Sums and Products of Roots

Reading the sum and product off the coefficients

Start with a monic quadratic, one whose leading coefficient is $1$, written $x^2 + bx + c$. Suppose it has roots $r$ and $s$. You already know from the last lessons that a quadratic with those roots can be written in factored form as $(x - r)(x - s)$. That form works because setting $x$ equal to either root makes one factor zero, and so makes the whole product zero. Expanding that product and comparing it with $x^2 + bx + c$ is the whole idea, and it pins down the sum and the product exactly.

The picture is just the two forms of the quadratic lined up one above the other. Whatever sits beneath the middle term of $x^2 + bx + c$ has to equal $b$, and whatever sits beneath the constant has to equal $c$.

When the leading coefficient is not 1

Everything so far assumed the quadratic started with a bare $x^2$. A general quadratic $ax^2 + bx + c = 0$ carries a leading coefficient $a$ that need not be $1$, with $a \ne 0$, or it would not be a quadratic at all. Two short routes handle it, and they agree.

The quick route is to divide. Dividing the equation $ax^2 + bx + c = 0$ through by $a$ does not move its roots, since scaling an equation by a nonzero number leaves the values that make it zero untouched. It produces the monic quadratic

whose middle coefficient is $\frac{b}{a}$ and whose constant is $\frac{c}{a}$. Reading the monic formulas with those in place gives $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$ at once. The second route factors instead, and it is worth seeing because it explains where the $a$ goes.

These two formulas hold for every quadratic, even one whose graph never crosses the axis and so has no real roots. An example is $x^2 + x + 1$, whose roots have sum $-1$ and product $1$. The two roots in that case are a kind of number introduced in a later chapter. So every example and exercise in this lesson stays with quadratics whose roots are real.

Building a quadratic from its roots

The formulas run just as well in reverse. If you are told the roots and asked for a quadratic, you do not need to expand anything from scratch. The sum and the product of those roots are all a monic quadratic needs. Because $(x - r)(x - s) = x^2 - (r + s)x + rs$, a monic quadratic with roots $r$ and $s$ is

The coefficient of $x$ is the negative of the sum, and the constant is the product with its own sign kept. That single template builds the quadratic from any pair of roots.

Recovering a missing root and checking a pair

The sum and product also work as shortcuts once you already know one root, or when someone hands you a proposed pair of roots to check. Both jobs use the same two totals.

To recover a missing root, use whichever total is easier. Suppose $x^2 - 10x + 21 = 0$ and you have spotted that $x = 3$ is a root. The two roots must add to $-b = 10$, so the other root is whatever is left after removing $3$ from that total:

The product gives the same answer as a cross-check, since the roots multiply to $c = 21$ and $21 \div 3 = 7$. Either total pins the second root without factoring the trinomial from scratch.

Checking a proposed pair works the same way, and it is quick because a correct pair has to pass both tests. To test whether $2$ and $5$ are the roots of $x^2 - 7x + 10$, compare their sum and product against the coefficients. The roots should sum to $-b = 7$ and multiply to $c = 10$, and $2 + 5 = 7$ with $2 \cdot 5 = 10$, so the pair passes both tests and is confirmed. A proposed pair like $1$ and $6$ would sum to $7$ but multiply to only $6$, failing the product test. That pair can therefore be ruled out at a glance, without any substitution.

The connection to factoring

This lesson quietly closes a loop opened two lessons ago. To factor a monic trinomial $x^2 + bx + c$ you searched for two numbers $p$ and $q$ with sum $b$ and product $c$, and wrote $(x + p)(x + q)$. Those two numbers are not the roots. The roots come from the sign flip in the zero-product step. Setting $(x + p)(x + q) = 0$ gives $x = -p$ or $x = -q$, so the roots are $-p$ and $-q$, the negatives of the factoring numbers.

Line the two searches up. The factoring numbers satisfy $p + q = b$ and $pq = c$. The roots $r = -p$ and $s = -q$ then satisfy

which are exactly the sum and product formulas from this lesson. The product is identical, $c$ either way, while the sum flips sign, $b$ for the factoring numbers and $-b$ for the roots. So the two numbers you hunt for when factoring a monic trinomial are the negatives of the roots. Both jobs share the product $c$ but carry opposite sums, and the reason for that is the one sign flip that turns a factor into a root.

Common mistakes