Reading the sum and product of the roots straight off the coefficients — Vieta's formulas.
You have spent four lessons learning to find roots — factoring, completing the square, the quadratic formula. Here is a surprise that feels almost like cheating: you can know the sum of the two roots and the product of the two roots without finding either one of them. They are hiding in plain sight, right there in the coefficients. Take x2 − 7x + 10 = 0. I will tell you the roots add up to 7 and multiply to 10 before you touch a pencil. (They are 2 and 5: indeed 2 + 5 = 7 and 2 · 5 = 10.)
By the end of this lesson you will be able to read the sum and product of the roots straight off a quadratic, find a second root from a first, evaluate clever expressions like x12 + x22 without solving, and build an equation backward from the roots you want. This shortcut is called Vieta's formulas. Throughout, keep the colors in mind: a is blue, b is teal, c is amber, and the two roots x1 and x2 are violet.
Where could a pattern like this even come from? The quadratic formula. For ax2 + bx + c = 0 the two roots are
x1 = −b + √Δ2a and x2 = −b − √Δ2a,
where Δ = b2 − 4ac is the discriminant from the last lesson. These two roots are twins: same numerator −b, same denominator 2a, differing only by +√Δ versus −√Δ. That symmetry is the whole trick. Watch what happens when we add them and when we multiply them.
Add them. The +√Δ and −√Δ cancel:
x1 + x2 = (−b + √Δ) + (−b − √Δ)2a = −2b2a = −ba
Multiply them. This is a difference of squares, (M + N)(M − N) = M2 − N2, with M = −b and N = √Δ:
x1 · x2 = (−b)2 − (√Δ)24a2 = b2 − (b2 − 4ac)4a2 = 4ac4a2 = ca
For any quadratic ax2 + bx + c = 0 with two roots,
x1 + x2 = −ba x1 · x2 = ca
And when the equation is monic (the leading coefficient a = 1, so it reads x2 + bx + c = 0), it is even cleaner: the sum is −b and the product is c.
Let us sanity-check on x2 − 7x + 10 = 0. Here a = 1, b = −7, c = 10. So the sum should be −b = −(−7) = 7 and the product should be c = 10. This equation factors as (x − 2)(x − 5) = 0, giving roots 2 and 5. And sure enough, 2 + 5 = 7 and 2 · 5 = 10. The coefficients knew all along.
The sum is −b/a, not b/a. People look at x2 − 7x + 10, see the −7, and announce "sum = −7." Wrong — the formula already has a minus built in, so the sum is −(−7) = +7. Read the coefficient with its sign, then flip it. The product, c/a, has no flip.
Dial up any quadratic. Watch the actual roots' sum and product match −b/a and c/a, every time.
Once you trust the two formulas, they become secret codes you can use to answer questions other people would solve the long way. The single most useful move: find the other root when you already know one.
Suppose you are told that x2 − 7x + 10 = 0 has 2 as a root, and you want the other. You have two independent shortcuts, and they should agree:
Through the sum. The roots add to 7, and one of them is 2, so the other is 7 − 2 = 5.
Through the product. The roots multiply to 10, and one of them is 2, so the other is 10 ÷ 2 = 5.
Both say 5. When the two routes agree, that is also a free check that your first root was correct.
You can also answer yes/no questions without ever solving. "Are the roots of x2 − 7x + 10 both positive?" Their product is +10 (so they have the same sign) and their sum is +7 (so that shared sign is positive) — yes, both positive, no solving required. A negative product would have meant one positive root and one negative root.
Here is a monic equation and one of its roots. Get the second root two different ways and see them agree.
Here is where Vieta earns its keep. A great many expressions in x1 and x2 are symmetric — they do not change if you swap the two roots — and every symmetric expression can be rebuilt out of just two ingredients: the sum S = x1 + x2 and the product P = x1 · x2. Since Vieta hands you S and P for free, you can evaluate these without ever knowing the roots themselves.
