Quadratic Equations
A quadratic equation is a polynomial equation of degree 2, written in standard form as ax² + bx + c = 0, where a ≠ 0. These equations have at most 2 solutions, which represent the x-intercepts of a parabola when graphed. The solutions can be found through factoring, completing the square, or the quadratic formula.
Why it matters
Quadratic equations model countless real-world scenarios involving area, projectile motion, and optimization. A baseball's height follows the equation h = -16t² + 64t + 5, where solutions determine when it hits the ground. In business, profit functions like P = -2x² + 100x - 800 help companies find break-even points at x = 10 and x = 40 units. Engineers use quadratics to design parabolic satellite dishes and suspension bridge cables. The Pythagorean theorem often leads to quadratics when finding unknown side lengths. Quadratics appear throughout algebra, precalculus, and calculus, forming the foundation for polynomial functions, conic sections, and optimization problems in advanced mathematics.
How to solve quadratic equations
Quadratic Equations
- Write in standard form: ax² + bx + c = 0.
- Factor, or use the quadratic formula: x = (-b ± √(b²-4ac)) / 2a.
- Check both solutions by substituting back.
Example: x² − 5x + 6 = 0 → (x−2)(x−3) = 0 → x = 2 or x = 3.
Worked examples
A square has an area of 25 cm². What is the side length?
Answer: x = 5 cm (side length must be positive)
- Interpret the context → Side length = x, so x² = 25 — A square's area equals side length squared. Since length must be positive, we only keep the positive root.
- Understand the equation → x² = 25 — We need to find a number that, when squared (multiplied by itself), gives us 25.
- Take the square root of both sides → x = ±√25 — When we take the square root, we must include BOTH the positive and negative root, because both (+a)² and (−a)² give a².
- Calculate √25 → √25 = 5 — Since 5 × 5 = 25, the square root of 25 is 5.
- Write both solutions → x = 5 or x = −5 — A quadratic equation can have up to 2 solutions. Here we have exactly 2.
- Verify both solutions → (5)² = 25 ✓, (−5)² = 25 ✓ — Substitute each value back into x² = 25 to confirm.
x² − 4x + 3 = 0
Answer: x = 1 or x = 3
- Write the equation in standard form → x² − 4x + 3 = 0 (a = 1, b = -4, c = 3) — Standard form is ax² + bx + c = 0. Identify a, b, and c.
- Find two numbers that multiply to c and add to b → Need: p × q = 3 and p + q = -4 → p = -3, q = -1 — We need two numbers whose product is 3 and whose sum is -4. Those are -3 and -1 because -3 × -1 = 3 and -3 + -1 = -4.
- Write the factored form → (x - 3)·(x - 1) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 3, x = 1 — If a × b = 0, then a = 0 or b = 0. Set each factor equal to zero and solve.
- Verify by substituting back → x = 3: 3² − 4·3 + 3 = 9 − 12 + 3 = 0 ✓ — Both solutions satisfy the original equation.
x² + 0x − 9 = 0
Answer: x = -3 or x = 3
- Write the equation in standard form → x² + 0x − 9 = 0 (a = 1, b = 0, c = -9) — Standard form is ax² + bx + c = 0. Identify a, b, and c.
- Find two numbers that multiply to c and add to b → Need: p × q = -9 and p + q = 0 → p = -3, q = 3 — We need two numbers whose product is -9 and whose sum is 0. Those are -3 and 3 because -3 × 3 = -9 and -3 + 3 = 0.
- Write the factored form → (x - 3)·(x + 3) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 3, x = -3 — If a × b = 0, then a = 0 or b = 0. Set each factor equal to zero and solve.
- Verify by substituting back → x = 3: 3² + 0·3 − 9 = 9 + 0 − 9 = 0 ✓ — Both solutions satisfy the original equation.
Common mistakes
- When factoring x² - 25 = 0, writing x = 5 instead of recognizing both solutions x = 5 and x = -5 from the difference of squares pattern
- In x² + 6x + 9 = 0, incorrectly factoring as (x + 3)(x + 3) = 0 but concluding there are two different solutions instead of one repeated solution x = -3
- Using the quadratic formula on 2x² - 8x + 6 = 0 and getting x = (8 ± √16)/4 = 3 or 1, forgetting that the discriminant calculation should be b² - 4ac = 64 - 48 = 16