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Quadratic Equation Solver

Linear, quadratic & 2×2 systems with textbook steps — discriminant, vertex, factored form, parabola graph & Vieta checks.

x² +x += 0

💡 Sum & product of roots are verified automatically (Vieta's formulas) — the same check your examiner uses.

Solution · two real roots · D = 1

x₁ = 3, x₂ = 2

Vertex

(2.5, -0.25)

Factored form

(x − 3)(x − 2)

Step-by-step working

1. 1x² − 5x + 6 = 0

2. Discriminant D = b² − 4ac = (-5)² − 4(1)(6) = 1

3. D > 0 → two real roots

4. x = (−b ± √D) ÷ 2a = (5 ± 1) ÷ 2

5. x₁ = 3, x₂ = 2

6. Sum check: x₁ + x₂ = 5 = −b/a ✓ · Product: x₁·x₂ = 6 = c/a ✓

📈 The parabola

Roots marked where the curve crosses the x-axis · diamond = vertex.

x = 3x = 2vertex (2.5, -0.25)

💡 The discriminant D = b² − 4ac decides everything before you solve: D > 0 two real roots, D = 0 one repeated root, D < 0 complex roots. Examiners love asking for D alone.

🧮Three solvers, every step shown

Not just the answer — the working, written like a good notebook. Solve linear equations (ax + b = c) with the move-across logic narrated, quadraticswith the discriminant analysed, both roots, vertex, factored form, a Vieta's-formulas self-check, and a live parabola graph with the roots and vertex marked — and 2×2 systemsby Cramer's rule, honestly detecting parallel and identical lines. Copy the whole working with one click.

📊Everything you'd want to know

  • Linear mode: solution, plus honest 'no solution' (contradiction) and 'every x' (identity) cases.
  • Quadratic mode: discriminant first, both roots (real or complex a ± bi), vertex, factored form when it's tidy.
  • A live parabola graph — roots as green dots on the x-axis, vertex as a diamond.
  • System mode: x and y by determinants, with parallel (no solution) and same-line (infinite) detection.
  • Every mode ends with a substitution check — the habit examiners reward.

🧮The maths

x = (−b ± √(b² − 4ac)) ÷ 2a
Cramer: x = (c₁b₂ − c₂b₁) ÷ (a₁b₂ − a₂b₁)

The discriminant D = b² − 4ac is the quadratic's fortune-teller: positive means two crossings of the x-axis, zero means the parabola kisses it at the vertex, negative means it never touches. For systems, a zero determinant means the lines don't cross once — they're parallel or identical, and the constants decide which.

x² − 5x + 6 = 0: D = 25 − 24 = 1 > 0, so x = (5 ± 1)/2 → x₁ = 3, x₂ = 2, factored as (x − 3)(x − 2). Check: 3 + 2 = 5 = −b/a ✓ and 3 × 2 = 6 = c/a ✓.

💡Exam habits worth stealing

  • Compute D before anything else — it tells you how many roots to expect and earns method marks alone.
  • Verify with Vieta's: sum of roots = −b/a, product = c/a. Two seconds, catches most slips.
  • For systems, check your (x, y) in the equation you didn't use to find it.
  • A 'no solution' answer is often correct — don't force numbers out of parallel lines.

💡 Frequently Asked Questions

How do I solve a quadratic equation step by step?+

Compute the discriminant D = b² − 4ac first, then apply x = (−b ± √D) ÷ 2a. For x² − 5x + 6 = 0: D = 1, so x = (5 ± 1)/2, giving 3 and 2. The solver writes out exactly these steps for your coefficients and verifies with sum/product checks.

What does the discriminant tell you?+

How many real roots exist before you solve: D > 0 means two, D = 0 means one repeated root (the parabola touches the axis at its vertex), D < 0 means none real — the roots are complex conjugates a ± bi. The solver reports D explicitly in every solution.

How do I solve two equations with two unknowns?+

By elimination or Cramer's rule: x = (c₁b₂ − c₂b₁) ÷ D and y = (a₁c₂ − a₂c₁) ÷ D where D = a₁b₂ − a₂b₁. If D = 0 the lines are parallel (no solution) or identical (infinite solutions) — the solver detects and explains both.

What are Vieta's formulas?+

For ax² + bx + c = 0, the roots always satisfy x₁ + x₂ = −b/a and x₁ × x₂ = c/a. They're the fastest way to check your roots — the solver runs this verification automatically on every quadratic.

Can this solver show complex roots?+

Yes — when D < 0 it reports the roots as a ± bi with both parts computed, and the graph shows why: the parabola never reaches the x-axis.

What is the vertex of a parabola and why does it matter?+

The turning point, at x = −b/2a. It's the minimum (a > 0) or maximum (a < 0) of the quadratic — the key to optimisation questions — and it's marked with a diamond on the solver's graph.

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