The core idea
A quadratic equation has the form ax²+bx+c = 0 with a ≠ 0. Factoring, completing the square and the quadratic formula solve the same problem in different ways. Roots, the vertex and the physical domain answer different questions about its model.
1. Know which question you are solving
The function q(x) = ax²+bx+c assigns an output to every real x. The equation q(x) = 0 asks which inputs produce zero. A graph of q is a parabola; roots are its intersections with the horizontal axis. Asking for q(2), asking when q(x) = 2, and asking for the greatest value of q are three different tasks. The coefficient a must be nonzero: otherwise the squared term disappears. When a is positive the parabola opens upward; when negative it opens downward. Equal steps in x do not usually produce equal changes in q(x). For q(x) = x², outputs at 0, 1, 2, 3 are 0, 1, 4, 9; first differences are 1, 3, 5. Constant second differences help recognise quadratic patterns in equally spaced data, without proving that every future value follows the same rule.
Sources: OpenStax: Quadratic Functions ↗
2. Factoring turns one product into alternatives
For x²−6x+5 = 0, find two numbers whose product is 5 and sum is −6: −1 and −5. Then (x−1)(x−5) = 0. A product of real numbers is zero only if at least one factor is zero, so x = 1 or x = 5. Substitute both into the original equation to verify them. This zero-product rule requires zero on one side; from uv = 10 you cannot conclude u = 10 or v = 10. Do not divide an equation by a variable without checking whether it might be zero. In x²−4x = 0, dividing by x loses the root x = 0. Factoring x(x−4) = 0 preserves both 0 and 4. Factoring is efficient when simple factors exist, but failure to spot them does not mean the equation has no solutions.
Sources: OpenStax: Quadratic Equations ↗
3. Completing the square reveals the vertex
The expression x²−6x+5 can be rewritten as (x−3)²−4: expand the square to check the equality. Because a square is nonnegative, its minimum value is −4 at x = 3. The vertex is therefore (3, −4), halfway between the two roots 1 and 5. More generally, ax²+bx+c = a(x−h)²+k, where h = −b/(2a) and k = q(h). If a > 0, k is a minimum; if a < 0, it is a maximum over all real inputs. To solve (x−3)²−4 = 0, move 4 and take both possibilities: x−3 = ±2. The principal square root √4 equals 2, but solving a squared equation requires considering both signs. A contextual domain may exclude the vertex, so the unrestricted extreme need not be attainable.
Sources: OpenStax: Quadratic Functions ↗ · OpenStax: Quadratic Equations ↗
4. The quadratic formula classifies possible roots
Completing the square for the general equation gives x = (−b ± √(b²−4ac))/(2a). The entire numerator is divided by 2a. The discriminant D = b²−4ac tells us whether the graph crosses the horizontal axis twice, touches it once, or never meets it in the real plane. D > 0 gives two distinct real roots; D = 0 gives one repeated real root; D < 0 gives no real roots. For x²−2x−1 = 0, D = 8 and x = 1 ± √2, approximately 2.414 and −0.414. Keep exact forms until rounding is needed. For x²+1 = 0, no real x works because x² cannot be −1. Complex numbers provide further solutions in a larger number system; that extension is beyond this lesson.
Sources: OpenStax: Quadratic Equations ↗
5. Worked case: recover dimensions from an area
A hypothetical rectangular garden has area 40 m² and is 3 m longer than it is wide. Let its width be x metres, so the length is x+3. The model is x(x+3) = 40 with x > 0. Rearranging gives x²+3x−40 = 0, which factors as (x+8)(x−5) = 0. The mathematical roots are −8 and 5. Width −8 m is outside the model’s domain, so the garden is 5 m wide and 8 m long. Check both the area, 5×8 = 40 m², and the difference, 8−5 = 3 m. Its perimeter is 26 m, a different quantity. The rejected root is not an algebra mistake; it solves the equation but not the physical problem. Units and constraints complete the solution.
Sources: OpenStax: Quadratic Equations ↗
6. Worked case: roots and maximum describe different events
Consider the invented vertical-motion model h(t) = −5t²+20t metres, where t is seconds after launch from ground level. It assumes constant downward acceleration of 10 m/s² and ignores air resistance. It is a mathematical example, not a measurement or an instruction to launch anything. Factoring gives h(t) = −5t(t−4), so ground level occurs at t = 0 and t = 4. During the flight use 0 ≤ t ≤ 4. Completing the square gives h(t) = −5(t−2)²+20: the greatest height is 20 m at t = 2 s. Height 15 m occurs twice, at t = 1 and t = 3, once rising and once falling. Extending the formula beyond 4 seconds predicts negative heights; that continuation does not describe this ground-stopping model.
Ground level and the highest point
Sources: OpenStax: Quadratic Functions ↗
PUT IT INTO PRACTICE
Practice: a quadratic through three forms
- Start with q(x) = 2x²−8x+6. Factor it and find every real root without dividing by x.
- Rewrite q in completed-square form. Identify the vertex, the axis of symmetry, and whether its extreme is a minimum or maximum.
- Now restrict x to [0, 1]. Compare the outputs at the endpoints and decide whether the unrestricted vertex remains available.
- Solution: q = 2(x−1)(x−3) = 2(x−2)²−2. Roots are 1 and 3; the vertex is (2, −2), a minimum. On [0, 1], the vertex is excluded and q decreases from 6 to 0, so its restricted minimum is 0 at x = 1. Restrictions can change the answer even when the formula stays identical.
Check your understanding
Why must a be nonzero?
Otherwise ax² disappears and the equation is no longer quadratic. A method that divides by a would also become invalid.
Why does x² = 9 have two real solutions?
Both 3 and −3 square to 9. The equation asks for every input, while √9 names only the principal nonnegative square root.
Does a negative discriminant mean no solution in any number system?
No. It means no real roots. Complex roots exist for such a quadratic, although they may not represent real physical lengths.
Why retain t = 0 in the motion case?
It represents the initial ground-level position and satisfies the equation. Dividing by t would silently remove that meaningful event.
Is the vertex always a valid model input?
No. Its horizontal coordinate must belong to the stated domain. A restriction can put the best attainable value at a boundary instead.
What makes a checked answer complete?
Substitute into the original equation, attach units, and check the original constraints. Satisfying rearranged algebra alone can overlook the actual question.