2.3 — The Quadratic Formula

One Derivation to Rule Them All

At the end of the last section, we made an observation: completing the square works for every quadratic, but the algebra gets repetitive. Each time, you move the constant, halve the middle coefficient, square it, add to both sides, take the square root, and solve. The same sequence of steps, over and over, with different numbers plugged in.

What if we did those steps just once — with letters instead of numbers? We'd get a formula that swallows any quadratic equation and spits out the solutions directly.

That's exactly what we're about to do. And unlike most formulas students are asked to memorize, this one won't be handed to you from nowhere. You're going to watch it emerge from a process you already understand.

Derive It Yourself

Here's the challenge. Start with the most general quadratic equation possible:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

Try this before reading on. You know how to complete the square from Section 2.2. Apply those same steps to this general equation. The algebra is messier because you're working with $a$, $b$, and $c$ instead of specific numbers — but the logic is identical. See how far you get.

Step-by-step derivation

**Step 1: Divide by $a$** so the leading coefficient is $1$. $$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$ **Step 2: Move the constant to the other side.** $$x^2 + \frac{b}{a}x = -\frac{c}{a}$$ **Step 3: Find the missing piece.** The coefficient of $x$ is $\frac{b}{a}$. Halve it: $\frac{b}{2a}$. Square it: $\frac{b^2}{4a^2}$. This is the corner square from Section 2.2 — just written with letters. **Step 4: Add it to both sides.** $$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$$ The left side is now a perfect square: $$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}$$ **Step 5: Simplify the right side.** Get a common denominator of $4a^2$: $$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$ **Step 6: Take the square root of both sides.** $$x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$ **Step 7: Solve for $x$.** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

And there it is:

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

This is the quadratic formula. Every step is one you've already done in Section 2.2. The only difference is that we used $a$, $b$, $c$ instead of specific numbers — so the result works for any quadratic equation.

Key insight: The quadratic formula isn't a magic spell. It's completing the square, done once in general. If you ever forget the formula, you can re-derive it from scratch in about a minute.

Using It Efficiently

The formula has several moving parts, so let's build good habits by working through an example carefully.

Solve $2x^2 - 7x + 3 = 0$.

Before plugging anything in, identify the three coefficients:

$$a = 2, \quad b = -7, \quad c = 3$$

Common mistake: Getting the sign of $b$ wrong. The formula says $-b$, and if $b$ is already negative, then $-b$ is positive. Be deliberate about this: write out $-(-7) = 7$ rather than trying to do it in your head.

Now substitute:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$$

$$x = \frac{7 \pm \sqrt{49 - 24}}{4}$$

$$x = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$$

Two solutions:

$$x = \frac{7 + 5}{4} = 3 \quad \text{or} \quad x = \frac{7 - 5}{4} = \frac{1}{2}$$

Verify it yourself. Plug $x = 3$ into $2x^2 - 7x + 3$: you get $18 - 21 + 3 = 0$. Plug in $x = \frac{1}{2}$: you get $\frac{1}{2} - \frac{7}{2} + 3 = 0$. Both check out.

A Practical Tip

Compute the discriminant — the expression under the square root — first, before doing anything else. Write it off to the side:

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

This keeps the rest of the calculation clean. For the example above, $\Delta = 49 - 24 = 25$, and once you know $\sqrt{25} = 5$, the rest is simple arithmetic.

Check your understanding

Solve using the quadratic formula: $x^2 + 2x - 8 = 0$

Start by identifying $a$, $b$, and $c$. Compute the discriminant first.

Check your answer

$a = 1$, $b = 2$, $c = -8$. Discriminant: $\Delta = 4 - 4(1)(-8) = 4 + 32 = 36$. $$x = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$$ $$x = \frac{4}{2} = 2 \quad \text{or} \quad x = \frac{-8}{2} = -4$$ Verify: $2^2 + 2(2) - 8 = 4 + 4 - 8 = 0$. $(-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0$. Both work. Note: this equation is also factorable as $(x + 4)(x - 2) = 0$. The quadratic formula gives the same answers — it's just more systematic.

The Discriminant: A Traffic Light for Solutions

Look at the formula again:

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

Everything outside the square root is predictable arithmetic. The entire character of the solutions depends on one expression:

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

This is called the discriminant, and it acts as a traffic light. Before you solve a single thing, it tells you what kind of answer to expect.

Green light: $\Delta > 0$ — Two distinct real roots

When $b^2 - 4ac$ is positive, the square root gives a real number, and the $\pm$ produces two different values. The parabola crosses the $x$-axis at two points.

Example: $x^2 - 5x + 6 = 0$

$$\Delta = 25 - 24 = 1 > 0$$

Two roots: $x = \frac{5 \pm 1}{2}$, giving $x = 3$ and $x = 2$. (These are the same roots we found by factoring in Section 2.1 — reassuring.)

