Quadratic Equations
Teaching quadratic equations often begins with students solving x² = 16 and wondering why there are two answers. The CCSS.HSA.REI and CCSS.HSA.SSE standards require students to solve quadratic equations using multiple methods, from factoring simple expressions to applying the quadratic formula for complex cases.
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Why it matters
Quadratic equations model countless real-world scenarios where relationships involve squared variables. Engineers use them to calculate projectile motion—a ball thrown upward follows the path h = -16t² + 32t + 6, where maximum height occurs at t = 1 second. Architects apply quadratics when designing parabolic arches that can span 200 feet while supporting thousands of pounds. Business owners use quadratic profit functions like P = -2x² + 80x - 350 to find that selling 20 units maximizes profit at $450. Even basic geometry problems involve quadratics: finding that a rectangle with perimeter 24 cm has maximum area 36 cm² when both dimensions equal 6 cm. Students encounter quadratics in physics (acceleration), economics (supply and demand curves), and computer graphics (parabolic animations).
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 9 cm². What is the side length?
Answer: x = 3 cm (side length must be positive)
- Interpret the context → Side length = x, so x² = 9 — A square's area equals side length squared. Since length must be positive, we only keep the positive root.
- Understand the equation → x² = 9 — We need to find a number that, when squared (multiplied by itself), gives us 9.
- Take the square root of both sides → x = ±√9 — 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 √9 → √9 = 3 — Since 3 × 3 = 9, the square root of 9 is 3.
- Write both solutions → x = 3 or x = −3 — A quadratic equation can have up to 2 solutions. Here we have exactly 2.
- Verify both solutions → (3)² = 9 ✓, (−3)² = 9 ✓ — Substitute each value back into x² = 9 to confirm.
x² − 4x + 4 = 0
Answer: x = 2 (double root)
- Write the equation in standard form → x² − 4x + 4 = 0 (a = 1, b = -4, c = 4) — 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 = 4 and p + q = -4 → p = -2, q = -2 — We need two numbers whose product is 4 and whose sum is -4. Those are -2 and -2 because -2 × -2 = 4 and -2 + -2 = -4.
- Write the factored form → (x - 2)^2 = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 2, x = 2 — If a × b = 0, then a = 0 or b = 0. Set each factor equal to zero and solve.
- Verify by substituting back → x = 2: 2² − 4·2 + 4 = 4 − 8 + 4 = 0 ✓ — Both solutions satisfy the original equation.
x² − 1x − 2 = 0
Answer: x = -1 or x = 2
- Write the equation in standard form → x² − 1x − 2 = 0 (a = 1, b = -1, c = -2) — 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 = -2 and p + q = -1 → p = -2, q = 1 — We need two numbers whose product is -2 and whose sum is -1. Those are -2 and 1 because -2 × 1 = -2 and -2 + 1 = -1.
- Write the factored form → (x - 2)·(x + 1) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 2, 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 = 2: 2² − 1·2 − 2 = 4 − 2 − 2 = 0 ✓ — Both solutions satisfy the original equation.
Common mistakes
- ✗Students solve x² = 25 and write only x = 5, forgetting the negative solution x = -5
- ✗When factoring x² + 7x + 12 = 0, students write (x + 3)(x + 4) = 0 but then solve x = 3 and x = 4 instead of x = -3 and x = -4
- ✗Students apply the quadratic formula to 2x² - 8x + 6 = 0 using a = 2, b = 8, c = 6, forgetting that b = -8
- ✗When checking solutions, students substitute x = 2 into x² - 5x + 6 and calculate 4 - 10 + 6 = 0 correctly, but miss that 2² - 5(2) + 6 actually equals 4 - 10 + 6
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