Quadratic discriminant:
number of roots without solving the equation

Where the discriminant comes from

A quadratic equation ax² + bx + c = 0 is solved by the quadratic formula. Look at that formula and you will see that the part under the square root sign, b² − 4ac, decides whether the answer exists as a real number. That part has its own name: the discriminant, usually written D or Δ.

You can therefore know the nature of the roots before taking any square root.

The three cases

The graph of y = ax² + bx + c is a parabola, and the roots are where it meets the x-axis. A positive D means it cuts through, a zero D means it just touches at the vertex, and a negative D means it floats entirely above or below the axis.

Three worked examples

A fourth case with a coefficient: 3x² − 2x − 1 = 0. D = 4 + 12 = 16, so the roots are (2 ± 4) ÷ 6, which are 1 and −1/3.

Where the arithmetic goes wrong

• Squaring a negative b: (−5)² is 25, not −25. Put brackets around negative values.
• Forgetting that the c term subtracts 4ac: if a and c are both negative the product is positive and −4ac is negative, which can flip the sign of D.
• Using the equation before it is rearranged. 2x² = 3x − 5 must become 2x² − 3x + 5 = 0, so b = −3 and c = 5.
• Leaving out the a in 4ac when a is not 1.

A sign slip in D changes the conclusion, not just the final number, so double-check this one value before using the formula.

Using D to find an unknown coefficient

The discriminant is useful in reverse. Suppose x² + kx + 9 = 0 must have exactly one root. Set D to zero: k² − 4 × 1 × 9 = 0, so k² = 36 and k = 6 or k = −6. Each value gives a perfect square: x² + 6x + 9 = (x + 3)² and x² − 6x + 9 = (x − 3)².

For two distinct real roots you want D > 0, so k² > 36, which means k > 6 or k < −6. For no real roots, −6 < k < 6.

The link to the vertex and the complex roots

The discriminant also tells you how far the parabola's turning point sits from the x-axis. The vertex height is −D ÷ (4a). For x² − 5x + 6, D = 1 and a = 1, so the vertex is at height −1/4, a quarter unit below the axis, at x = 5 ÷ 2 = 2.5. That is why the curve dips just under the axis and crosses it at x = 2 and x = 3.

If a is positive and D is negative, the vertex is above the axis and the curve never comes down to it. If a is negative the shape is flipped, and a negative D then means the whole curve lies below the axis.

When D is negative the roots are not absent, they are complex. For 2x² + 3x + 5 = 0 they are (−3 ± i√31) ÷ 4, about −0.75 ± 1.39i. In most school and practical problems, the verdict is simply that no real solution exists, and that is the answer to give.

A quick test before factorising

Before you try to factorise, compute D. If it is not a perfect square, the quadratic will not factor into neat whole-number brackets, and you can save time by going straight to the formula. For x² + x + 1, D = 1 − 4 = −3 and there is nothing to factorise over the reals. For x² + 2x − 4, D = 4 + 16 = 20, which is not a perfect square, so the roots are −1 ± √5 and do not come from simple factors.

What it means in a real problem

A ball thrown upward follows h = −5t² + 20t, with h in metres and t in seconds (using g ≈ 10 m/s² for simplicity). Will it reach 25 m? Set −5t² + 20t = 25, or 5t² − 20t + 25 = 0. Here D = 400 − 500 = −100, which is negative, so there is no time at which the height is 25 m. The ball never gets there; its maximum is 20 m at t = 2 s.

The same reasoning decides whether a cost curve ever breaks even, whether a business's break-even condition has a solution, or whether two curves intersect. A discriminant calculator confirms the sign instantly, and the full roots can follow if they exist.