Yellow light: $\Delta = 0$ — One repeated root

When the discriminant is exactly zero, the $\pm$ doesn't matter — adding or subtracting zero gives the same result. There's only one solution:

$$x = \frac{-b}{2a}$$

The parabola just touches the $x$-axis at its vertex and bounces back.

Example: $x^2 - 6x + 9 = 0$

$$\Delta = 36 - 36 = 0$$

One root: $x = \frac{6}{2} = 3$. (This is the perfect square trinomial $(x - 3)^2 = 0$ from Section 2.1.)

Red light: $\Delta < 0$ — No real roots

When $b^2 - 4ac$ is negative, you'd need the square root of a negative number. That doesn't exist among the real numbers. The quadratic equation has no real solutions.

Geometrically, the parabola never touches the $x$-axis — it floats entirely above it (if $a > 0$) or entirely below it (if $a < 0$).

Example: $x^2 + 2x + 5 = 0$

$$\Delta = 4 - 20 = -16 < 0$$

No real roots. The parabola $y = x^2 + 2x + 5$ has its vertex at $(-1, 4)$, which is above the $x$-axis. Since it opens upward, it never comes down far enough to cross.

Why "no real roots"? We say "real" because there's a larger number system — the complex numbers — where $\sqrt{-16}$ does have meaning. In that world, every quadratic has exactly two roots, always. But that's a story for another course.

The Discriminant and the Parabola

There's a satisfying way to see all three cases at once. Think about a parabola opening upward:

• As you shift the parabola down (decreasing $c$), the vertex drops, and eventually the curve crosses the $x$-axis. The discriminant goes from negative to zero to positive.
• As you shift it up (increasing $c$), the vertex rises, and the curve lifts away from the axis. The discriminant goes from positive to zero to negative.

The discriminant measures, in a precise algebraic sense, how far the vertex is from the $x$-axis relative to the shape of the parabola. It's the bridge between the algebra (the formula) and the geometry (the graph).

Connection to vertex form: From Section 2.2, we know the vertex is at $\left(-\frac{b}{2a},\; c - \frac{b^2}{4a}\right)$. The $y$-coordinate of the vertex is $c - \frac{b^2}{4a} = \frac{4ac - b^2}{4a} = \frac{-\Delta}{4a}$. When $a > 0$, the vertex is below the axis exactly when $\Delta > 0$, on the axis when $\Delta = 0$, and above it when $\Delta < 0$. The discriminant and the vertex are telling the same story.

Check your understanding

Without solving, determine how many real solutions each equation has:

(a) $3x^2 - x + 4 = 0$

(b) $x^2 - 10x + 25 = 0$

(c) $2x^2 + x - 6 = 0$

Check your answer

(a) $\Delta = 1 - 48 = -47 < 0$. **No real solutions.** The parabola doesn't touch the $x$-axis. (b) $\Delta = 100 - 100 = 0$. **One repeated root.** The parabola just touches the axis. (It's the perfect square $(x - 5)^2 = 0$.) (c) $\Delta = 1 + 48 = 49 > 0$. **Two distinct real roots.** And since $49$ is a perfect square, the roots are rational — this equation is factorable.

When to Use Which Method

You now have three tools for solving quadratics. Here's how to choose:

Method	Best when...	Limitation
Factoring	Coefficients are small and roots are integers	Fails for non-integer roots
Completing the square	You need vertex form, or want to understand the geometry	Algebra gets heavy when $a \neq 1$
Quadratic formula	You want a guaranteed answer for any equation	Overkill for simple cases; doesn't reveal structure as clearly

The quadratic formula always works, so you might wonder: why ever bother with the other methods?

Because speed and understanding aren't the same thing. Factoring a simple quadratic by sight is faster than plugging into a formula. And completing the square reveals the vertex — something the quadratic formula buries inside its arithmetic. Each method has its own strengths.

A good analogy: The quadratic formula is like a GPS navigator. It always gets you to the destination. But factoring is like knowing the neighborhood — it's faster when you recognize where you are. And completing the square is like reading a map — it shows you the terrain, not just the route.

Reflection

Let's trace the arc of this chapter.

We started with factoring (Section 2.1): fast and elegant, but limited to "nice" roots. Then we developed completing the square (Section 2.2): universal but laborious, with the bonus of revealing vertex form. Now the quadratic formula (this section) packages completing the square into a single reusable expression.

Each method builds on the one before it:

• Factoring uses the zero-product property.
• Completing the square uses geometry to reshape equations.
• The quadratic formula is completing the square done once, in general.

And the discriminant $\Delta = b^2 - 4ac$ ties everything together. It doesn't just tell you how many roots there are — it tells you why. A positive discriminant means the parabola reaches the axis. A zero discriminant means it barely touches. A negative discriminant means it never arrives. The algebra and the geometry agree, as they always should.

