What do airplane wings, noise-cancelling headphones, and phone signal reception all have in common? They all rely on equations that sometimes have no real solutions—but still have answers. Those answers live in the world of complex numbers, and today you will learn how to find them when solving 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 real numbers and \(a \neq 0\).
The solutions (or roots) are the values of \(x\) that make the equation true. One powerful way to understand how many solutions a quadratic has, and what type they are, is to use the discriminant.
The discriminant is the expression
\[ D = b^2 - 4ac \]
The discriminant connects to the graph of the quadratic, as shown in [Figure 1], because it tells us how the parabola meets the \(x\)-axis.
The sign of \(D\) tells us what kind of solutions we get:
Graphically, if \(D < 0\), the parabola does not cross the \(x\)-axis at all. The solutions are still there, but they are complex numbers, not real numbers.
Later, when we talk about real-world applications, we will come back to the idea that “no real solution” on the graph does not mean “no useful information.” The equations still describe real phenomena; we just need complex numbers to handle them.
You probably learned that you cannot take the square root of a negative number in the real-number system. For example, there is no real number \(x\) such that
\(x^2 = -9\)
because squaring any real number makes it nonnegative.
To handle square roots of negative numbers, mathematicians introduced a new number, called the imaginary unit and denoted by \(i\):
\(i^2 = -1\)
From this definition, we get
In general, for any positive real number \(k\), we have
\[ \sqrt{-k} = i\sqrt{k} \]
A complex number has the form
\(a + bi\)
where \(a\) and \(b\) are real numbers. The part \(a\) is called the real part and \(bi\) is the imaginary part. Real numbers are just complex numbers with \(b = 0\) (for example, \(5 = 5 + 0i\)).
When we solve quadratic equations with real coefficients and get a negative discriminant, we naturally end up with complex numbers of the form \(a + bi\).
The most direct way to solve any quadratic equation is to use the quadratic formula. For \(ax^2 + bx + c = 0\), the solutions are
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The expression under the square root, \(b^2 - 4ac\), is the discriminant \(D\). The way we handle the square root depends on whether \(D\) is positive, zero, or negative. The flow of decisions in this process is summarized in [Figure 2].
If \(D < 0\), then we write \(D = -k\) where \(k > 0\). Then
\[ \sqrt{D} = \sqrt{-k} = i\sqrt{k} \]
So the solutions become
\[ x = \frac{-b \pm i\sqrt{k}}{2a} \]
This is exactly the form \(a + bi\) (or \(a - bi\)), which is a complex number.
As you practice, you will see that the main difference between real and complex solutions using the quadratic formula is how you simplify the square root when \(D < 0\).
Now let us solve some quadratic equations where the discriminant is negative, leading to complex solutions.
Example 1: A simple quadratic with complex roots
Solve \(x^2 + 4x + 5 = 0\).
Step 1: Identify \(a\), \(b\), and \(c\).
Here \(a = 1\), \(b = 4\), and \(c = 5\).
Step 2: Compute the discriminant.
\[ D = b^2 - 4ac = 4^2 - 4(1)(5) = 16 - 20 = -4 \]
Since \(D = -4 < 0\), the solutions will be complex, and the graph of the parabola does not cross the \(x\)-axis, like the third parabola in [Figure 1].
Step 3: Use the quadratic formula.
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-4}}{2(1)} \]
Step 4: Simplify the square root of the negative number.
\[ \sqrt{-4} = \sqrt{4 \cdot (-1)} = 2\sqrt{-1} = 2i \]
So
\[ x = \frac{-4 \pm 2i}{2} \]
Step 5: Simplify the fraction.
\[ x = -2 \pm i \]
So the two complex solutions are \(x = -2 + i\) and \(x = -2 - i\).
Example 2: Larger coefficients, factoring under the root
Solve \(3x^2 - 2x + 5 = 0\).
Step 1: Identify \(a\), \(b\), and \(c\).
Here \(a = 3\), \(b = -2\), and \(c = 5\).
Step 2: Compute the discriminant.
\[ D = b^2 - 4ac = (-2)^2 - 4(3)(5) = 4 - 60 = -56 \]
Again, \(D < 0\), so we expect complex solutions.
Step 3: Apply the quadratic formula.
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-2) \pm \sqrt{-56}}{2(3)} = \frac{2 \pm \sqrt{-56}}{6} \]
Step 4: Simplify \(\sqrt{-56}\).
