Quadratic Formula
x = (−b ± √(b²−4ac))/(2a) solves ax²+bx+c=0. Discriminant Δ=b²−4ac: positive → two real roots, zero → one (double) root, negative → two complex conjugate roots.
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Formula works for all quadratics — always gives correct roots. When Δ<0, roots are (−b±i√|Δ|)/(2a) — complex conjugates. Common mistake: forgetting the ± gives only one root.
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Why: The quadratic formula solves any quadratic equation — no factoring or guessing. Derived by completing the square. Works for real and complex roots. Foundational in algebra and applied math.
How: Step 1: Identify a, b, c. Step 2: Compute Δ=b²−4ac. Step 3: √Δ (real if Δ≥0, i√|Δ| if Δ<0). Step 4: x=(−b±√Δ)/(2a). Simplify radicals. Decimal: approximate.
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📐 Examples — Click to Load
Coefficients: ax² + bx + c = 0
Formula Components (-b, √Δ, 2a)
Formula Breakdown
For educational and informational purposes only. Verify with a qualified professional.
🧮 Fascinating Math Facts
x²−2=0: Δ=8, x=±√2. x²+1=0: Δ=−4, x=±i.
— Examples
Formula derived by completing the square on ax²+bx+c=0.
— Derivation
📋 Step-by-Step Formula Application
For ax² + bx + c = 0: (1) Identify a, b, c. (2) Compute Δ = b² - 4ac. (3) Substitute into x = (-b ± √Δ)/(2a). (4) Use + for x₁ and - for x₂. Simplify radicals when possible.
🔀 When Formula vs Factoring
Use factoring when the quadratic has nice integer roots (e.g. x²-5x+6 = (x-2)(x-3)).Use the formula when factoring is difficult (irrational roots, large coefficients, or when you need a guaranteed method). The formula always works for any quadratic with a ≠ 0.
📜 History
Babylonians (c. 2000 BCE) solved specific quadratic problems. al-Khwarizmi (9th century) gave systematic algebraic methods in his book "Al-Kitab al-Mukhtasar fi Hisab al-Jabr wal-Muqabala" (the origin of algebra). The modern formula x = (-b ± √(b²-4ac))/(2a) was derived by completing the square.
⚠️ Common Mistakes
- Forgetting the ± — there are two roots: one with + and one with -.
- Wrong sign on b: the formula uses -b, so if b is negative, -b is positive.
- Dividing only the √Δ by 2a: the entire numerator (-b ± √Δ) is divided by 2a.
- When Δ < 0, writing √Δ as a real number — use i√|Δ| for complex roots.
- Not simplifying radicals: e.g. √8 = 2√2.
📐 Graphical Interpretation
The roots are the x-intercepts of the parabola y = ax² + bx + c. When Δ > 0, the parabola crosses the x-axis twice. When Δ = 0, it touches at the vertex. When Δ < 0, there are no real x-intercepts (the parabola stays above or below the axis).
🔢 Exact vs Decimal
Exact form keeps radicals (e.g. (3 + √5)/2) for precision. Decimal form gives approximate values. Use exact when the result is used in further algebra; use decimal for practical applications.
❓ FAQ
What is the quadratic formula?
x = (-b ± √(b²-4ac))/(2a). It gives both roots of ax²+bx+c=0 in one expression.
Why does the formula always work?
It is derived by completing the square on ax²+bx+c=0. The derivation is valid for any a≠0, b, c.
When do I get complex roots?
When Δ = b²-4ac < 0. Then √Δ = i√|Δ|, and the roots are (-b ± i√|Δ|)/(2a).
How do I simplify √(b²-4ac)?
Factor out perfect squares: e.g. √12 = 2√3. The calculator shows simplified radical form when possible.
Who invented the quadratic formula?
Ancient Babylonians solved specific cases. al-Khwarizmi (9th century) gave systematic methods. The modern form came from completing the square.
📝 Worked Examples
📌 Summary
The quadratic formula x = (-b ± √(b²-4ac))/(2a) solves any quadratic equation. Compute the discriminant first to know if roots are real or complex. Use exact form for algebra, decimal for applications. The formula is derived by completing the square and always works when a≠0.
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