In Section 5.7 we solved some quadratic equations in cases where the numbers were chosen
nicely, so that it was easy enough to factor the polynomials by trial and error. In this chapter
we’ll learn how to solve quadratic equations with a more general method (the quadratic
formula), which works even if the numbers are messy.
The quadratic formula is based on a technique called completing the square. We’ll
introduce this in Section 8.1 and use it to solve some equations, and then in Section 8.2
we’ll show how it helps in making plots. In Section 8.3 we’ll use the technique to derive the
quadratic formula, which we’ll then apply to many examples and exercises. A critical part
of the quadratic formula is the discriminant, which we’ll discuss in Section 8.4, where we’ll
learn how it can be used to find the maximum or minimum of a function. The discriminant
leads us into the realm of imaginary numbers and complex numbers, which are the subjects
of Section 8.5.
Chapter 8
Quadratic equations
From Algebra: For the Enthusiastic Beginner (Draft version, July 2024)
© David Morin, www.davidmorin.physics.fas.harvard.edu
In Section 5.7 we solved some quadratic equations in cases where the numbers were chosen
nicely, so that it was easy enough to factor the polynomials by trial and error. In this chapter
we’ll learn how to solve quadratic equations with a more general method (the quadratic
formula), which works even if the numbers are messy.
The quadratic formula is based on a technique called completing the square. We’ll
introduce this in Section 8.1 and use it to solve some equations, and then in Section 8.2
we’ll show how it helps in making plots. In Section 8.3 we’ll use the technique to derive the
quadratic for mula , which we’ll then apply to many examples and exercises. A critical part
of the quadratic formula is the discriminant, which we’ll discuss in Section 8.4, where we’ll
learn how it can be used to find the maximum or minimum of a function. The discriminant
leads us into the realm of imaginary numbers and complex numbers, which are the subjects
of Section 8.5.
8.1 Completing the square
The quadratic equations we solved in Section 5.7 were ones where we could guess what the
factoring was. As you recall from Section 4.2.2, factoring into binomials involves finding
two numbers whose product and sum (or perhaps a sum of the form 𝑥 + 2𝑦, etc.) take on
particular values.
But what about equations where the numbers aren’t chosen nicely, and guessing won’t
get the job done? For these cases, we need to use the quadratic formula. And to understand
where this formula comes from, we first need to learn the technique of completing the square.
In the end, the quadratic formula is simply a general expression (that is, one involving letters)
for the result you get when you complete the square. So let’s see what this technique entails.
We’ll introduce it by looking at four quadratic equations, each one slightly more involved
than the previous. The coefficients in these equations will involve numbers. We’ll apply the
technique to letters in Section 8.3.
1. Consider the equation,
𝑥
2
= 4. (8.1)
476
8.1. Completing the square 477
How can we solve this for 𝑥? Well, we need to undo the squaring operation, so we should
take the square root of both sides of the equation. This gives
𝑥 = ±
√
4 =⇒ 𝑥 = ±2. (8.2)
Note the “±” sign. Both −2 and 2 are solutions, so it would be incorrect to say only that
𝑥 = 2.
Always remember to include the “±” when taking a square root! Forgetting it
is a common error.
2. Here’s another equation:
(𝑥 − 3)
2
= 4. (8.3)
How can we solve this one? What you don’t want to do is expand the square on the lefthand
side to obtain 𝑥
2
− 6𝑥 + 9 = 4. Actually, for this problem this wouldn’t be a terrible idea,
because subtracting 4 from both sides would leave you with the polynomial 𝑥
2
− 6𝑥 + 5,
which can be factored quickly into (𝑥 − 5)(𝑥 − 1), yielding the solutions of 5 and 1. But
in general, expanding the square isn’t what you want to do, because unless the numbers
are chosen nicely, you won’t be able to factor the polynomial by trial and error.
