Math Core

Earlier in this unit, a negative discriminant meant "no real solutions" and the work stopped there. Now that you can take square roots of negative numbers, you can keep going. Every quadratic equation has solutions in the complex numbers, and the methods you already know find them without any changes.

Square roots of negatives, put to work

An equation such as $x^2 = -9$ now has two solutions, because both $3i$ and $-3i$ square to $-9$ :

x^2 = -9 \quad\Longrightarrow\quad x = \pm\sqrt{-9} = \pm 3i.

The square root method works exactly as before, with $\sqrt{-a} = i\sqrt{a}$ in the last step.

Worked example: Square roots

Solve $(x - 2)^2 + 9 = 0$ .

Isolate the square, then take square roots:

(x - 2)^2 = -9 \quad\Longrightarrow\quad x - 2 = \pm 3i \quad\Longrightarrow\quad x = 2 \pm 3i.

Check $x = 2 + 3i$ : then $x - 2 = 3i$ , and $(3i)^2 + 9 = -9 + 9 = 0$ .

The quadratic formula with a negative discriminant

When $b^2 - 4ac < 0$ , write the square root of the discriminant using $i$ , then simplify into standard form $a + bi$ .

Worked example: Using the formula

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

Here $a = 1$ , $b = -4$ , $c = 13$ , so the discriminant is $16 - 52 = -36$ .

x = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i.

Notice that this equation and the previous one have the same solutions. That's no coincidence: expanding $(x - 2)^2 + 9$ gives $x^2 - 4x + 13$ . Completing the square on $x^2 - 4x + 13$ leads straight back to $(x - 2)^2 = -9$ .

Worked example: A leading coefficient

Solve $2x^2 + 2x + 5 = 0$ .

The discriminant is $2^2 - 4(2)(5) = 4 - 40 = -36$ .

x = \frac{-2 \pm \sqrt{-36}}{2(2)} = \frac{-2 \pm 6i}{4} = -\frac{1}{2} \pm \frac{3}{2}i.

Divide both the real part and the imaginary part by the denominator $4$ .

Common mistake

When simplifying $\dfrac{-2 \pm 6i}{4}$ , it is tempting to cancel only part of the numerator and write $-2 \pm \dfrac{3}{2}i$ . The whole numerator is divided by $4$ , so each term must be divided: $-\dfrac{2}{4} \pm \dfrac{6}{4}i = -\dfrac{1}{2} \pm \dfrac{3}{2}i$ . Writing the answer as a real part plus an imaginary part forces you to handle both.

Complex solutions come in conjugate pairs

In every example so far, the two solutions were $p + qi$ and $p - qi$ , a pair of complex conjugates. The quadratic formula shows why. When $D < 0$ ,

x = -\frac{b}{2a} \pm \frac{\sqrt{-D}}{2a}\, i,

so both solutions share the real part $-\dfrac{b}{2a}$ and have opposite imaginary parts.

The discriminant, completed

For $ax^2 + bx + c = 0$ with real coefficients and $D = b^2 - 4ac$ :

$D > 0$ : two different real solutions.
$D = 0$ : one real solution (a double root).
$D < 0$ : two nonreal complex solutions, which are conjugates $p \pm qi$ with $p = -\dfrac{b}{2a}$ .

The real part of the pair is $-\dfrac{b}{2a}$ , the x-coordinate of the vertex. So even when the parabola misses the x-axis, its complex zeros are "centered" on the axis of symmetry.

y = x² - 4x + 13 never meets the x-axis. Its zeros are the complex numbers 2 ± 3i, whose real part 2 matches the axis of symmetry.Open in grapher →

Working backward from complex solutions

If a quadratic with real coefficients has one nonreal solution, the conjugate must be the other. Knowing both solutions, you can rebuild the equation. The fastest route uses the sum and product of the solutions.

For $x^2 + bx + c = 0$ with solutions $r$ and $s$ , the factored form $(x - r)(x - s) = x^2 - (r + s)x + rs$ shows that

b = -(r + s), \qquad c = rs.

For conjugates, $r + s = 2p$ and $rs = p^2 + q^2$ , both real, just as the previous lesson's tip promised.

Worked example: Building a quadratic

Write a quadratic equation $x^2 + bx + c = 0$ with a solution $3 + 2i$ .

The other solution must be $3 - 2i$ . Their sum is $6$ and their product is $3^2 + 2^2 = 13$ . So the equation is

x^2 - 6x + 13 = 0.

Check with the formula: $x = \dfrac{6 \pm \sqrt{36 - 52}}{2} = \dfrac{6 \pm 4i}{2} = 3 \pm 2i$ .

Tip

Use the sum and product to check any pair of complex solutions. For $2x^2 + 2x + 5 = 0$ , the solutions $-\dfrac{1}{2} \pm \dfrac{3}{2}i$ should add to $-\dfrac{b}{a} = -1$ and multiply to $\dfrac{c}{a} = \dfrac{5}{2}$ . Indeed, the sum is $-1$ , and the product is $\dfrac{1}{4} + \dfrac{9}{4} = \dfrac{10}{4} = \dfrac{5}{2}$ .

Practice

For problems whose solutions are $a \pm bi$ , enter $(a, b)$ using the positive value of $b$ .

Practice 1

Solve $x^2 + 36 = 0$ .

Practice 2

The solutions of $(x + 1)^2 = -16$ are $a \pm bi$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 3

The solutions of $x^2 - 6x + 10 = 0$ are $a \pm bi$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 4

What kind of solutions does $x^2 + 2x + 7 = 0$ have?

Practice 5

The solutions of $2x^2 - 4x + 5 = 0$ are $a \pm bi$ . Enter $(a, b)$ . (Type a square root as sqrt(6).)

Enter a point like (2, -3)

Practice 6

The solutions of $5x^2 - 2x + 1 = 0$ are $a \pm bi$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 7

The equation $x^2 + bx + c = 0$ has real coefficients and one solution $4 - i$ . What is $c$ ?

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

Practice 8

For which values of $k$ does $x^2 + 6x + k = 0$ have two nonreal solutions? Write an inequality in $k$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5