Sum and Product of Roots

Intuition

This chapter has been one long push in a single direction. You started from standard form $ax^2 + bx + c$ and worked toward the factored form $a(x - r_1)(x - r_2)$ , because factored form is where the roots become visible. Every method you learned, factoring, completing the square, the quadratic formula, was a way of getting there. This lesson runs the map the other way. Expand $a(x - r_1)(x - r_2)$ and line the result up against $ax^2 + bx + c$ , and the coefficients $b$ and $c$ turn out to be built directly from the two roots. You can read the sum of the roots and their product straight off the coefficients. And here is the punchline of the whole chapter: you read both of those totals without ever finding the roots. You will know what two numbers add to and multiply to before you know either number.

Learning goals

Read $r_1 + r_2 = -\tfrac{b}{a}$ and $r_1 r_2 = \tfrac{c}{a}$
Derive them by expanding the factored form and matching
Build a quadratic from roots, up to a nonzero scalar
Reduce a symmetric function of the roots to sum and product
Link $(r_1 - r_2)^2$ to $\Delta$ as the real-root test
State the converse with the condition $s^2 - 4p \ge 0$

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.

Try it on a quadratic whose roots you can already see, $(x - 2)(x - 3)$ . Multiplying it out gives

(x - 2)(x - 3) = x^2 - 5x + 6.

The roots $2$ and $3$ add to $5$ , and the middle coefficient is $-5$ , its negative. The roots multiply to $6$ , and that is exactly the constant term. That is not a coincidence of this one example. The same pattern holds for every quadratic with real roots, and the proof below is just this multiplication done with letters instead of numbers.

Why $r_1 + r_2 = -\dfrac{b}{a}$ and $r_1 r_2 = \dfrac{c}{a}$ #

Suppose the quadratic $ax^2 + bx + c$ with $a \ne 0$ has real roots $r_1$ and $r_2$ . Because the roots are real, the quadratic factors over the real numbers, and it is $a$ times the product of the two factors those roots produce:

ax^2 + bx + c = a(x - r_1)(x - r_2).

The leading $a$ is there so the squared terms match, since $(x - r_1)(x - r_2)$ on its own starts with a bare $x^2$ . Expand the right side:

a(x - r_1)(x - r_2) = a\left(x^2 - (r_1 + r_2)x + r_1 r_2\right) = ax^2 - a(r_1 + r_2)x + a\,r_1 r_2.

Two polynomials are equal for every $x$ only when their coefficients agree one at a time. The $x^2$ terms already match. Line up the coefficient of $x$ and the constant term:

b = -a(r_1 + r_2) \qquad \text{and} \qquad c = a\,r_1 r_2.

Divide each equation by $a$ , which is legal because $a \ne 0$ :

r_1 + r_2 = -\frac{b}{a} \qquad \text{and} \qquad r_1 r_2 = \frac{c}{a}.

Notice the hypothesis the argument used: the factorization $a(x - r_1)(x - r_2)$ exists because the roots are real, that is, because the discriminant satisfies $\Delta = b^2 - 4ac \ge 0$ . When $a = 1$ these collapse to $r_1 + r_2 = -b$ and $r_1 r_2 = c$ . The minus sign lives on the sum alone, and it comes from the two $-r_1$ and $-r_2$ inside the factors.

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.

Expanding the factored form and stacking it above standard form. Matching the middle coefficients gives the sum of the roots, and matching the constants gives the product, with no solving.

The matching argument above is complete on its own. There is also a second route to the same two totals, starting straight from the quadratic formula instead of from matching coefficients.

Worked example 1 Read the sum and product of the roots of $2x^2 - 11x + 5$ without solving

Read the coefficients: $a = 2$ , $b = -11$ , $c = 5$ . This lesson’s formulas need real roots, so first check the discriminant:

\Delta = b^2 - 4ac = (-11)^2 - 4(2)(5) = 121 - 40 = 81 \ge 0.

