A quadratic equation, ax² + bx + c = 0, has a short formula and a long list of ways to get it wrong. The most useful single number in the whole process is the one most people compute in passing and then ignore: the discriminant, Δ = b² − 4ac. It tells you how many real solutions there are, whether they will be neat fractions or square roots, and in a word problem whether the situation described is even possible. This article works through the formula, then two real-world examples, then the mistakes that cost marks. Every number below was produced by the equation solver and checked by hand.
The formula, and where the discriminant sits in it
The quadratic formula is
x = (−b ± √(b² − 4ac)) / (2a)
The part under the square root is the discriminant. Because a square root of a negative number is not a real number, and the square root of zero is zero, Δ sorts every quadratic into three cases:
| Δ | Real roots | What the parabola y = ax² + bx + c does |
|---|---|---|
| positive | two different | crosses the x-axis twice |
| zero | one, counted twice | touches the x-axis at its vertex |
| negative | none (two complex roots) | stays entirely above or below the axis |
A quick way to see all three is to keep a and b fixed and change c. Take x² − 4x + c:
- c = 3: Δ = 16 − 12 = 4. Two roots, x = (4 ± 2)/2, so x = 3 and x = 1. The equation factors as (x − 3)(x − 1) = 0.
- c = 4: Δ = 16 − 16 = 0. One double root, x = 4/2 = 2. It factors as (x − 2)² = 0.
- c = 5: Δ = 16 − 20 = −4. No real roots; the complex roots are x = (4 ± 2i)/2 = 2 ± i.
Raising c lifts the parabola. At c = 4 its lowest point sits exactly on the axis, and above that it never comes back down.
A second reading: perfect squares and surds
Δ also tells you what kind of answer to expect when a, b and c are whole numbers. If Δ is a perfect square (0, 1, 4, 9, 16, …) the roots are fractions and the quadratic factors over the integers. If Δ is positive but not a perfect square, the roots involve a square root that cannot be removed.
For x² + 2x − 2 = 0, Δ = 4 + 8 = 12. Twelve is not a perfect square, but it has a square factor: 12 = 4 × 3, so √12 = 2√3. Then
x = (−2 ± 2√3) / 2 = −1 ± √3
which is about 0.7320508076 and −2.732050808. Simplifying the surd before dividing is what makes the 2s cancel; leaving √12 in place gives a correct but untidy answer that many teachers mark down.
Worked example 1: a ball thrown upwards
A ball is thrown straight up at 20 m/s from a hand 1.5 m above the ground. Ignoring air resistance and taking g = 9.8 m/s², its height after t seconds is
h(t) = −4.9t² + 20t + 1.5
When does it land? Set h = 0: a = −4.9, b = 20, c = 1.5. Δ = 20² − 4 × (−4.9) × 1.5 = 400 + 29.4 = 429.4. Positive, so two real roots: t = (−20 ± √429.4) / (−9.8), which gives t ≈ 4.155 s and t ≈ −0.074 s. The negative root is real algebra but not physics: it is when the ball would have left the ground if it had been launched from below. The answer is about 4.16 seconds.
When is it 15 m high? Set h = 15, which moves 15 to the left: −4.9t² + 20t − 13.5 = 0. Δ = 400 − 4 × 4.9 × 13.5 = 400 − 264.6 = 135.4. Two roots: t ≈ 0.853 s on the way up and t ≈ 3.228 s on the way down.
Does it ever reach 25 m? −4.9t² + 20t − 23.5 = 0 gives Δ = 400 − 460.6 = −60.6. Negative: no real time works, so the ball never gets that high. You did not need to finish the formula to know it.
The vertex confirms this: the peak is at t = −b/(2a) = 20/9.8 ≈ 2.041 s, at a height of 1.5 + 20²/(4 × 4.9) ≈ 21.91 m. The discriminant for a target height H is zero exactly when H is that peak.
Worked example 2: a garden with 20 m of fence
You have 20 m of fencing for a rectangular bed. If one side is x metres, the other is 10 − x, and the area is x(10 − x).
- Area 21 m²: x(10 − x) = 21 becomes x² − 10x + 21 = 0. Δ = 100 − 84 = 16, a perfect square, so the roots are whole: x = (10 ± 4)/2 = 7 or 3. The bed is 7 m by 3 m (both roots describe the same rectangle, turned round).
- Area 25 m²: x² − 10x + 25 = 0, Δ = 0, so x = 5: a 5 m square, the largest area this fence can enclose.
- Area 30 m²: x² − 10x + 30 = 0, Δ = −20. Impossible with 20 m of fence; the complex roots 5 ± i√5 are the algebra's way of saying no.
Common mistakes
- Losing the sign of b. The formula starts with −b. If b = −10, then −b = +10. Writing "−10" here is the single most common slip. Putting brackets round every substituted value, (−10)² − 4(1)(21), prevents it, and also stops −10² being read as −100.
- Dividing only part of the numerator by 2a. In (−2 ± 2√3)/2 both terms are divided: −1 ± √3, not −2 ± √3.
- Forgetting the ±. A quadratic with Δ > 0 has two roots. Taking only the + sign loses half the answer, which in the garden example means losing the 3 m side.
- Dividing by 2 instead of 2a. When a ≠ 1 the denominator is 2a. For the ball, 2a = −9.8, and the negative sign matters.
- Using the formula when a = 0. Then there is no x² term and the equation is linear: bx + c = 0, so x = −c/b. The formula would divide by zero.
- Rejecting a root without a reason, or keeping one that makes no sense. A negative time or length is discarded because of the context, not because the algebra is wrong. Say why.
- Not checking. The roots of ax² + bx + c add up to −b/a and multiply to c/a. For x² − 10x + 21: 7 + 3 = 10 and 7 × 3 = 21. That takes five seconds and catches most arithmetic errors.
Using the solver to check your work
Type a, b and c into the Quadratic tab of the equation solver. It shows Δ with your values substituted, simplifies the square root, gives exact roots (fractions, surds or complex numbers) next to their decimals, and adds the vertex, the factored form when there is one, and the sum-and-product check. Decimal coefficients such as −4.9 are read exactly as −49/10, which is why the ball's roots appear as 100/49 ± √10735/49 before the decimals. To see the parabola itself, plot it with the function grapher.