What you do want to do with Eq. (8.3) is again take the square root of both sides, as we
did in Eq. (8.2). This gives
𝑥 − 3 = ±
√
4 =⇒ 𝑥 − 3 = ±2 =⇒ 𝑥 = 3 ± 2 =⇒ 𝑥 = 5 or 1. (8.4)
The only difference between this equation and Eq. (8.2) is that after taking the square root,
we’re now left with 𝑥 −3 on the lefthand side, instead of just 𝑥. But that’s not much of an
issue. We simply need to add 3 to both sides to isolate (solve for) 𝑥. And then we need to
evaluate the two answers stemming from the “±” sign.
Again, don’t forget the “±” sign! A quadratic equation has two solutions in general, and
you will obtain only one solution if you forget the “±.” However, in the special case where
the righthand side of Eq. (8.3) is zero, there is in fact only one solution, because you end
up with 𝑥 = 3 ± 0.
3. Here’s a third equation:
𝑥
2
− 6𝑥 + 9 = 4. (8.5)
How do we solve this? As mentioned above, you could subtract 4 from both sides and
then factor the polynomial. But that method works here only because the numbers were
chosen nicely. The better strategy is to note that the lefthand side is a perfect square. That
is, it is the square of the binomial 𝑥 − 3. So we can rewrite the equation as (𝑥 − 3)
2
= 4,
which we already solved above.
In short, you don’t want to expand a square if you’re given one, but you do want to write
an expression as a square if you can. That is, you want to “un-expand” it. You can then
simply take the square root and follow the steps in Eq. (8.4).
478 Chapter 8. Quadratic equations
4. Here’s our four th and final equation, one that illustrates the “completing the square”
technique we’ve been building up to:
𝑥
2
− 6𝑥 + 5 = 0. (8.6)
How can we solve this? Again, we could quickly factor it into (𝑥 −5)(𝑥 −1) = 0, but let’s
pretend that the numbers weren’t chosen so nicely.
We solved all of the above cases by using the fact that the lefthand side was a square.
But the lefthand side of Eq. (8.6) isn’t a square. However, we can make it be a square by
turning the 5 into a 9. Of course, we can’t just arbitrarily erase the 5 and write down a 9.
But what we can do is add 4 to both sides. This turns the 5 into a 9 (and also puts a 4 on
the righthand side). So we have
(𝑥
2
− 6𝑥 + 5) + 4 = 4 =⇒ 𝑥
2
− 6𝑥 + 9 = 4 =⇒ (𝑥 − 3)
2
= 4, (8.7)
which we already solved above. Adding 4 to both sides here is just another example
of performing the same (useful) operation on both sides of an equation, as we did all
throughout Chapter 5.
The process of adding 4 to both sides in Eq. (8.7) is called “completing the square.” The
𝑥
2
− 6𝑥 terms on the lefthand side are part of a square, and the missing piece is a +9. So
by creating a +9 (by adding 4 to both sides) to yield 𝑥
2
− 6𝑥 + 9, we completed the square.
This process could also reasonably be called “forming the square,” or “creating the square,”
or something like that.
“Completing the square” means modifying the constant term (by performing
the same operation on both sides of the equation), so that when it is added to
the 𝑥
2
and 𝑥 terms, the result is a square.
As we mentioned a number of times in Chapter 5, there is an infinite number of common
operations we can perform on both sides of any given equation. But nearly all of them are
useless. For example, we could add 17 to both sides of Eq. (8.6) to obtain 𝑥
2
−6𝑥 +22 = 17.
This is an entirely true equation, but it doesn’t help us at all. Adding 4 to both sides is the
special operation that gives us something useful on the lefthand side, namely a perfect square.
The general form of the square of a binomial (assuming the coefficient of 𝑥
2
is 1) is
(𝑥 + 𝑚)
2
= 𝑥
2
+ (2𝑚)𝑥 + 𝑚
2
; the above equations had 𝑚 = −3. Therefore,
When completing the square (assuming the coefficient of 𝑥
2
is 1), the constant
term (the 𝑚
2
) is always obtained by taking half of the coefficient of 𝑥 (half of
2𝑚 is 𝑚) and squaring it.