The workhorse identity comes from the perfect square. Since (x1 + x2)2 = x12 + 2x1x2 + x22, subtract the middle to peel off the squares:
x12 + x22 = S2 − 2P
1x1 + 1x2 = x2 + x1x1x2 = SP
(x1 − x2)2 = S2 − 4P
x12x2 + x1x22 = x1x2(x1 + x2) = P · S
Try them all with our example, where S = 7 and P = 10 (roots 2 and 5):
x12 + x22 = 72 − 2(10) = 49 − 20 = 29. Check: 22 + 52 = 4 + 25 = 29. ✓
1/x1 + 1/x2 = S/P = 7/10 = 0.7. Check: 1/2 + 1/5 = 0.5 + 0.2 = 0.7. ✓
(x1 − x2)2 = 72 − 4(10) = 49 − 40 = 9, so |x1 − x2| = 3. Check: 5 − 2 = 3. ✓
Notice that last identity does double duty as a reality check. Because (x1 − x2)2 is a real square, it can never be negative when the roots are real. So if you ever compute S2 − 4P and get a negative number, that is the tell that the roots are not real. (Compare it to Δ = b2 − 4ac from the last lesson — for a monic equation they are the very same quantity.)
Set only the sum S and product P. Every symmetric expression falls out — no roots needed.
Every formula so far ran forward: equation in, sum and product out. Now run it backward. If you want a quadratic whose roots are exactly r and s, you already know its sum and product — so you can write the equation down directly. The monic quadratic with roots r and s is
x2 − (r + s)x + (r · s) = 0
The minus in front of the sum is Vieta's −b/a rule worn on the outside: the coefficient of x is the negative of the sum. Let us build the equation with roots 3 and −5:
Sum r + s = 3 + (−5) = −2. Product r · s = 3 · (−5) = −15.
Plug in: x2 − (−2)x + (−15) = x2 + 2x − 15 = 0.
Check by factoring: (x − 3)(x + 5) = 0 → roots 3 and −5. ✓
This is exactly how the surprises in exercise problems are manufactured. Want a quadratic whose roots are the irrational pair 2 + √3 and 2 − √3? You never have to multiply scary radicals out the long way — their sum is 4 and their product is (2 + √3)(2 − √3) = 4 − 3 = 1, so the equation is simply x2 − 4x + 1 = 0. Clean.
Choose two roots r and s. The matching monic quadratic and its factored form appear.
Because the two roots share the numerator −b and the denominator 2a and differ only by ±√Δ, adding them cancels the radical to give the sum = −b/a, and multiplying them is a difference of squares that collapses to the product = c/a — and for a monic equation that is just sum −b, product c. With those two numbers, S and P, in hand you can find a second root from a first, decide signs without solving, rebuild any symmetric expression (S2 − 2P, S/P, S2 − 4P), and run the whole thing backward to manufacture x2 − (r + s)x + rs = 0 from whatever roots you please. The single thing to guard is that lone minus sign in the sum.
We have been assuming the discriminant is positive and the roots are real. Next we follow the algebra past that wall: when b2 − 4ac < 0 there are still two roots, but they live in the complex numbers — and, beautifully, Vieta's sum and product survive the trip completely intact.
Work each one out first, then open the answer to check your thinking.
Tap the answer you think is right.
Vieta's formulas — sum of roots = −b/a, product = c/a — are the heart of this lesson. They align with Common Core A-APR.B (the relationship between the zeros and the coefficients of a polynomial) and are a staple of Algebra II and competition preparation. The misconception to hunt down relentlessly is the sign of the sum: students see x2 − 7x + 10 and say the sum is −7 instead of +7. Drill "read the coefficient with its sign, then negate it." The symmetric-function identities (S2 − 2P, S/P, S2 − 4P) are reusable far beyond this chapter; the pedagogical key is to keep everything anchored to the two numbers S and P, so students stop reaching for the individual roots they do not actually need.