Type a quadratic and it is solved the way a teacher would choose – factoring when the numbers allow (x² − 5x + 6 = (x − 2)(x − 3)), otherwise the quadratic formula or completing the square. 2x² + 3x − 4 = 0 shows the discriminant, exact and decimal roots, complex roots when it is negative, and a graph with the vertex marked.
The quadratic formula
For ax^2 + bx + c = 0 with a \ne 0:
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}The number under the root, the discriminant D = b^2 - 4ac, tells you what kind of answers to expect:
| Discriminant | Solutions | Example |
|---|---|---|
D > 0, a perfect square | Two rational solutions – it factors | x^2 - 5x + 6 = 0: D = 1 |
D > 0, not a square | Two irrational solutions with a square root | x^2 - 6x + 4 = 0: D = 20 |
D = 0 | One repeated solution | x^2 + 6x + 9 = 0 |
D < 0 | No real solutions (two complex ones) | x^2 + 2x + 5 = 0: D = -16 |
Example: 2x^2 + 3x - 4 = 0, so a = 2, b = 3, c = −4.
D = 3^2 - 4 \cdot 2 \cdot (-4) = 9 + 32 = 41,\qquad x = \frac{-3 \pm \sqrt{41}}{4} \approx 0.851 \text{ or } -2.35141 isn’t a perfect square, so this one can’t be factored with whole numbers – the formula is the way. Open it in the solver, which also draws the parabola and marks both crossings and the vertex.
Which method is best?
- Square-root method – when there is no x-term (
3x^2 - 27 = 0) or the equation is already a square ((x - 3)^2 = 16). - Factor out x – when there is no constant term (
x^2 = 5x). Never divide by x: that loses the solution x = 0. - Factoring – when the discriminant is a perfect square. With a = 1, find two numbers that multiply to c and add to b; otherwise use the AC method (split the middle term and group).
- Quadratic formula – always works; the best choice when the answers involve square roots.
- Completing the square – always works too, and shows where the formula comes from; it also gives the vertex form
a(x - h)^2 + k.
Completing the square, briefly
Solve x^2 - 6x + 4 = 0
\begin{aligned} x^2 - 6x &= -4 \\ x^2 - 6x + 9 &= -4 + 9 && \text{half of } -6 \text{ is } -3,\ (-3)^2 = 9 \\ (x - 3)^2 &= 5 \\ x - 3 &= \pm\sqrt{5} \\ x &= 3 \pm \sqrt{5} \end{aligned}A quick check: sum and product of the roots
For ax^2 + bx + c = 0 the two solutions always add up to -\tfrac{b}{a} and multiply to \tfrac{c}{a}. For 2x² + 3x − 4 = 0 they must add to −3/2 and multiply to −2 – and indeed 0.851 + (−2.351) = −1.5 and 0.851 × (−2.351) ≈ −2. It’s a fast way to catch a sign slip, and it works in reverse: two numbers that add to 5 and multiply to 6 are the roots of x² − 5x + 6 = 0.
Quadratics in word problems
- Area: a rectangle is 3 m longer than it is wide and has an area of 40 m². With width w, w(w + 3) = 40, so w² + 3w − 40 = 0 gives w = 5 or w = −8. A width can’t be negative, so the room is 5 m by 8 m.
- Things thrown upwards: a ball’s height is h = 20t − 5t² metres after t seconds. It is 15 m up when 20t − 5t² = 15: t = 1 s on the way up and t = 3 s on the way down. The highest point is halfway between, at t = 2 s.
- Profit: when price changes affect how many you sell, revenue is often a quadratic in the price – its vertex gives the price that earns the most.
Always check answers against the story – a length, a time or a number of people can’t be negative, so one of the two solutions is often rejected.
To factor a quadratic expression without solving it, use the factoring calculator.
Frequently asked questions
How do I know if a quadratic can be factored?
With whole numbers, exactly when the discriminant b² − 4ac is a perfect square (0, 1, 4, 9, 16 …).
Can a quadratic have no solution?
It always has two solutions counting complex numbers, but when b² − 4ac is negative there is no real solution – the parabola does not cross the x-axis.
Why not divide both sides by x?
Because x might be 0. Dividing x² = 5x by x gives only x = 5 and loses x = 0. Factor instead: x(x − 5) = 0.