Quadratic Formula and Vertex

• Apply the quadratic formula x = (−b ± √(b² − 4ac)) / (2a) to any ax² + bx + c = 0
• Interpret the discriminant b² − 4ac: positive → two real roots, zero → one real root, negative → no real roots
• Convert y = ax² + bx + c into vertex form y = a(x − h)² + k by completing the square
• Identify the maximum or minimum value of a quadratic from vertex form without graphing
• Recognize when a quadratic has a max (a < 0) versus a min (a > 0)

The quadratic formula solves any quadratic equation ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a). It always works, even when factoring fails, so it is the safety net on ACT Math. Use it when the quadratic does not factor cleanly (integer roots). For example, x² − 3x + 1 = 0 has no integer factors, so the formula is the fastest path. The piece under the square root, b² − 4ac, is called the discriminant. Its sign tells you how many real solutions the equation has, without actually solving: if b² − 4ac > 0, there are two distinct real roots; if b² − 4ac = 0, there is exactly one real root (a repeated root, and the parabola just touches the x-axis); if b² − 4ac < 0, there are no real roots (the parabola never touches the x-axis; the solutions are non-real complex numbers). The ACT sometimes asks 'For what values of c does x² + 4x + c = 0 have two real solutions?'. This is a discriminant question. Set b² − 4ac > 0 and solve: 16 − 4c > 0, so c < 4. Vertex form y = a(x − h)² + k makes the vertex obvious: it is the point (h, k). To convert from standard form y = ax² + bx + c to vertex form, complete the square: take half of b, square it, add and subtract it. For example y = x² − 8x + 12 = (x² − 8x + 16) − 16 + 12 = (x − 4)² − 4, so the vertex is (4, −4). Or (faster on the ACT) find the vertex x-coordinate using −b/(2a), then plug back in to get the y-coordinate. The vertex is the minimum value of y if a > 0 (parabola opens up) and the maximum if a < 0 (parabola opens down). Problems like 'What is the maximum height of a thrown ball described by h(t) = −16t² + 64t + 5?' are vertex questions. Find t = −64/(2·−16) = 2, then h(2) = −16(4) + 128 + 5 = 69 feet. In meters the model is h(t) = −4.9t² + vt + h₀: an object launched at 29.4 m/s from a 120 m platform peaks at t = 29.4/9.8 = 3 seconds, at h(3) = −44.1 + 88.2 + 120 = 164.1 m. Sum and product of the solutions: for ax² + bx + c = 0 the two solutions add to −b/a and multiply to c/a, which answers many 'find b' or 'sum of the roots' questions without solving: for x² − 6x + n = 0 to have solutions 8 and −2, n = 8 × (−2) = −16 (and 8 + (−2) = 6 matches the −6x). To build a quadratic from its solutions r and s, write y = a(x − r)(x − s); the x-intercepts are r and s, and the vertex sits halfway between them at x = (r + s)/2. When the discriminant is negative, the formula gives complex solutions of the form p ± qi.

• Try to factor first. If it works in under 20 seconds, great. If not, switch to the quadratic formula without hesitating.
• If a question asks for the number of solutions but not the solutions themselves, use the discriminant. It's much faster.
• Vertex x-coordinate is −b/(2a). Memorize this and skip completing the square under time pressure.
• For max-height or max-profit problems, the answer is the y-value of the vertex, not the x-value. Read the question.
• A parabola opens up when a > 0 (has a min) and opens down when a < 0 (has a max). Check the sign of a first.