Sum and Product of Roots

Running the chapter’s map backwards

You have three forms for the same quadratic, and each answers a question at sight. Standard form $ax^2 + bx + c$ hands you the $y$-intercept $c$ and, through the sign of $a$, the end behavior. Vertex form $a(x - h)^2 + k$ hands you the vertex, the axis, and the range. Factored form $a(x - r_1)(x - r_2)$ hands you the zeros $r_1$ and $r_2$. Every earlier lesson moved from standard form toward factored form. Now go backward. Multiply the factored form out and see what the coefficients have to be.

The point worth repeating is what did not happen. Nowhere did you solve for $r_1$ or $r_2$. You matched coefficients, and the sum and product fell out of the letters. This is the reverse of every other lesson in the chapter, and it is why this lesson closes it.

A second derivation, straight from the formula

The matching argument is complete on its own, but there is a second route that is a pleasure to see. The quadratic formula names the two roots outright:

Add them and multiply them directly, and the same two totals appear.

Constructing a quadratic from its roots

The formulas run just as well in reverse. Given the two roots, you can write a quadratic without expanding anything, because a monic quadratic (one whose leading coefficient is $1$) with roots $r_1$ and $r_2$ is exactly

The coefficient of $x$ is the negative of the sum, and the constant is the product with its own sign. There is a caveat that matters. Scaling by any nonzero number leaves the roots untouched, since $a(x - r_1)(x - r_2)$ is zero at exactly $r_1$ and $r_2$ for every $a \ne 0$. So the roots do not determine the quadratic, only its shape. The whole family $a(x - r_1)(x - r_2)$ shares those roots, and you get a single answer only by fixing the leading coefficient. Choosing $a = 1$ gives the monic one above; that is the standard choice, but it is a choice.

Finding the other root

When one root is handed to you, the sum or the product recovers the other one in a single step, with no factoring. Both totals work, and using both is a free check.

Symmetric functions of the roots

Here is where the method earns its keep. Many quantities built from both roots can be rewritten using only the sum and the product, so you can evaluate them without ever finding a single root. The key identity comes from expanding a square:

The sum of the squares is the square of the sum minus twice the product. Two more follow the same way. The sum of the reciprocals needs $c \ne 0$ so that neither root is zero:

And the squared difference of the roots, which will pay off in a moment:

Look again at that last line. For a monic quadratic, $(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2 = 20$, which is exactly the discriminant $\Delta = 20$. That is no accident. In general,

This ties the discriminant from the previous lesson back to the roots themselves, and it explains why $\Delta \ge 0$ is precisely the condition for real roots. For real $r_1$ and $r_2$, the quantity $(r_1 - r_2)^2$ is a real square, so it cannot be negative, which forces $\Delta = a^2 (r_1 - r_2)^2 \ge 0$. The two roots are real exactly when the number the discriminant measures, their squared gap, is not negative. Geometrically, the two roots sit symmetrically about the axis of symmetry $x = -\frac{b}{2a}$, and $\Delta$ measures how far apart they spread.

The converse, stated honestly

The forward story says: real roots give a sum $-\frac{b}{a}$ and a product $\frac{c}{a}$. The natural question is the converse. If someone hands you a target sum $s$ and product $p$, are there real numbers with that sum and product? Not always, and the exact condition is worth stating with its hypothesis attached, because the careless version is false.

The hypothesis $s^2 - 4p \ge 0$ is the whole point, and it is exactly what the following slogan drops. That slogan is tempting but wrong: “any two numbers with sum $s$ and product $p$ are the roots of $x^2 - sx + p$.” Real numbers with sum $5$ and product $4$ exist, since $5^2 - 4(4) = 9 \ge 0$; they are $1$ and $4$. But no real numbers have sum $5$ and product $7$, because $5^2 - 4(7) = -3 < 0$. Writing down $x^2 - 5x + 7$ does not conjure them into existence.

Checking a solution without solving again

The same two totals verify an answer you found by another method, which is a fast and independent check. Suppose you solved $2x^2 - 7x + 3 = 0$ and claim the roots are $3$ and $\tfrac{1}{2}$. The roots must sum to $-\frac{b}{a} = \frac{7}{2}$ and multiply to $\frac{c}{a} = \frac{3}{2}$. Your pair gives $3 + \tfrac{1}{2} = \tfrac{7}{2}$ and $3 \cdot \tfrac{1}{2} = \tfrac{3}{2}$, matching both totals, so the solution is confirmed. A pair that matches only the sum, or only the product, is wrong; a correct pair passes both tests at once.

Closing the chapter

Stand back and look at the whole chapter through its three forms. Standard form $ax^2 + bx + c$ shows the $y$-intercept and end behavior. Vertex form $a(x - h)^2 + k$ shows the vertex, axis, and range. Factored form $a(x - r_1)(x - r_2)$ shows the zeros. Choosing a form is choosing which question you can answer without work. This lesson took the last step, expanding factored form back into standard form, and here is what that step found. Standard form was quietly carrying the sum and product of its roots the whole time, folded into $-\frac{b}{a}$ and $\frac{c}{a}$, even though the roots are invisible in it.

One honest boundary remains. Every derivation here assumed the roots were real, so that the factorization $a(x - r_1)(x - r_2)$ used real numbers. These identities actually survive into the next chapter, where even a negative discriminant has roots and the same sum $-\frac{b}{a}$ and product $\frac{c}{a}$ still hold. That is the story of complex numbers, and it begins the chapter after this one.

Common mistakes