to get 8t² − 16t − 3 = 0). Point 2 — Discriminant and formula application: Using original coefficients a = −16, b = 32, c = 6: Discriminant D = b² − 4ac = (32)² − 4(−16)(6) = 1024 + 384 = 1408. Then t = (−b ± √D) / (2a) = (−32 ± √1408) / (2 × −16) = (−32 ± √1408) / (−32). Note that √1408 = √(64 × 22) = 8√22 ≈ 37.52. Point 3 — Correct final answer with rejection of negative time: Evaluating both solutions: t = (−32 + 37.52) / (−32) = 5.52 / (−32) ≈ −0.17 seconds (rejected, because time cannot be negative). t = (−32 − 37.52) / (−32) = (−69.52) / (−32) ≈ 2.17 seconds. The ball hits the ground at approximately t ≈ 2.17 seconds. Full credit requires showing that the negative time solution is rejected because time must be positive, and arriving at the correct positive answer of approximately 2.17 seconds. Note: Students who use the equivalent form 16t² − 32t − 6 = 0 (a = 16, b = −32, c = −6) will get: t = (32 ± √1408) / 32. Then t = (32 + 37.52)/32 ≈ 69.52/32 ≈ 2.17 or t = (32 − 37.52)/32 ≈ −5.52/32 ≈ −0.17. Same results; full credit.

SUMMARY

Summary — Solving Quadratic Equations

Summary

A quadratic equation takes the form ax² + bx + c = 0 and typically has two solutions. You can solve by three methods: the square root method (when there is no bx term), factoring (when two numbers multiply to c and add to b), or the quadratic formula (which works every time and is provided on your GED formula sheet). Always start by putting the equation in standard form and always check your answers by plugging them back in.

The discriminant (b² − 4ac) tells you how many real solutions to expect: positive means two, zero means one, and negative means none. On applied word problems, remember to reject solutions that don't make sense in context—you cannot have a negative length or a negative time. Practice choosing the right method quickly: if factoring isn't obvious within 30 seconds, switch to the quadratic formula. Confidence with quadratics puts you in a strong position for the 55% of the GED that covers algebraic problem solving.