Discriminant in Quadratic Equations - Formula & Examples

What Is the Discriminant in Math?

For a quadratic equation $ax^2 + bx + c = 0$ (with $a \neq 0$), the discriminant is the expression:

$$\Delta = b^2 - 4ac$$

It appears as the piece under the radical in the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a}$$

The discriminant's sign determines how many real solutions the equation has — without you doing any actual solving.

The Discriminant Formula

For $ax^2 + bx + c = 0$:

$$\Delta = b^2 - 4ac$$

It's a single number you compute from the three coefficients. Always check that the equation is in standard form ($ax^2 + bx + c = 0$) before reading off $a, b, c$.

Worked example. Find the discriminant of $2x^2 - 5x + 1 = 0$.

Here $a = 2, b = -5, c = 1$:

$$\Delta = (-5)^2 - 4(2)(1) = 25 - 8 = 17$$

Since $\Delta > 0$, the equation has two distinct real roots — no need to solve to know this.

What Are the Three Cases?

The sign of $\Delta$ determines the nature of the roots:

Case 1: $\Delta > 0$ — Two Distinct Real Roots

The quadratic formula gives $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$ — two different real numbers because $\sqrt{\Delta}$ is a positive real.

Case 2: $\Delta = 0$ — One Repeated Root

The $\pm \sqrt{\Delta} = 0$, so the formula collapses to $x = -\dfrac{b}{2a}$ — a single value, counted twice. The parabola's vertex sits on the x-axis.

Case 3: $\Delta < 0$ — No Real Roots

The square root of a negative number isn't a real number — it's an imaginary number. The two roots are complex conjugates: $x = \dfrac{-b \pm i\sqrt{|\Delta|}}{2a}$. Geometrically, the parabola doesn't intersect the x-axis at all.

Three Worked Examples — Quick, Standard, Stretch

Quick — Identify the Case

For $x^2 + 4x + 4 = 0$, find the discriminant and describe the roots.

$a = 1, b = 4, c = 4$: $\Delta = 16 - 16 = 0$. One repeated real root at $x = -2$. (Check: $(x+2)^2 = 0$.)

Standard — Use the Discriminant to Find a Parameter

For what value of $k$ does $x^2 + kx + 9 = 0$ have one repeated root?

One repeated root means $\Delta = 0$. So $k^2 - 4(1)(9) = 0$, giving $k^2 = 36$ and $k = \pm 6$.

Check. For $k = 6$: $x^2 + 6x + 9 = (x+3)^2 = 0$ ✓ — repeated root at $x = -3$. For $k = -6$: $x^2 - 6x + 9 = (x-3)^2 = 0$ ✓.

Stretch — Discriminant and the Range of Values

For what values of $k$ does $kx^2 + 3x + 2 = 0$ have two distinct real roots?

Two distinct real roots means $\Delta > 0$:

$$9 - 8k > 0 \implies k < \tfrac{9}{8}$$

Also need $a \neq 0$, so $k \neq 0$.

Answer: $k < \tfrac{9}{8}$ and $k \neq 0$, i.e., $k \in (-\infty, 0) \cup (0, \tfrac{9}{8})$.

Why Does the Discriminant Matter? (The Real-World GROUND)

"To find when something happens, find where it equals zero." — physicist's heuristic.

The discriminant isn't just an algebraic curiosity. It tells you whether a real-world equation has solutions in the real world.

• Physics — projectile motion. A ball thrown up reaches height $h(t) = -\tfrac{1}{2}gt^2 + v_0 t + h_0$. Setting $h(t) = 0$ asks "when does the ball land?" The discriminant tells you whether it lands at all (and if so, when).

• Engineering — stability analysis. Many engineering problems reduce to "does this quadratic have real roots?" — vibration frequencies, control system stability, structural analysis. The discriminant gives the answer in one expression.

• Economics — equilibrium. Quadratic supply-demand equations have an equilibrium price only when the discriminant is non-negative.

• Computer graphics — ray tracing. Determining whether a ray hits a sphere reduces to a quadratic in distance; the discriminant decides hit vs miss.

The discriminant appears in Al-Khwarizmi's 820 CE Al-Jabr — though without modern notation. The modern symbol $\Delta$ comes from late-19th-century algebraic notation, building on François Viète's 16th-century symbolic algebra.

A Worked Example

For what value of $k$ does $x^2 + (k-2)x + 4 = 0$ have real and equal roots?

The intuitive (wrong) approach. A student writes $\Delta = (k-2)^2 - 4 \cdot 4 = 0$ and immediately gets $k = 2 \pm 4 = 6$ or $-2$.

Why it fails. The student forgot that $(k-2)^2 = 16$ has two solutions — $k - 2 = \pm 4$. Only checking one branch misses one of the answers.

The correct method. Real and equal roots means $\Delta = 0$:

$$(k-2)^2 - 16 = 0 \implies (k-2)^2 = 16 \implies k - 2 = \pm 4$$

So $k = 6$ or $k = -2$. Both work — both produce a quadratic with discriminant zero.

Check. For $k = 6$: equation becomes $x^2 + 4x + 4 = (x+2)^2 = 0$, repeated root at $-2$ ✓. For $k = -2$: equation becomes $x^2 - 4x + 4 = (x-2)^2 = 0$, repeated root at $2$ ✓.

What Are the Most Common Mistakes With the Discriminant?

Mistake 1: Sign errors when $b$ is negative

Where it slips in: Writing $b^2$ as a negative number.

Don't do this: For $x^2 - 5x + 6 = 0$, computing $b^2 = -25$.

The correct way: $b^2 = (-5)^2 = 25$. Squaring always produces a non-negative result, regardless of the sign of $b$.

Mistake 2: Forgetting to put the equation in standard form first

Where it slips in: Reading $a, b, c$ from $3x^2 + 2 = 5x$ as if it were standard form.

Don't do this: $\Delta = 0^2 - 4(3)(2) = -24$ for $3x^2 + 2 = 5x$.

The correct way: Rearrange first: $3x^2 - 5x + 2 = 0$, so $a = 3, b = -5, c = 2$, and $\Delta = 25 - 24 = 1$.

Mistake 3: Confusing "no real roots" with "no roots"

Where it slips in: A negative discriminant means no real roots — but the equation does have two roots, which are complex.

Don't do this: Stating $x^2 + 1 = 0$ has "no solution."

The correct way: Over the real numbers, no solution; over the complex numbers, two solutions $x = \pm i$.

Key Takeaways

• The discriminant $\Delta = b^2 - 4ac$ is the piece under the square root in the quadratic formula.

• Sign reveals nature of roots: $\Delta > 0$ → two real, $\Delta = 0$ → one repeated, $\Delta < 0$ → two complex.

• Geometric picture: the discriminant counts how many times the parabola crosses the x-axis.

• Standard form first — rearrange to $ax^2 + bx + c = 0$ before reading off $a, b, c$.

• Generalises to higher degrees — every polynomial has a discriminant, though the formula grows with degree.

A Practical Next Step

Try these three before moving on to the quadratic formula and complex roots.

1. Find the discriminant of $3x^2 + 4x + 2 = 0$. How many real roots?

2. For what value of $k$ does $x^2 + kx + 25 = 0$ have one repeated root?

3. For what values of $k$ does $x^2 + 2x + k = 0$ have no real roots?

If problem 3 felt tricky, $\Delta < 0$ means $4 - 4k < 0$, so $k > 1$. Want a Bhanzu trainer to walk through more discriminant problems? Book a free demo class — online globally.