Sums and Products of Roots

Reading the sum and product off the coefficients

Look back at $x^2 - 7x + 10 = (x - 2)(x - 5)$: the roots $2$ and $5$ summed to $7$ and multiplied to $10$. Here is why that always happens. Start with a monic quadratic, one whose leading coefficient is $1$, written $x^2 + bx + c$, and 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.

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.

Both derivations above started from a real factorization $a(x - r)(x - s)$, so they take $r$ and $s$ to be two real roots, whether those two roots are different numbers or the same number repeated. In this lesson, and in every example and exercise it contains, $r$ and $s$ always mean a real root pair.

One more thing has to be true before any of this works: the equation has to already be in standard form, $ax^2 + bx + c = 0$, with every term moved to one side. An equation like $3x^2 + 12 = 11x$ is not there yet. Move every term to one side first, $3x^2 - 11x + 12 = 0$, and only then read off $a = 3$, $b = -11$, $c = 12$. Reading coefficients from the wrong side of the equation is a quiet way to flip a sign.

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, the sum always works. 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 check, since the roots multiply to $c = 21$ and $21 \div 3 = 7$. That division worked because the known root, $3$, was not zero. The product route breaks down exactly when the known root is $0$, since dividing by $0$ is never allowed; the sum route has no such gap, which is why it comes first.

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

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. Both jobs share the same product $c$; only the sum flips sign, from $b$ for the factoring numbers to $-b$ for the roots, and that one sign flip is the whole difference between a factor and a root.

Common mistakes