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Quadratic Equation: ax² + bx + c = 0

Roots: x = (-b ± √(b²-4ac))/(2a). Discriminant Δ = b²-4ac: Δ>0 two real roots, Δ=0 one double root, Δ<0 two complex roots. Vertex at x = -b/(2a).

Concept Fundamentals
(-b±√Δ)/(2a)
Formula
b²-4ac
Δ
-b/(2a)
Vertex
2 real roots
Δ>0

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Δ>0: two distinct real roots. Δ=0: one repeated root. Δ<0: two complex conjugates. Vertex at axis of symmetry. Sum of roots = -b/a, product = c/a (Vieta).

Key quantities
(-b±√Δ)/(2a)
Formula
Key relation
b²-4ac
Δ
Key relation
-b/(2a)
Vertex
Key relation
2 real roots
Δ>0
Key relation

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Why: Quadratics model projectiles, profit, area. Discriminant tells root type. Vertex gives max/min. Completing the square or formula yields roots.

How: Compute Δ = b²-4ac. If Δ≥0: x = (-b±√Δ)/(2a). If Δ<0: complex roots with ±i√|Δ|. Vertex: x = -b/(2a), y = f(x).

Δ>0: two distinct real roots. Δ=0: one repeated root.Δ<0: two complex conjugates. Vertex at axis of symmetry.
Sources:Khan AcademyNCTM Standards

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Solve QuadraticEnter a, b, c

Coefficients (ax² + bx + c = 0)

quadratic_solve.sh
SOLVED
$ solve ax²+bx+c=0 where a=1, b=-3, c=2
Discriminant
1
x₁
2
x₂
1
Type
real
Vertex
(1.5, -0.25)
Axis of Symmetry
x = 1.5
Quadratic Equation Calculator
1x² + -3x + 2 = 0
x₁ = 2, x₂ = 1
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Discriminant & Vertex

Solution Type

📐 Step-by-Step Breakdown

SETUP
Discriminant Δ
b² - 4ac = -3² - 4(1)(2) = 1
Solution Type
Two distinct real roots
RESULT
x₁
2
x₂
1
Vertex
(1.5, -0.25)
ext{Axis}: x = -b/(2a)
Axis of Symmetry
x = 1.5

For educational and informational purposes only. Verify with a qualified professional.

🧮 Fascinating Math Facts

📐

x = (-b±√(b²-4ac))/(2a)

— Quadratic formula

Δ

Δ = b²-4ac determines root type

— Discriminant

📋 Key Takeaways

  • • Quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)
  • • Discriminant Δ = b² - 4ac determines solution type
  • • Δ > 0: two real roots; Δ = 0: one repeated root; Δ < 0: complex conjugate roots
  • • Vertex at x = -b/(2a); axis of symmetry is vertical through vertex

💡 Did You Know?

📐The quadratic formula works for any ax² + bx + c = 0 with a ≠ 0Source: Algebra
🎯The vertex of the parabola is always at x = -b/(2a)Source: Geometry
📊Δ = 0 means the parabola touches the x-axis at exactly one pointSource: Graphing
🔢Sum of roots = -b/a, product of roots = c/a (Vieta's formulas)Source: Number Theory
⚡Projectile motion follows a quadratic path; height is a quadratic in timeSource: Physics
📉If a &lt; 0, the parabola opens downward; if a &gt; 0, upwardSource: Graphing

📖 How It Works

For ax² + bx + c = 0, compute Δ = b² - 4ac. If Δ ≥ 0, use x = (-b ± √Δ)/(2a). If Δ < 0, the roots are complex: real part = -b/(2a), imaginary part = ±√|Δ|/(2a).

📝 Worked Example: x² - 5x + 6 = 0

Step 1: Δ = 25 - 24 = 1 > 0 → two real roots

Step 2: x = (5 ± 1) / 2 = 3 or 2

Result: x₁ = 3, x₂ = 2

⚠️ Common Mistakes to Avoid

  • a = 0: Then it's linear, not quadratic. Use linear methods.
  • Wrong sign in formula: x = (-b ± √Δ)/(2a). The -b is critical.
  • Forgetting 2a: The denominator is 2a, not 2.

🎯 Expert Tips

💡 Vieta Verification

x₁ + x₂ = -b/a and x₁ × x₂ = c/a. Check your roots.

💡 Factoring Shortcut

If roots are integers, try factoring first (e.g., x²-5x+6 = (x-2)(x-3)).

❓ FAQ

What if a = 0?

Then the equation is linear (bx + c = 0), not quadratic. Use linear methods.

Can I get exact roots?

For integer coefficients and perfect-square discriminant, yes. Otherwise you get decimals or radicals.

What is the vertex?

The vertex is at x = -b/(2a), y = f(-b/(2a)) = c - b²/(4a).

Why complex roots?

When Δ < 0, √Δ is imaginary. The roots are conjugates: p ± qi.

How do I verify?

Substitute each root into ax² + bx + c; the result should be 0.

⚠️ Disclaimer: This calculator provides numerical solutions. For exact symbolic form with radicals, use a computer algebra system.

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