First, factor \(56\) into a product that includes a perfect square:
\[ 56 = 7 \cdot 8 \]\n\[ 56 = 7 \cdot 4 \cdot 2 = 28 \cdot 2 \]
A convenient factorization is \(56 = 4 \cdot 14\). Then
\[ \sqrt{-56} = \sqrt{-1 \cdot 56} = \sqrt{-1 \cdot 4 \cdot 14} = \sqrt{4} \cdot \sqrt{-14} = 2\sqrt{-14} \]
Now rewrite \(\sqrt{-14}\):
\[ \sqrt{-14} = i\sqrt{14} \]
So
\[ \sqrt{-56} = 2i\sqrt{14} \]
Step 5: Substitute back and simplify.
\[ x = \frac{2 \pm 2i\sqrt{14}}{6} \]
Factor \(2\) out of the numerator:
\[ x = \frac{2(1 \pm i\sqrt{14})}{6} = \frac{1 \pm i\sqrt{14}}{3} \]
The two complex solutions are \(x = \dfrac{1}{3} + \dfrac{\sqrt{14}}{3}i\) and \(x = \dfrac{1}{3} - \dfrac{\sqrt{14}}{3}i\).
Example 3: Simplifying first, then solving
Solve \(2x^2 + 8x + 18 = 0\).
Step 1: Simplify the equation if possible.
All coefficients are divisible by \(2\), so divide the whole equation by \(2\):
\[ x^2 + 4x + 9 = 0 \]
Now \(a = 1\), \(b = 4\), \(c = 9\).
Step 2: Compute the discriminant.
\[ D = b^2 - 4ac = 4^2 - 4(1)(9) = 16 - 36 = -20 \]
Step 3: Apply the quadratic formula.
\[ x = \frac{-4 \pm \sqrt{-20}}{2(1)} = \frac{-4 \pm \sqrt{-20}}{2} \]
Step 4: Simplify \(\sqrt{-20}\).
Use \(20 = 4 \cdot 5\):
\[ \sqrt{-20} = \sqrt{-1 \cdot 4 \cdot 5} = 2\sqrt{-5} = 2i\sqrt{5} \]
Step 5: Substitute and simplify.
\[ x = \frac{-4 \pm 2i\sqrt{5}}{2} = -2 \pm i\sqrt{5} \]
So the solutions are \(x = -2 + i\sqrt{5}\) and \(x = -2 - i\sqrt{5}\).
The quadratic formula actually comes from a method called completing the square. This method also works when the solutions are complex. Let us walk through one example to see how complex numbers appear naturally.
Solve \(x^2 + 6x + 13 = 0\) by completing the square.
Step 1: Move the constant term to the other side.
\[ x^2 + 6x = -13 \]
Step 2: Take half of the coefficient of \(x\) and square it.
The coefficient of \(x\) is \(6\). Half of that is \(3\), and its square is \(9\).
Step 3: Add this square to both sides.
\[ x^2 + 6x + 9 = -13 + 9 \]\[ x^2 + 6x + 9 = -4 \]
Step 4: Rewrite the left side as a perfect square.
\[ (x + 3)^2 = -4 \]
Step 5: Take the square root of both sides.
\[ x + 3 = \pm \sqrt{-4} \]
We already know \(\sqrt{-4} = 2i\), so
\[ x + 3 = \pm 2i \]
Step 6: Solve for \(x\).
\[ x = -3 \pm 2i \]
This matches exactly what we would get using the quadratic formula. Completing the square gives the same complex solutions, just through a different path.
In every example above, the complex solutions came in pairs that looked like \(a + bi\) and \(a - bi\). These are called complex conjugates.
If a quadratic equation has real coefficients (which ours all do) and it has a non-real complex solution \(a + bi\), then \(a - bi\) is automatically a solution as well. This always happens because of how addition, subtraction, and multiplication interact with conjugates.
For example, in Example 1 we found solutions \(x = -2 + i\) and \(x = -2 - i\). We can factor the quadratic using these roots:
\[ x^2 + 4x + 5 = (x - (-2 + i))(x - (-2 - i)) \]\[ x^2 + 4x + 5 = (x + 2 - i)(x + 2 + i) \]
If we expanded the product on the right, we would get back the original quadratic with real coefficients. The imaginary parts cancel each other out in the middle term when you expand, which is why the final coefficients remain real.
This pattern is important in higher math and engineering: when you solve real-coefficient equations and get complex numbers, they show up as conjugate pairs, just like we saw with the quadratic formula branching to \(a \pm bi\) in [Figure 2].
You might wonder: if these solutions are not real numbers, how can they describe anything real?
Here are some key areas where quadratic equations with complex solutions appear:
In all these cases, “no real solution” does not mean the system is meaningless. Instead, the complex solutions give deeper information about how something moves, oscillates, or changes over time.