Since $\Delta \ge 0$ , apply the formulas directly:

r_1 + r_2 = -\frac{b}{a} = -\frac{-11}{2} = \frac{11}{2}, \qquad r_1 r_2 = \frac{c}{a} = \frac{5}{2}.

No solving was needed. As a check, this quadratic happens to factor as $(2x - 1)(x - 5)$ , with roots $\tfrac{1}{2}$ and $5$ , and indeed $\tfrac{1}{2} + 5 = \tfrac{11}{2}$ while $\tfrac{1}{2} \cdot 5 = \tfrac{5}{2}$ .

Check your understanding

$5x^2 + 12x - 9$ has real roots. Without solving, find their sum and product.

A sum

-\tfrac{12}{5}

, product

-\tfrac{9}{5}

B sum

\tfrac{12}{5}

, product

-\tfrac{9}{5}

C sum

-\tfrac{12}{5}

, product

\tfrac{9}{5}

D sum

\tfrac{9}{5}

, product

-\tfrac{12}{5}

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

x^2 - (r_1 + r_2)x + r_1 r_2.

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 fix the factors and the zeros, but not the leading coefficient $a$ , which is free to be anything nonzero. The whole family $a(x - r_1)(x - r_2)$ shares those roots, even though changing $a$ changes the parabola itself, its width and, if $a$ turns negative, which way it opens. You get a single quadratic only by fixing the leading coefficient. Choosing $a = 1$ gives the monic one above; that is the standard choice, but it is a choice.

Worked example 2 Build a quadratic with roots $\tfrac{3}{2}$ and $-4$ and integer coefficients

Start with the sum and product of the two roots:

r_1 + r_2 = \tfrac{3}{2} + (-4) = -\tfrac{5}{2}, \qquad r_1 r_2 = \tfrac{3}{2} \cdot (-4) = -6.

Drop these into the monic template $x^2 - (r_1 + r_2)x + r_1 r_2$ , watching that the middle coefficient is the negative of the sum:

x^2 - \left(-\tfrac{5}{2}\right)x + (-6) = x^2 + \tfrac{5}{2}x - 6.

That is a correct answer. To clear the fraction, multiply every term by $2$ , which scales the quadratic without moving its roots:

2\left(x^2 + \tfrac{5}{2}x - 6\right) = 2x^2 + 5x - 12.

Both $x^2 + \tfrac{5}{2}x - 6$ and $2x^2 + 5x - 12$ have roots $\tfrac{3}{2}$ and $-4$ ; they are the same quadratic up to the scalar $2$ . Factoring the integer version back to $(2x - 3)(x + 4)$ confirms the roots.

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.

Worked example 3 One root of $3x^2 - 14x + 8 = 0$ is $4$ . Find the other root.

The roots add to $-\frac{b}{a} = \frac{14}{3}$ and multiply to $\frac{c}{a} = \frac{8}{3}$ . Call the known root $r_1 = 4$ and use the sum to peel off the other:

r_2 = (r_1 + r_2) - r_1 = \frac{14}{3} - 4 = \frac{14 - 12}{3} = \frac{2}{3}.

The product confirms it, since the roots multiply to $\tfrac{8}{3}$ and $\tfrac{8}{3} \div 4 = \tfrac{2}{3}$ . The other root is $\tfrac{2}{3}$ .

Check your understanding

Which monic quadratic has roots $6$ and $-1$ ?

x^2 - 5x - 6

x^2 + 5x - 6

x^2 - 5x + 6

x^2 - 7x - 6

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:

(r_1 + r_2)^2 = r_1^2 + 2 r_1 r_2 + r_2^2 \implies r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2.

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:

\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_2 + r_1}{r_1 r_2} = \frac{r_1 + r_2}{r_1 r_2}.

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

(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2.

