Introduction to Quadratic Equations

A quadratic equation is any equation that can be written in the form:

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

where $a$, $b$, and $c$ are constants and $a \neq 0$. The $x^2$ term is what makes it quadratic—"quadratic" comes from the Latin word for square.

These equations are everywhere in the real world: the path of a thrown ball, the shape of a satellite dish, the design of suspension bridges, even calculating profit in business often involves quadratic equations.

Standard Form

The standard form of a quadratic equation is $ax^2 + bx + c = 0$.

Examples:

$x^2 + 5x + 6 = 0$ (here $a = 1$, $b = 5$, $c = 6$)
$2x^2 - 7x + 3 = 0$ (here $a = 2$, $b = -7$, $c = 3$)
$x^2 - 9 = 0$ (here $a = 1$, $b = 0$, $c = -9$)

Notice that $b$ or $c$ can be zero, but $a$ cannot be zero (or else you wouldn't have an $x^2$ term, and it wouldn't be quadratic).

Graphing Quadratic Functions

If you replace the $= 0$ with $= y$, you get a quadratic function: $y = ax^2 + bx + c$.

The graph of a quadratic function is a parabola—a U-shaped curve.

If $a > 0$, the parabola opens upward (like a smile).

If $a < 0$, the parabola opens downward (like a frown).

The vertex is the highest or lowest point on the parabola. For an upward-opening parabola, it's the minimum point. For a downward-opening parabola, it's the maximum point.

Example: The graph of $y = x^2$ is an upward-opening parabola with vertex at $(0, 0)$.

Example: The graph of $y = -x^2 + 4$ is a downward-opening parabola with vertex at $(0, 4)$.

Solutions and Roots

The solutions (or roots or zeros) of a quadratic equation are the values of $x$ that make the equation true.

Graphically, these are the $x$-intercepts—the points where the parabola crosses the $x$-axis.

A quadratic equation can have:

Two real solutions (the parabola crosses the $x$-axis twice)
One real solution (the parabola just touches the $x$-axis at the vertex)
No real solutions (the parabola doesn't touch the $x$-axis at all)

Example: $x^2 - 5x + 6 = 0$ has two solutions: $x = 2$ and $x = 3$.

Check: $2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓

Check: $3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓

The Quadratic Formula

The quadratic formula is a reliable method for solving any quadratic equation. It works even when factoring is difficult or impossible.

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

The $\pm$ symbol means you get two solutions: one using plus, one using minus.

Example: Solve $x^2 + 5x + 6 = 0$ using the quadratic formula.

Identify: $a = 1$, $b = 5$, $c = 6$

$$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$$

$$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$$

$$x = \frac{-5 \pm \sqrt{1}}{2}$$

$$x = \frac{-5 \pm 1}{2}$$

Two solutions: $$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$$

$$x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$$

So the solutions are $x = -2$ and $x = -3$.

Example: Solve $2x^2 - 3x - 5 = 0$.

Here $a = 2$, $b = -3$, $c = -5$

$$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}$$

$$x = \frac{3 \pm \sqrt{9 + 40}}{4}$$

$$x = \frac{3 \pm \sqrt{49}}{4}$$

$$x = \frac{3 \pm 7}{4}$$

$$x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}$$ or $$x = \frac{3 - 7}{4} = \frac{-4}{4} = -1$$

Solutions: $x = \frac{5}{2}$ and $x = -1$

The Discriminant

The expression under the square root in the quadratic formula, $b^2 - 4ac$, is called the discriminant. It tells you how many real solutions the equation has:

If $b^2 - 4ac > 0$, there are two real solutions
If $b^2 - 4ac = 0$, there is one real solution
If $b^2 - 4ac < 0$, there are no real solutions (the solutions involve imaginary numbers)

Example: How many real solutions does $x^2 + 2x + 5 = 0$ have?

Calculate the discriminant: $$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$$

Since $-16 < 0$, there are no real solutions. The parabola $y = x^2 + 2x + 5$ doesn't cross the $x$-axis.

Solving by Factoring (Preview)

In Algebra II, you'll learn more about factoring quadratic expressions. When a quadratic can be factored, it's often the fastest way to solve it.

Example: $x^2 + 5x + 6 = 0$ factors as $(x + 2)(x + 3) = 0$.

If a product equals zero, at least one factor must be zero:

Either $x + 2 = 0$ (so $x = -2$)

Or $x + 3 = 0$ (so $x = -3$)

Same solutions we got with the quadratic formula!

Real-World Example

A ball is thrown upward with an initial velocity of 20 meters per second from a height of 2 meters. Its height $h$ (in meters) after $t$ seconds is given by:

$$h = -5t^2 + 20t + 2$$

When does the ball hit the ground?

Set $h = 0$: $$-5t^2 + 20t + 2 = 0$$

Use the quadratic formula with $a = -5$, $b = 20$, $c = 2$:

$$t = \frac{-20 \pm \sqrt{20^2 - 4(-5)(2)}}{2(-5)}$$

$$t = \frac{-20 \pm \sqrt{400 + 40}}{-10}$$

$$t = \frac{-20 \pm \sqrt{440}}{-10}$$

$$t = \frac{-20 \pm 20.98}{-10}$$

$$t = \frac{-20 + 20.98}{-10} \approx -0.098$$ or $$t = \frac{-20 - 20.98}{-10} \approx 4.098$$

The negative time doesn't make sense (time can't be negative), so the ball hits the ground after about 4.1 seconds.

Test Yourself

Which of these is a quadratic equation?
- $3x + 5 = 0$
- $x^2 - 4x + 1 = 0$
- $x^3 + 2x = 0$
Solve using the quadratic formula: $x^2 + 3x - 4 = 0$
Find the discriminant of $x^2 - 6x + 9 = 0$. How many solutions does it have?
Solve: $x^2 = 16$ (Hint: Rewrite as $x^2 - 16 = 0$)

Solutions:

$x^2 - 4x + 1 = 0$ is quadratic (has $x^2$ term)
$a = 1$, $b = 3$, $c = -4$. $x = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}$. So $x = 1$ or $x = -4$.
Discriminant: $(-6)^2 - 4(1)(9) = 36 - 36 = 0$. One solution (the parabola touches the $x$-axis once).
$x = \pm 4$ (two solutions: 4 and -4)

What's Next?

This is just an introduction to quadratics. In Algebra II, you'll learn more sophisticated factoring techniques, complete the square, work with the vertex form of parabolas, and solve more complex quadratic equations. You'll also see how quadratics connect to many other areas of math.

Mistakes to Avoid

Forgetting the $\pm$ symbol in the quadratic formula costs you half your answer. The formula gives you two solutions (or tells you they're identical if the discriminant is zero). Don't drop that plus-minus sign.

Sign errors when substituting into the formula are incredibly common. If $b = -5$, then $-b = 5$, not $-5$. Pay close attention to those negatives, especially when squaring $b$.

Dividing only part of the numerator by $2a$ is incorrect. The entire expression $-b \pm \sqrt{b^2 - 4ac}$ sits in the numerator, and all of it gets divided by $2a$. You can't split it up.

Not simplifying radicals or fractions in your final answer makes your work look incomplete. If you get $\frac{6}{2}$, simplify it to 3. If you get $\sqrt{8}$, simplify it to $2\sqrt{2}$.

Thinking equations like $x^2 = 9$ have only one solution misses half the picture. Both $x = 3$ and $x = -3$ work because $(-3)^2 = 9$. Always consider both the positive and negative square roots.