Four solving methods, the discriminant, vertex form, and worked applications — everything you need to master quadratics.
ax² + bx + c = 0
Most common form. Identifies a, b, c directly for the quadratic formula and discriminant.
a(x − h)² + k = 0
Reveals the vertex (h, k) and axis of symmetry x = h immediately. Best for graphing.
a(x − r₁)(x − r₂) = 0
Exposes the roots r₁ and r₂ directly. Apply zero product property to solve.
Best when: roots are rational integers and coefficients are small.
Best when: b = 0, giving ax² + c = 0, or when the equation is already in the form (x − h)² = k.
Converts to vertex form. Works on any quadratic and derives the quadratic formula.
Always works for any quadratic ax² + bx + c = 0 with a ≠ 0.
Identify a, b, c from standard form, substitute, and simplify carefully. Simplify the radical before splitting into + and − cases.
The discriminant is the expression under the radical in the quadratic formula. It determines the number and type of solutions without solving the full equation.
| Discriminant Value | Number of Solutions | Type | Graph Behavior |
|---|---|---|---|
| b² − 4ac > 0 | Two solutions | Distinct real roots | Parabola crosses x-axis twice |
| b² − 4ac = 0 | One solution | Repeated (double) root | Parabola touches x-axis at vertex |
| b² − 4ac < 0 | No real solutions | Two complex roots: ±i√|Δ| / (2a) | Parabola does not cross x-axis |
Starting with ax² + bx + c = 0:
Move constant right
ax² + bx = −c
Divide by a (if a ≠ 1)
x² + (b/a)x = −c/a — leading coefficient must be 1 before completing the square
Add (b/2a)² to both sides
x² + (b/a)x + (b/2a)² = −c/a + (b/2a)² — this is the 'completing' step
Factor left as perfect square
(x + b/2a)² = (b² − 4ac) / (4a²)
Take ±√ of both sides
x + b/2a = ±√(b² − 4ac) / (2a)
Solve for x
x = (−b ± √(b² − 4ac)) / (2a) — this is exactly the quadratic formula!
y = a(x − h)² + k. The vertex is at (h, k). Note the sign: (x − h) means the vertex is at x = h, not x = −h.
x = h = −b/(2a). The parabola is symmetric about this vertical line. Find h from standard form using −b/(2a), then plug in to find k.
If a > 0, parabola opens up → vertex is a minimum. If a < 0, parabola opens down → vertex is a maximum. |a| controls width (larger = narrower).
Standard → Vertex: complete the square or use h = −b/(2a), k = f(h). Vertex → Standard: expand a(x − h)² + k and collect like terms.
f(x) = 2x² − 8x + 3 → a = 2, b = −8, c = 3
Step 1: Axis of symmetry: x = −b/(2a) = −(−8)/(2·2) = 8/4 = 2
Step 2: k = f(2) = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5
Vertex = (2, −5). Parabola opens up (a > 0) → minimum at (2, −5).
Vertex form: f(x) = 2(x − 2)² − 5
3x² + 5x − 2 = 0 → a = 3, b = 5, c = −2
ac = (3)(−2) = −6. Need two numbers: multiply to −6, add to 5 → 6 and −1
Rewrite: 3x² + 6x − x − 2 = 0
Group: 3x(x + 2) − 1(x + 2) = 0
Factor: (3x − 1)(x + 2) = 0
x = 1/3 or x = −2
Verify with quadratic formula: discriminant = 5² − 4(3)(−2) = 25 + 24 = 49
x = (−5 ± √49) / (2·3) = (−5 ± 7) / 6 → x = 2/6 = 1/3 or x = −12/6 = −2 ✓
h(t) = −16t² + 80t + 6 (height in feet, t in seconds)
Part A: Maximum height
a = −16, b = 80 → t at vertex = −b/(2a) = −80/(−32) = 2.5 sec
h(2.5) = −16(2.5)² + 80(2.5) + 6 = −16(6.25) + 200 + 6 = −100 + 206 = 106
Maximum height = 106 feet at t = 2.5 seconds
Part B: Time to land (h = 0)
−16t² + 80t + 6 = 0 → 16t² − 80t − 6 = 0 → 8t² − 40t − 3 = 0
Discriminant = (−40)² − 4(8)(−3) = 1600 + 96 = 1696
t = (40 ± √1696) / 16 = (40 ± 41.18) / 16
t = 81.18/16 ≈ 5.07 (reject negative root)
Lands at approximately t ≈ 5.07 seconds
To factor ax² + bx + c = 0: find two numbers that multiply to ac and add to b. Rewrite the middle term using those numbers, then factor by grouping. Finally apply the zero product property — if (x − r₁)(x − r₂) = 0, then x = r₁ or x = r₂. Example: x² + 5x + 6 = 0 → find two numbers that multiply to 6 and add to 5: they are 2 and 3 → (x + 2)(x + 3) = 0 → x = −2 or x = −3.
Use the quadratic formula x = (−b ± √(b² − 4ac)) / (2a) when: (1) the quadratic does not factor over the integers, (2) the coefficients are large or messy, (3) you need an exact decimal answer, or (4) the discriminant is negative (complex roots). Factoring is faster when the roots are small integers, but the quadratic formula always works for any quadratic.
The discriminant is b² − 4ac (the expression under the square root in the quadratic formula). If b² − 4ac > 0, there are two distinct real solutions. If b² − 4ac = 0, there is exactly one real solution (a repeated root) and the parabola just touches the x-axis. If b² − 4ac < 0, there are no real solutions — the two solutions are complex (imaginary) numbers of the form (−b ± i√|b²−4ac|) / (2a).
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