What changed: In Section 2.1, we could only solve equations with integer roots. In Section 2.2, we could solve any equation, but the process was long. Now we have a formula that solves any quadratic equation in a single step — and a diagnostic tool (the discriminant) that tells us the answer's character before we even compute it.

Practice

Problem 1 — Concrete. Solve using the quadratic formula:

(a) $x^2 - 4x - 21 = 0$

(b) $3x^2 + 2x - 1 = 0$

(c) $x^2 + 6x + 9 = 0$

Solutions

(a) $a = 1$, $b = -4$, $c = -21$. $\Delta = 16 + 84 = 100$. $$x = \frac{4 \pm 10}{2} \implies x = 7 \text{ or } x = -3$$ (b) $a = 3$, $b = 2$, $c = -1$. $\Delta = 4 + 12 = 16$. $$x = \frac{-2 \pm 4}{6} \implies x = \frac{1}{3} \text{ or } x = -1$$ (c) $a = 1$, $b = 6$, $c = 9$. $\Delta = 36 - 36 = 0$. $$x = \frac{-6}{2} = -3 \quad \text{(repeated root)}$$ This is $(x + 3)^2 = 0$ — the discriminant confirms it's a perfect square trinomial.

Problem 2 — Discriminant. Without fully solving, classify each equation (two roots, one root, or no real roots), and state whether the roots are rational or irrational:

(a) $5x^2 + 3x - 2 = 0$

(b) $x^2 + x + 1 = 0$

(c) $x^2 - 3x + 1 = 0$

Solutions

(a) $\Delta = 9 + 40 = 49$. Positive and a perfect square. **Two distinct rational roots.** (b) $\Delta = 1 - 4 = -3$. Negative. **No real roots.** (c) $\Delta = 9 - 4 = 5$. Positive but not a perfect square. **Two distinct irrational roots.** (The solutions involve $\sqrt{5}$, which can't be simplified to a fraction.)

Problem 3 — Connection. Solve $x^2 + 3x + 1 = 0$ using the quadratic formula. (This is the equation from Section 2.1 that we couldn't factor.)

Solution

$a = 1$, $b = 3$, $c = 1$. $\Delta = 9 - 4 = 5$. $$x = \frac{-3 \pm \sqrt{5}}{2}$$ So $x = \frac{-3 + \sqrt{5}}{2} \approx -0.382$ or $x = \frac{-3 - \sqrt{5}}{2} \approx -2.618$. These are the roots that factoring couldn't find — they're irrational. But the quadratic formula handles them without hesitation. The equation that stumped us in Section 2.1 is now solved.

Problem 4 — Debug. A student applies the quadratic formula to $x^2 + 6x + 5 = 0$ and writes:

"$a = 1$, $b = 6$, $c = 5$. So $x = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2}$, giving $x = 5$ or $x = 1$."

Both answers are wrong. Find the error.

Solution

The student forgot the negative sign on $b$. The formula says $-b$, not $b$. Since $b = 6$, the numerator should start with $-6$, not $6$. Correct calculation: $$x = \frac{-6 \pm \sqrt{36 - 20}}{2} = \frac{-6 \pm 4}{2}$$ $$x = \frac{-6 + 4}{2} = -1 \quad \text{or} \quad x = \frac{-6 - 4}{2} = -5$$ Verify: $(-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0$. $(-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0$. Both work. **Lesson:** The $-b$ at the front of the formula is the most common source of sign errors. Always write $-(b)$ explicitly before simplifying — especially when $b$ is positive.

Problem 5 — Thinking deeper. The discriminant of $x^2 + bx + 1 = 0$ is $b^2 - 4$. For what values of $b$ does this equation have (a) two real roots, (b) exactly one root, (c) no real roots?

Solution

(a) Two real roots when $\Delta > 0$: $b^2 - 4 > 0$, so $b^2 > 4$, meaning $b > 2$ or $b < -2$. (b) One repeated root when $\Delta = 0$: $b^2 = 4$, so $b = 2$ or $b = -2$. (c) No real roots when $\Delta < 0$: $b^2 < 4$, so $-2 < b < 2$. This is worth visualizing. As $b$ varies from $0$ outward, the parabola $y = x^2 + bx + 1$ tilts and its vertex drops. At $b = \pm 2$, the vertex just barely touches the $x$-axis. Beyond that, the parabola breaks through and crosses twice. The discriminant quantifies this transition precisely.

What's Next

You now have a complete toolkit for solving any quadratic equation, and a way to predict the nature of its solutions before solving. In the next chapter, we'll shift perspective: instead of asking where a parabola crosses the axis, we'll study the parabola itself — its shape, its transformations, and what the different forms of a quadratic ($standard$, $vertex$, $factored$) each reveal about the graph.