How to Solve Quadratic Equations by Formula

How to Solve Quadratic Equations by Formula

Introduction

"Master the quadratic formula, and you’ll unlock everything from projectile motion in physics to profit maximization in business—plus, you’ll solve 90% of algebra exam questions in under 60 seconds."

What You Need To Know First

Before diving into the quadratic formula, ensure you understand: 1. Standard form of a quadratic equation: ax² + bx + c = 0 (where a, b, and c are numbers, and a ≠ 0). 2. Square roots and simplifying radicals: You’ll need to simplify expressions like √(50) into 5√2. 3. Substitution: Plugging numbers into formulas (e.g., replacing a, b, and c with actual values).

If any of these are unclear, review them first—this guide assumes you’re solid on them.

Key Vocabulary

Formulas To Know

1. The Quadratic Formula

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

What each variable means: - a: Coefficient of x² (must not be zero). - b: Coefficient of x. - c: Constant term (the number with no x). - ±: Means "plus or minus"—you’ll get two solutions (unless the discriminant is zero).

MEMORISE THIS? ✅ YES. It’s the most important formula in algebra. Write it on your cheat sheet now.

2. The Discriminant

Formula: [ D = b^2 - 4ac ]

What it tells you: - D > 0: Two real and distinct roots (e.g., x=2 and x=3). - D = 0: One real repeated root (e.g., x=4 only). - D < 0: Two complex roots (e.g., x = 2 ± 3i). (Not needed for most K12 exams, but good to know.)

MEMORISE THIS? ❌ No—just understand what it means. The formula is given on most exam sheets.

Step-by-Step Method

Follow these steps exactly for every quadratic equation.

1. Write the equation in standard form: ax² + bx + c = 0.
2. Move all terms to one side (e.g., 2x² = 5x – 3 → 2x² – 5x + 3 = 0).

3. Combine like terms.

4. Identify a, b, and c.

5. a = coefficient of x².
6. b = coefficient of x.

7. c = constant term.

8. Calculate the discriminant (D): D = b² – 4ac.

9. Plug in a, b, and c.

10. Simplify (e.g., D = (-5)² – 4(2)(3) = 25 – 24 = 1).

11. Plug into the quadratic formula:
    [ x = \frac{-b \pm \sqrt{D}}{2a} ]

12. Replace b, D, and a with your numbers.

13. Always write the ± symbol—it gives two answers!

14. Simplify the square root (if possible).

15. Example: √8 = 2√2.

16. Split into two solutions:

17. x = (–b + √D) / (2a)

18. x = (–b – √D) / (2a)

19. Simplify fractions (if possible).

20. Example: x = (6 ± 2)/4 → x = 8/4 = 2 or x = 4/4 = 1.

21. Write the final answer in curly braces (for two solutions) or as a single value (if D=0).

22. Example: x = {2, 1} or x = 3.

Worked Example Using the Steps

Problem: Solve 3x² – 5x + 1 = 0.

1. Standard form: Already in ax² + bx + c = 0 form.
2. Identify coefficients:
3. a = 3, b = –5, c = 1.
4. Discriminant:
   D = b² – 4ac = (–5)² – 4(3)(1) = 25 – 12 = 13.
5. Quadratic formula:
   [ x = \frac{-(-5) \pm \sqrt{13}}{2(3)} = \frac{5 \pm \sqrt{13}}{6} ]
6. Simplify square root: √13 cannot be simplified further.
7. Split solutions:
8. x = (5 + √13)/6
9. x = (5 – √13)/6
10. Simplify fractions: Already simplified.
11. Final answer:
    [ x = \frac{5 \pm \sqrt{13}}{6} ]

Worked Examples

Example 1 – Basic (No Tricks)

Problem: Solve x² – 6x + 8 = 0.

Solution: 1. Standard form: Already correct. 2. a = 1, b = –6, c = 8. 3. Discriminant: D = (–6)² – 4(1)(8) = 36 – 32 = 4. 4. Quadratic formula:
[ x = \frac{-(-6) \pm \sqrt{4}}{2(1)} = \frac{6 \pm 2}{2} ] 5. Split solutions:
- x = (6 + 2)/2 = 8/2 = 4
- x = (6 – 2)/2 = 4/2 = 2 6. Final answer: x = {4, 2}.

What we did and why: - We followed the steps in order to avoid mistakes. - The discriminant was a perfect square (4), so the roots were rational and whole numbers.

Example 2 – Medium (Fractional Coefficients)

Problem: Solve 2x² + 3x – 1 = 0.

Solution: 1. Standard form: Already correct. 2. a = 2, b = 3, c = –1. 3. Discriminant: D = 3² – 4(2)(–1) = 9 + 8 = 17. 4. Quadratic formula:
[ x = \frac{-3 \pm \sqrt{17}}{4} ] 5. Split solutions:
- x = (–3 + √17)/4
- x = (–3 – √17)/4 6. Final answer: x = { (–3 + √17)/4, (–3 – √17)/4 }.

What we did and why: - The discriminant (17) wasn’t a perfect square, so the roots were irrational. - We didn’t round—exact form is required in exams.

Example 3 – Exam Style (Disguised Equation)

Problem: The height (h) of a ball in meters after t seconds is given by h = –5t² + 20t + 1. When does the ball hit the ground? (Hint: h = 0 when it hits the ground.)

Solution: 1. Set h = 0: –5t² + 20t + 1 = 0. 2. Rewrite in standard form: 5t² – 20t – 1 = 0 (multiply both sides by –1). 3. a = 5, b = –20, c = –1. 4. Discriminant: D = (–20)² – 4(5)(–1) = 400 + 20 = 420. 5. Quadratic formula:
[ t = \frac{-(-20) \pm \sqrt{420}}{2(5)} = \frac{20 \pm \sqrt{420}}{10} ] 6. Simplify √420:
420 = 4 × 105 = 4 × 5 × 21 = 4 × 5 × 3 × 7
√420 = √(4 × 105) = 2√105. 7. Final answer:
[ t = \frac{20 \pm 2\sqrt{105}}{10} = \frac{10 \pm \sqrt{105}}{5} ]
- Since time can’t be negative, we take the positive root:
[ t = \frac{10 + \sqrt{105}}{5} \approx 4.05 \text{ seconds} ]

What we did and why: - We translated the word problem into an equation (h = 0). - We multiplied by –1 to make a positive (easier to work with). - We rejected the negative time because it’s not physically meaningful.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, listen up—this is your last-minute quadratic formula crash course. Here’s what you must remember:

1. Standard form is king: ax² + bx + c = 0. If it’s not in this form, rearrange it first.
2. Circle a, b, and c before plugging into the formula. Double-check signs—especially b.
3. Discriminant first: D = b² – 4ac. If D is negative, you’re done (no real roots). If D=0, one answer. If D>0, two answers.
4. Write the ± symbol—it’s two solutions unless D=0.
5. Simplify the square root (e.g., √8 = 2√2) and reduce fractions (e.g., 6/4 = 3/2).
6. Watch for traps: Equations not set to zero, a=0, or hidden fractions. Read the question twice.

Now go practice three problems—one basic, one with fractions, and one word problem. You’ve got this. Good luck!