Math Core

A quadratic equation is an equation that can be written as $ax^2 + bx + c = 0$ with $a \ne 0$ . Its solutions are the x-intercepts of the parabola $y = ax^2 + bx + c$ . In the last unit you learned to factor trinomials; now that skill pays off, because a factored quadratic can be solved almost instantly.

The zero product property

If you multiply two numbers and get $0$ , what can you say about them? At least one of them must be $0$ : $5 \cdot 3$ isn't zero, $(-2)(7)$ isn't zero, but $0 \cdot 7$ is. This simple fact is the whole method.

Zero product property

If $A \cdot B = 0$ , then $A = 0$ or $B = 0$ (or both).

So if an equation says $(x - 4)(x + 7) = 0$ , one of the two factors must be zero:

x - 4 = 0 \quad\text{or}\quad x + 7 = 0 \quad\Longrightarrow\quad x = 4 \quad\text{or}\quad x = -7.

Check: $(4 - 4)(4 + 7) = 0 \cdot 11 = 0$ and $(-7 - 4)(-7 + 7) = (-11) \cdot 0 = 0$ . Both work.

The property only works for zero. If $A \cdot B = 6$ , the factors could be $2$ and $3$ , or $1$ and $6$ , or $-12$ and $-\tfrac{1}{2}$ , or infinitely many other pairs, so you learn nothing about either factor on its own.

Worked example: Equations already factored

Solve each equation.

$(x + 3)(x - 8) = 0$
$3x(2x - 5) = 0$

Solutions.

$x + 3 = 0$ or $x - 8 = 0$ , so $x = -3$ or $x = 8$ .
The factors are $3x$ and $2x - 5$ . Setting $3x = 0$ gives $x = 0$ . Setting $2x - 5 = 0$ gives $x = \dfrac{5}{2}$ . The solutions are $0$ and $\dfrac{5}{2}$ .

Solving by factoring, step by step

Rewrite the equation in standard form, with $0$ on one side.
Factor the other side completely (GCF first, then the trinomial).
Set each factor containing $x$ equal to $0$ .
Solve each small equation.
Check each solution in the original equation.

Worked example: A trinomial

Solve $x^2 + 2x - 15 = 0$ .

Look for two numbers that multiply to $-15$ and add to $2$ : they are $5$ and $-3$ .

\begin{aligned} x^2 + 2x - 15 &= 0 \\ (x + 5)(x - 3) &= 0 \\ x + 5 = 0 \quad &\text{or} \quad x - 3 = 0 \\ x = -5 \quad &\text{or} \quad x = 3 \end{aligned}

These are exactly the x-intercepts of $y = x^2 + 2x - 15$ . The vertex sits halfway between them, at $x = -1$ .

The solutions of x² + 2x - 15 = 0 are the x-intercepts of y = x² + 2x - 15.Open in grapher →

Worked example: Rearrange first

Solve $x^2 = 5x + 14$ .

The equation is not in standard form yet. Subtract $5x$ and $14$ from both sides:

\begin{aligned} x^2 - 5x - 14 &= 0 \\ (x - 7)(x + 2) &= 0 \\ x = 7 \quad &\text{or} \quad x = -2 \end{aligned}

Check $x = 7$ : $49 = 35 + 14$ . ✓ Check $x = -2$ : $4 = -10 + 14$ . ✓

Common mistake

Two traps to avoid:

Setting factors equal to a nonzero number. $(x - 2)(x + 1) = 4$ does not mean $x - 2 = 4$ . Expand, move the $4$ over to get $x^2 - x - 6 = 0$ , and factor again: $(x - 3)(x + 2) = 0$ .
Dividing by $x$ . From $x^2 = 6x$ , dividing by $x$ gives only $x = 6$ and loses the solution $x = 0$ . Instead write $x^2 - 6x = 0$ , so $x(x - 6) = 0$ and $x = 0$ or $x = 6$ .

When $a$ is not 1

Use the factoring methods from the last unit, then finish the same way.

Worked example: A leading coefficient

Solve $2x^2 + 7x - 4 = 0$ .

For the $ac$ method: $ac = 2(-4) = -8$ , and the pair $8$ and $-1$ multiplies to $-8$ and adds to $7$ . Split the middle term and group:

\begin{aligned} 2x^2 + 8x - x - 4 &= 0 \\ 2x(x + 4) - 1(x + 4) &= 0 \\ (2x - 1)(x + 4) &= 0 \end{aligned}

So $2x - 1 = 0$ or $x + 4 = 0$ , giving $x = \dfrac{1}{2}$ or $x = -4$ .

One solution: a double root

Sometimes both factors are the same. Solve $x^2 - 10x + 25 = 0$ :

(x - 5)^2 = 0 \quad\Longrightarrow\quad x = 5.

There is only one solution, called a double root. On the graph, the parabola $y = x^2 - 10x + 25$ doesn't cross the $x$ -axis; it just touches it at its vertex, $(5, 0)$ .

A perfect square trinomial has one x-intercept: the parabola touches the axis at its vertex.Open in grapher →

Tip

Word problems often produce one solution that doesn't make sense, such as a negative length or a negative time. Solve the equation completely, then keep only the answers that fit the situation.