Math 20-1 · Quadratic equations
Solving quadratics and systems
Three ways to solve the same equation, and knowing which to reach for. Factoring when it works, the formula when it does not, and the discriminant to tell you before you start whether there is anything to find.
- 1. The words, first
- 2. Factoring
- 3. Solving, and what the discriminant tells you
- 4. Systems with a quadratic in them
- 5. What costs marks
The words, first
The idea: A quadratic function and a quadratic equation are different questions about the same expression: where is the curve, versus where does it hit zero.
| Word | What it means |
|---|---|
| Root / zero / x-intercept | Three names for the same thing: a value of x making the expression zero, which is where the graph crosses the x-axis. |
| Zero product property | If a product is zero, at least one factor is zero. This is the whole reason factoring solves equations. |
| Greatest common factor | The largest factor shared by every term. Always taken out first, before any other factoring technique. |
| Difference of squares | a² − b² = (a − b)(a + b). A sum of squares does not factor over the real numbers. |
| Quadratic formula | x = [−b ± √(b² − 4ac)]/(2a). It works on every quadratic, factorable or not. |
| Discriminant | b² − 4ac, the part under the root. Positive means two real roots, zero means one, negative means none. |
| System of equations | Two or more equations solved together. A solution is a point satisfying all of them at once. |
| Substitution | Replacing one variable using one equation so the other has only one variable left. |
Factoring
The idea: Common factor first. Then look for a pattern. Then, if it is not obvious, stop and use the formula.
Order of attack.
- Common factor: 3x² + 6x = 3x(x + 2). Missing this makes everything after it harder.
- Difference of squares: x² − 16 = (x − 4)(x + 4).
- Trinomial with a = 1: find two numbers multiplying to c and adding to b. For x² + 7x + 12, that is 3 and 4.
- Trinomial with a ≠ 1: trial with checking, or decomposition. For 2x² + 7x + 3, (2x + 1)(x + 3) expands to 2x² + 7x + 3 ✓, while (2x + 3)(x + 1) gives a middle term of 5x.
Patterns in disguise. (x + 1)² − 9 is a difference of squares with a = x + 1 and b = 3, so it is (x − 2)(x + 4). Expanding first also works and lands in the same place.
Always expand to check. It takes ten seconds and catches every sign slip.
Solving, and what the discriminant tells you
The idea: Compute b² − 4ac before you commit. It costs one line and tells you what kind of answer to expect — or whether to stop.
b² − 4ac > 0 → two distinct real roots · = 0 → one real root (the vertex sits on the axis) · < 0 → no real roots
By factoring. x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3. The roots come out with the opposite signs to the numbers in the brackets.
By formula. x² + 4x − 1 = 0 has b² − 4ac = 16 + 4 = 20, so
x = (−4 ± √20)/2 = (−4 ± 2√5)/2 = −2 ± √5
The last step is where marks go: √20 must be simplified to 2√5, and then both terms in the numerator are divided by 2. Dividing only one is the classic error.
By square root. If there is no x term, just take the root — and take both: x² = 49 gives x = ±7.
Systems with a quadratic in them
The idea: Substitute to get one equation in one variable. What comes out is a quadratic, and the discriminant now tells you how many times the two graphs meet.
Worked example: y = x² − 4 and y = x + 2.
- Set them equal: x² − 4 = x + 2.
- Rearrange: x² − x − 6 = 0.
- Factor: (x − 3)(x + 2) = 0, so x = 3 or x = −2.
- Substitute back into the simpler equation: y = 5 and y = 0.
The solutions are the points (3, 5) and (−2, 0). A system's answer is a point, so stopping at the x-values is only half the question.
How many solutions are possible? A line can miss a parabola, touch it once as a tangent, or cut it twice — so 0, 1 or 2. Two parabolas can also meet in 0, 1 or 2 points, or coincide entirely.
What costs marks
The idea: Four things, all of them small and all of them common.
- Not setting the equation to zero before factoring. The zero product property needs a zero.
- Dividing only part of the numerator in the quadratic formula.
- Forgetting the ± when taking a square root.
- Giving only the x-coordinates as the solution to a system.