MISSION · Algebra
The Quadratic Formula and the Discriminant
The method that works when factoring won't
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
Every quadratic \(ax^2 + bx + c = 0\) is solved by one formula, and one piece of that formula tells you how many solutions there are before you finish.
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]When to reach for it
Factoring is faster when it works. It works when the numbers are friendly, which in real problems they usually aren’t. The formula never fails, never needs a hunt for factor pairs, and handles irrational and decimal answers without complaint.
A sensible habit: spend thirty seconds looking for factors, then stop and use the formula.
Read the coefficients off carefully first. In \(3x^2 - 5x + 2 = 0\), \(a = 3\), \(b = -5\), \(c = 2\) — and that minus sign belongs to \(b\). Losing it is the single most common source of a wrong answer here, because \(-b\) then comes out with the wrong sign and everything after it follows.
The discriminant
The part under the root, \(b^2 - 4ac\), is called the discriminant, and it decides everything about the answers:
| \(b^2 - 4ac\) | Solutions | The parabola |
|---|---|---|
| Positive | Two real solutions | Crosses the \(x\)-axis twice |
| Zero | One repeated solution | Touches the axis at its vertex |
| Negative | No real solutions | Misses the axis entirely |
The reason is the square root. A positive number has two roots, giving the \(\pm\) two different values. Zero has one, so the \(\pm\) adds nothing. A negative number has no real square root at all, so the formula produces nothing real — the parabola sits entirely above or below the axis.
That lets you answer “how many times does this cross?” with one subtraction, without solving anything.
Where the formula comes from
It isn’t handed down. It’s what you get by completing the square on the general quadratic — taking \(ax^2 + bx + c\), forcing the first two terms into a perfect square, and solving the result for \(x\). The \(\pm\) appears exactly where you take the square root of both sides, since both roots count.
Doing that derivation once is worth the twenty minutes. It converts the formula from something memorised into something you could rebuild if you forgot it.
Reading the answers back
The two solutions are the \(x\)-intercepts of \(y = ax^2 + bx + c\), and they sit symmetrically either side of the vertex. Their midpoint is \(x = -\frac{b}{2a}\), which is the axis of symmetry — visible in the formula as the part outside the root.
So the formula really says: start at the axis of symmetry, then step out by \(\frac{\sqrt{b^2-4ac}}{2a}\) in both directions.
Common mix-ups
- Losing the sign of \(b\). If \(b = -5\) then \(-b = +5\).
- Dividing only part of the numerator by \(2a\). The whole top, including \(-b\), is divided.
- Forgetting to rearrange first. The formula needs the equation set to zero.
- Reading a negative discriminant as an error. It’s an answer: no real solutions.
Check yourself
Solve \(2x^2 + 3x - 5 = 0\), and say what the discriminant told you before you started.
\(b^2 - 4ac = 9 + 40 = 49\) — positive, and a perfect square, so two rational solutions. The formula gives \(x = \frac{-3 \pm 7}{4}\), so \(x = 1\) or \(x = -2.5\).
LEVEL 3 · REFLECT
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