Math Core

Completing the square works on every quadratic, but it takes several steps each time. If you complete the square once on the general equation $ax^2 + bx + c = 0$ , you get a formula that solves any quadratic just by substituting $a$ , $b$ and $c$ . A small piece of that formula, the discriminant, even tells you how many solutions to expect before you find them.

Where the formula comes from

Follow the same steps as in the last lesson, but with letters instead of numbers. Start with $ax^2 + bx + c = 0$ , where $a \ne 0$ .

\begin{aligned} x^2 + \frac{b}{a}x &= -\frac{c}{a} && \text{divide by } a \text{, move the constant} \\ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} &= \frac{b^2}{4a^2} - \frac{c}{a} && \text{add } \left(\tfrac{b}{2a}\right)^2 \text{ to both sides} \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} && \text{common denominator on the right} \\ x + \frac{b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a} && \text{square roots} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{aligned}

You don't need to reproduce this every time, but it shows why the formula is true: it is just completing the square, done once and for all.

The quadratic formula

The solutions of $ax^2 + bx + c = 0$ , where $a \ne 0$ , are

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

The equation must be in standard form (equal to $0$ ) before you read off $a$ , $b$ and $c$ .

Worked example: Rational solutions

Solve $x^2 - 3x - 10 = 0$ with the quadratic formula.

Here $a = 1$ , $b = -3$ and $c = -10$ . First compute the part under the radical:

b^2 - 4ac = (-3)^2 - 4(1)(-10) = 9 + 40 = 49.

Then

x = \frac{-(-3) \pm \sqrt{49}}{2(1)} = \frac{3 \pm 7}{2}.

So $x = \dfrac{10}{2} = 5$ or $x = \dfrac{-4}{2} = -2$ . (Factoring as $(x - 5)(x + 2)$ gives the same answers.)

Worked example: Irrational solutions

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

Here $a = 2$ , $b = 4$ , $c = -3$ , and $b^2 - 4ac = 16 - 4(2)(-3) = 16 + 24 = 40$ .

x = \frac{-4 \pm \sqrt{40}}{4} = \frac{-4 \pm 2\sqrt{10}}{4} = \frac{-2 \pm \sqrt{10}}{2}.

The last step divides every term of the numerator and the denominator by $2$ . The solutions are about $0.58$ and $-2.58$ .

Common mistake

Watch the signs, especially when $b$ is negative.

$-b$ means "the opposite of $b$ ." If $b = -3$ , then $-b = 3$ .
$b^2$ is never negative: $(-3)^2 = 9$ , not $-9$ . Use parentheses when you substitute.
The fraction bar goes under the whole numerator, $-b \pm \sqrt{b^2 - 4ac}$ , not just the square root. When you simplify, divide every term by the same number.

The discriminant

The expression under the square root decides what kind of solutions you get.

Definition

Discriminant

The discriminant of $ax^2 + bx + c = 0$ is $D = b^2 - 4ac$ .

If $D > 0$ , there are two real solutions ( $-b + \sqrt{D}$ and $-b - \sqrt{D}$ are different).
If $D = 0$ , there is exactly one real solution, $x = -\dfrac{b}{2a}$ (adding or subtracting $0$ gives the same number).
If $D < 0$ , there are no real solutions (a negative number has no real square root).

Graphically, the discriminant counts the x-intercepts of $y = ax^2 + bx + c$ .

y = x² - 4 (D = 16) crosses the x-axis twice at -2 and 2. y = x² (D = 0) touches it once at its vertex. y = x² + 3 (D = -12) never reaches it.Open in grapher →

When $D$ is a perfect square ( $0, 1, 4, 9, 16, \ldots$ ), the solutions are rational and the quadratic can be factored over the integers. When $D$ is positive but not a perfect square, the solutions are irrational.

Worked example: Counting solutions

Use the discriminant to find the number of real solutions.

$3x^2 - 5x + 4 = 0$
$4x^2 + 12x + 9 = 0$
$x^2 + x - 1 = 0$

Solutions.

$D = (-5)^2 - 4(3)(4) = 25 - 48 = -23$ . Negative, so no real solutions.
$D = 12^2 - 4(4)(9) = 144 - 144 = 0$ . Exactly one solution. (Indeed, $4x^2 + 12x + 9 = (2x + 3)^2$ .)
$D = 1^2 - 4(1)(-1) = 1 + 4 = 5$ . Positive, so two solutions, and since $5$ isn't a perfect square they are irrational.

Choosing a method

You now have four ways to solve a quadratic equation. All of them give the same answers, so choose the quickest.

the equation looks like	good method
no $x$ term, or $(x - h)^2 = d$	square roots
a trinomial that factors easily	factoring
$a = 1$ and $b$ is even	completing the square
anything else	the quadratic formula

Tip

The quadratic formula always works, so it's the safe choice when you don't spot a factorization within a few seconds. Computing $D$ first is a good habit: if $D < 0$ you can stop, and if $D$ is a perfect square you know factoring would have worked too.

Worked example: A throw that never reaches 35 feet

A ball is thrown upward, and its height in feet after $t$ seconds is $h(t) = -16t^2 + 40t + 5$ .

When does it land? Set $h(t) = 0$ and multiply by $-1$ : $16t^2 - 40t - 5 = 0$ . Then $D = (-40)^2 - 4(16)(-5) = 1600 + 320 = 1920$ , and

t = \frac{40 \pm \sqrt{1920}}{32} \approx \frac{40 \pm 43.82}{32}.

The solutions are about $2.62$ and $-0.12$ . Time can't be negative, so the ball lands after about $2.62$ seconds.

Does it ever reach $35$ feet? Set $h(t) = 35$ : $-16t^2 + 40t - 30 = 0$ , or $16t^2 - 40t + 30 = 0$ . The discriminant is $1600 - 4(16)(30) = 1600 - 1920 = -320$ . It's negative, so there is no solution: the ball never gets that high. (Its vertex is at $t = 1.25$ , where $h = 30$ feet.)

Practice

Practice 1

Solve $2x^2 - 7x + 3 = 0$ using the quadratic formula.

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

Practice 2

Find the discriminant of $3x^2 - 4x - 2 = 0$ .

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

Practice 3

How many real solutions does $5x^2 - 2x + 1 = 0$ have?

Practice 4

Solve $x^2 - 6x + 4 = 0$ .

Practice 5

Solve $3x^2 + 2x - 2 = 0$ . What is the larger solution? Give an exact answer, such as (2 + sqrt(3))/5.

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

Practice 6

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

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

Practice 7

Solve $x^2 + 4x - 7 = 0$ . Round each solution to the nearest hundredth.

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

Practice 8

A ball's height in feet after $t$ seconds is $h(t) = -16t^2 + 24t + 6$ . After how many seconds does it hit the ground? Round to the nearest hundredth.

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