Learning Objectives
Choose factoring, square roots, completing the square, or the quadratic formula.
Use the discriminant to count real solutions.
Connect standard, factored, and vertex forms with graph features.
Prerequisites: factoring, square roots, and linear equations. A quadratic function has the form f(x) = ax² + bx + c with a ≠ 0. Its graph is a parabola.
1. Choose a Method from the Equation's Form
Equation form | Convenient method |
|---|---|
(x − p)(x − q) = 0 | Zero-product property |
(x − h)² = k | Square roots |
ax² + bx + c = 0 | Factoring or quadratic formula |
A quadratic needing vertex form | Complete the square |
Worked Example: Isolate a Square
Solve (x − 3)² = 16. Taking square roots gives x − 3 = 4 or x − 3 = −4.
Therefore x = 7 or x = −1. Both values are 4 units from 3, so both give a square of 16.
2. The Quadratic Formula Applies When a ≠ 0
For ax² + bx + c = 0, the solutions are x = [−b ± √(b² − 4ac)]/(2a). Put the equation into zero form before identifying coefficients, and include each coefficient's sign.
Worked Example: Use Signed Coefficients
Solve 2x² + x − 3 = 0. Here a = 2, b = 1, and c = −3.
The discriminant is 1² − 4(2)(−3) = 25. Hence x = (−1 ± 5)/4.
The two solutions are 1 and −3/2. Factoring as (2x + 3)(x − 1) provides an independent check.
3. The Discriminant Counts Real Roots
For real coefficients with a ≠ 0, D = b² − 4ac determines the number of distinct real roots. If D > 0, there are two. If D = 0, there is one repeated real root. If D < 0, there are no real roots, though there are two nonreal complex roots.
Graphically, these cases mean the parabola crosses the x-axis twice, touches it at its vertex, or misses it. This lesson counts real solutions unless complex solutions are explicitly requested.
4. Complete the Square to Find the Vertex
Worked Example: Rewrite a Quadratic
Rewrite f(x) = x² − 6x + 5 in vertex form. Half the x coefficient is −3, whose square is 9.
Add and subtract 9: f(x) = (x² − 6x + 9) − 9 + 5 = (x − 3)² − 4.
The vertex is (3, −4), the axis of symmetry is x = 3, and the parabola opens upward. Its minimum value is −4.
Factoring the original expression as (x − 1)(x − 5) also reveals x-intercepts at 1 and 5, symmetric about x = 3.
In vertex form a(x − h)² + k, the vertex is (h, k). A positive a gives upward opening and a negative a gives downward opening. In standard form, the y-intercept is (0, c). Different forms make different features easy to read.
Common Mistakes
Do not drop the negative square-root branch when solving a squared equation.
The denominator 2a divides the entire numerator of the quadratic formula.
Vertex form x − h places the vertex at input h, not −h.
5. Practice
1. Solve x² − 5x + 6 = 0.
2. How many real solutions does x² + 2x + 5 = 0 have?
3. Find the vertex and maximum or minimum of f(x) = −2(x + 1)² + 7.
4. Complete the square for x² + 4x − 1 and use it to solve x² + 4x − 1 = 0.
Worked Solutions
1. (x − 2)(x − 3) = 0, so x = 2 or x = 3.
2. D = 4 − 20 = −16, so there are no real solutions.
3. The vertex is (−1, 7). The negative leading coefficient gives a maximum value of 7.
4. x² + 4x − 1 = (x + 2)² − 5. Setting this equal to zero gives (x + 2)² = 5, so x = −2 ± √5.
6. Summary
- Choose a solving method that fits the visible structure.
- The discriminant counts distinct real roots.
- Vertex form reveals the turning point; factored form reveals real zeros.
- Check solutions and use symmetry to verify graph features.