Math Core

You already know four ways to solve a quadratic equation: factoring, square roots, completing the square and the quadratic formula. In Algebra 2 the goal is fluency: looking at an equation, choosing the fastest method, predicting how many solutions to expect, and recognizing equations that are quadratics in disguise.

First, get it into shape

Every method except square roots needs the equation in the form $ax^2 + bx + c = 0$ . Before you do anything else, expand, collect all terms on one side, and simplify.

2x(x - 3) = x + 4 \quad\Longrightarrow\quad 2x^2 - 6x = x + 4 \quad\Longrightarrow\quad 2x^2 - 7x - 4 = 0.

If every coefficient shares a common factor, divide it out. If there are fractions, multiply through by a common denominator. Smaller whole-number coefficients mean less arithmetic and fewer mistakes.

Common mistake

Never divide both sides by an expression containing $x$ . From $x^2 = 5x$ , dividing by $x$ gives only $x = 5$ and loses the solution $x = 0$ . Instead, write $x^2 - 5x = 0$ , factor as $x(x - 5) = 0$ , and keep both solutions.

Choosing a method

the equation looks like	best method
$a(x - h)^2 = d$ , or no $x$ -term at all	square roots
easy to factor over the integers	factoring
$a = 1$ and $b$ is even	completing the square works cleanly
anything else	quadratic formula

The quadratic formula always works, so it is never wrong to use it. It is often just slower than the alternatives.

Worked example: Square roots

Solve $3(x - 2)^2 = 27$ .

Divide by $3$ to get $(x - 2)^2 = 9$ . The numbers whose square is $9$ are $3$ and $-3$ , so

x - 2 = \pm 3 \quad\Longrightarrow\quad x = 5 \ \text{or}\ x = -1.

Worked example: Factoring with a leading coefficient

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

Look for two numbers with product $2 \times (-4) = -8$ and sum $-7$ : they are $-8$ and $1$ . 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 $x = -\dfrac{1}{2}$ or $x = 4$ .

Worked example: The quadratic formula

Solve $3x^2 - 4x - 2 = 0$ .

Nothing factors nicely, so use $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 3$ , $b = -4$ , $c = -2$ :

x = \frac{4 \pm \sqrt{16 + 24}}{6} = \frac{4 \pm \sqrt{40}}{6} = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}.

To simplify, factor $\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}$ , then divide every term of the numerator and the denominator by $2$ .

The discriminant predicts the answer

The expression under the square root in the quadratic formula, $b^2 - 4ac$ , is called the discriminant. You can compute it before solving to learn what kind of answer is coming.

The discriminant

For $ax^2 + bx + c = 0$ with real coefficients, let $D = b^2 - 4ac$ .

$D > 0$ : two different real solutions. The parabola $y = ax^2 + bx + c$ crosses the x-axis twice.
$D = 0$ : exactly one real solution, $x = -\dfrac{b}{2a}$ . The parabola touches the x-axis at its vertex.
$D < 0$ : no real solutions. The parabola misses the x-axis entirely.

If $D$ is a perfect square (and $a$ , $b$ , $c$ are integers), the solutions are rational, which means the quadratic factors over the integers.

Three parabolas with the same axis: y = x² - 2x - 3 crosses the x-axis twice, y = x² - 2x + 1 touches it once, and y = x² - 2x + 4 never reaches it.Open in grapher →

The case $D < 0$ is the interesting one. There are no real solutions, but that is not the end of the story. Over the next three lessons you'll build a larger number system in which every quadratic equation has solutions.

Tip

The discriminant is a quick check on factoring. If you've spent a minute hunting for factors of $x^2 + 5x + 3$ with no luck, compute $D = 25 - 12 = 13$ . Since $13$ is not a perfect square, the quadratic doesn't factor over the integers. Switch to the formula.

Equations in quadratic form

Some higher-degree equations are quadratics wearing a disguise. In $x^4 - 13x^2 + 36 = 0$ , the exponent $4$ is exactly twice the exponent $2$ . If you let $u = x^2$ , then $x^4 = u^2$ , and the equation becomes an ordinary quadratic in $u$ .

Worked example: Substitution

Solve $x^4 - 13x^2 + 36 = 0$ .

Let $u = x^2$ :

u^2 - 13u + 36 = 0 \quad\Longrightarrow\quad (u - 4)(u - 9) = 0 \quad\Longrightarrow\quad u = 4 \ \text{or}\ u = 9.

Now go back to $x$ . If $x^2 = 4$ , then $x = \pm 2$ . If $x^2 = 9$ , then $x = \pm 3$ . The equation has four solutions: $-3, -2, 2, 3$ .

Always finish by converting back to the original variable. Stopping at $u = 4$ and $u = 9$ answers a different question. The same trick works on equations such as $(x + 1)^2 - 5(x + 1) + 6 = 0$ with $u = x + 1$ .

Practice

Practice 1

Solve $2x(x - 3) = x + 4$ .

Separate answers with commas, e.g. 2, -5

Practice 2

Solve $4(x + 1)^2 = 36$ .

Separate answers with commas, e.g. 2, -5

Practice 3

Solve $6x^2 + x - 2 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 4

Solve $x^2 - 4x - 1 = 0$ . Give exact answers (type a square root as sqrt(5)).

Separate answers with commas, e.g. 2, -5

Practice 5

Find the discriminant of $2x^2 - 3x + 5 = 0$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Which equation has exactly one real solution?

Practice 7

Find all real solutions of $x^4 - 5x^2 + 4 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 8

For what positive value of $k$ does $x^2 + kx + 16 = 0$ have exactly one real solution?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.