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.

Quick check: without solving the equation, how many real solutions does \(x^2 - 4x + 4 = 0\) have? Show answerDiscriminant: \((-4)^2 - 4(1)(4) = 16 - 16 = 0\). Since the discriminant equals zero, there is exactly one real solution — the parabola just touches the \(x\)-axis at its vertex.

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 the quadratic formula gave above.

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

1. Which of these is a quadratic equation?

   • \(3x + 5 = 0\)
   • \(x^2 - 4x + 1 = 0\)
   • \(x^3 + 2x = 0\) Show answer\(x^2 - 4x + 1 = 0\) is quadratic — it has an \(x^2\) term. The first is linear (no \(x^2\)); the third is cubic (\(x^3\) is the highest power).

2. Solve using the quadratic formula: \(x^2 + 3x - 4 = 0\) Show answer\(a = 1\), \(b = 3\), \(c = -4\). \(x = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}\). So \(x = 1\) or \(x = -4\).

3. Find the discriminant of \(x^2 - 6x + 9 = 0\). How many solutions does it have? Show answerDiscriminant: \((-6)^2 - 4(1)(9) = 36 - 36 = 0\). One solution — the parabola touches the \(x\)-axis exactly once.

4. Solve: \(x^2 = 16\) (Hint: rewrite as \(x^2 - 16 = 0\)) Show answer\(x = \pm 4\) — two solutions: \(x = 4\) and \(x = -4\). Don't forget the negative root!

What's Next?

This is the introduction. In Algebra II, you'll meet more sophisticated factoring techniques, completing the square, vertex form of parabolas, and the broader connections between quadratics and other parts of math. The dedicated quadratic formula lesson goes deeper on technique, and parabolas covers the geometry of the graph.

A few patterns to watch for as you practice. The \(\pm\) in the formula always produces two solutions (or one repeated solution if the discriminant is zero); dropping it costs you half the answer. Sign errors are the most common slip: if \(b = -5\), then \(-b = 5\). When squaring \(b\), the sign disappears either way — \((-5)^2 = 25\), same as \(5^2\). The entire numerator \(-b \pm \sqrt{b^2 - 4ac}\) gets divided by \(2a\); you can't split the division up. Simplify radicals and fractions in your final answer: \(\frac{6}{2}\) becomes 3, \(\sqrt{8}\) becomes \(2\sqrt{2}\). And equations like \(x^2 = 9\) have two solutions, \(x = 3\) and \(x = -3\) — always take both roots.