Worked example 4 For the roots of $x^2 - 6x + 4$ , evaluate $r_1^2 + r_2^2$ , $\tfrac{1}{r_1} + \tfrac{1}{r_2}$ , and $(r_1 - r_2)^2$

This quadratic is monic, so $r_1 + r_2 = -(-6) = 6$ and $r_1 r_2 = 4$ . Its discriminant is $\Delta = 36 - 16 = 20 \ge 0$ , so the roots are real (they are the irrational numbers $3 \pm \sqrt{5}$ , though you will never need to write that). Feed the two totals into each identity:

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2 = 6^2 - 2(4) = 36 - 8 = 28.

\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{6}{4} = \frac{3}{2}.

(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2 = 36 - 16 = 20.

Every answer is a clean rational number even though the roots are irrational. That is the power of working with symmetric combinations: the messy parts of the roots cancel.

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,

(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2 = \left(-\frac{b}{a}\right)^2 - 4 \cdot \frac{c}{a} = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} = \frac{\Delta}{a^2}.

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. One direction is right here: 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 other direction is the quadratic formula from the previous lesson: when $\Delta \ge 0$ , $\sqrt{\Delta}$ is a real number, so $\frac{-b \pm \sqrt{\Delta}}{2a}$ names two real roots. The two roots are real exactly when the quantity $a^2 (r_1 - r_2)^2$ , which is what $\Delta$ turns out to be, is not negative. Geometrically, the two roots sit symmetrically about the axis of symmetry $x = -\frac{b}{2a}$ , and the gap between them is $\sqrt{\Delta}/|a|$ : for a fixed leading coefficient $a$ , a larger discriminant means roots that sit farther apart.

The two roots sit symmetrically about the axis of symmetry at x equals negative b over 2a. Their sum fixes the midpoint, and the discriminant fixes how far apart they spread.

Check your understanding

For $2x^2 - 4x - 3$ , are the roots real, and if so what is $(r_1 - r_2)^2$ ?

A Real;

(r_1 - r_2)^2 = 10

B Real;

(r_1 - r_2)^2 = 40

C Not real, since

\Delta = -8 < 0

D Real;

(r_1 - r_2)^2 = 20

Only symmetric combinations are determined

“Symmetric” is doing real work in that heading. A combination of the roots is symmetric when swapping $r_1$ and $r_2$ leaves it unchanged, and only symmetric combinations are pinned down by the coefficients. The sum $r_1 + r_2$ , the product $r_1 r_2$ , and $r_1^2 + r_2^2$ are all symmetric, so each is fixed by $-\frac{b}{a}$ and $\frac{c}{a}$ . But $r_1 - r_2$ is not symmetric: swap the roots and it flips sign. The coefficients cannot decide that sign, which is why they determine $(r_1 - r_2)^2 = \Delta/a^2$ but never $r_1 - r_2$ itself. The method tells you everything symmetric about the roots and, honestly, nothing else.

Check your understanding

The roots of $x^2 - 8x + 2$ are real. Find $r_1^2 + r_2^2$ without solving.

60

64

68

62

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.

Real numbers with sum $s$ and product $p$ exist if and only if $s^2 - 4p \ge 0$ #

Let $s$ and $p$ be real numbers. We prove both directions.

First suppose real numbers $r_1$ and $r_2$ have sum $s$ and product $p$ . Then, by the squared-difference identity,

(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2 = s^2 - 4p.

The left side is the square of a real number, so it cannot be negative, which gives $s^2 - 4p \ge 0$ .

Now the other direction. Suppose $s^2 - 4p \ge 0$ , and consider the monic quadratic $x^2 - sx + p$ . Its discriminant is $(-s)^2 - 4(1)(p) = s^2 - 4p \ge 0$ , so it has real roots

r_1, r_2 = \frac{s \pm \sqrt{s^2 - 4p}}{2}.

By the sum and product formulas applied to this monic quadratic, $r_1 + r_2 = s$ and $r_1 r_2 = p$ . So two real numbers with the required sum and product exist, and they are the roots of $x^2 - sx + p$ .

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: “pick any real numbers $s$ and $p$ , and $x^2 - sx + p$ has two real numbers with that sum and product.” 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.

Worked example 5 Do two real numbers have sum $6$ and product $10$ ? Sum $6$ and product $8$ ?

Test each pair with the condition $s^2 - 4p \ge 0$ . For sum $6$ and product $10$ :

s^2 - 4p = 6^2 - 4(10) = 36 - 40 = -4 < 0.

The condition fails, so no real numbers add to $6$ and multiply to $10$ . For sum $6$ and product $8$ :

s^2 - 4p = 6^2 - 4(8) = 36 - 32 = 4 \ge 0.

The condition holds, so such numbers exist. They are the roots of $x^2 - 6x + 8 = (x - 2)(x - 4)$ , namely $2$ and $4$ , which indeed sum to $6$ and multiply to $8$ .

Check your understanding

Do two real numbers exist with sum $s = 4$ and product $p = 5$ ?

A No;

4^2 - 4(5) = -4 < 0

B Yes; they are the roots of

x^2 - 4x + 5

C No;

4^2 - 4(5) = -36 < 0

D Yes;

5^2 - 4(4) = 9 \ge 0

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

This lesson took the last step in the chapter’s map: expanding factored form back into standard form, and lining the two up. 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

Dropping the minus sign on the sum. The roots add to $-\frac{b}{a}$ , not $\frac{b}{a}$ . For $x^2 - 7x + 10$ the middle coefficient is $-7$ , so the sum is $-(-7) = 7$ . The product keeps the sign of $c$ and takes no extra minus.
Forgetting to divide by $a$ . For $ax^2 + bx + c$ the sum is $-\frac{b}{a}$ and the product is $\frac{c}{a}$ . Reading off $-b$ and $c$ only works when $a = 1$ .
Thinking the roots determine the quadratic. Roots $r_1$ and $r_2$ fix the quadratic only up to a nonzero scalar, since $a(x - r_1)(x - r_2)$ has those roots for every $a \ne 0$ . State “monic” or fix a leading coefficient to get a single answer.
Squaring the sum and stopping. The sum of squares is $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2$ , not $(r_1 + r_2)^2$ . The cross term $2 r_1 r_2$ must be subtracted.
Believing every $s$ and $p$ come from real numbers. Real numbers with sum $s$ and product $p$ exist only when $s^2 - 4p \ge 0$ . Sum $5$ with product $7$ has no real pair.
Expecting $r_1 - r_2$ from the coefficients. Only symmetric combinations are determined by $-\frac{b}{a}$ and $\frac{c}{a}$ . The difference $r_1 - r_2$ is not symmetric; the coefficients fix $(r_1 - r_2)^2 = \Delta/a^2$ but not its sign.

Key takeaways

For $ax^2 + bx + c$ with $a \ne 0$ and real roots ( $\Delta \ge 0$ ): $r_1 + r_2 = -\frac{b}{a}$ and $r_1 r_2 = \frac{c}{a}$ , read straight from the coefficients with no solving.
These come from expanding $a(x - r_1)(x - r_2) = ax^2 - a(r_1 + r_2)x + a\,r_1 r_2$ and matching coefficients, and are confirmed by adding and multiplying the two quadratic-formula roots.
Build from roots with the monic template $x^2 - (r_1 + r_2)x + r_1 r_2$ ; the quadratic is determined only up to a nonzero scalar, so fix the leading coefficient to pin it down.
Symmetric functions reduce to the sum and product: $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2$ , $\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}$ , and $(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2 = \frac{\Delta}{a^2}$ .
That last identity ties the roots to the discriminant: $(r_1 - r_2)^2 \ge 0$ for real roots forces $\Delta \ge 0$ , so the discriminant is exactly the real-root test.
Converse: real numbers with sum $s$ and product $p$ exist if and only if $s^2 - 4p \ge 0$ , and then they are the roots of $x^2 - sx + p$ .

Sum and Product of Roots

Learning goals

Running the chapter’s map backwards

Constructing a quadratic from its roots

Finding the other root

Symmetric functions of the roots

The converse, stated honestly

Checking a solution without solving again

Closing the chapter

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice