Lesson 6 of 12 · Algebra Foundations
Roots, Coefficients & Vieta's Formulas
You almost never need to know what a polynomial's roots actually are — only how they relate to each other. Vieta's formulas hand you the sum and product of the roots directly from the coefficients, and from there, nearly any symmetric question about the roots follows in one or two lines.
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Learning objectives
- State and apply Vieta's formulas for quadratics: sum and product of the roots from the coefficients.
- Extend Vieta's formulas to cubics: sum, pairwise sum, and product of the roots.
- Compute symmetric expressions in the roots — sums of squares, sums of reciprocals, sums of cubes, and the difference of the roots — without ever solving for the roots themselves.
- Recognize disguised Vieta's applications: problems that don't mention "roots" at all but reduce to exactly this technique.
The core idea
For a quadratic ax² + bx + c = 0 with roots r and s, factoring gives a(x − r)(x − s) = 0. Expanding that and matching coefficients with the original equation is exactly where Vieta's formulas come from — they're not a separate fact to memorize so much as factoring run in reverse.
The practical payoff: any question phrased as "the roots satisfy ___, find ___" almost always wants you to express the target quantity in terms of the sum and product of the roots (or, for cubics, the sum, pairwise sum, and product) — not to solve the equation and substitute actual numbers.
Vieta's formulas — quadratic
r + s = −b/a, rs = c/a
For ax² + bx + c = 0 with roots r, s. The most common error is forgetting to divide by a when a ≠ 1.
Vieta's formulas — cubic
p+q+r = −b/a, pq+qr+rp = c/a, pqr = −d/a
For ax³ + bx² + cx + d = 0 with roots p, q, r. Note the alternating signs: sum is negative b/a, pairwise sum is positive c/a, product is negative d/a.
Difference of roots (quadratic)
(r − s)² = (r + s)² − 4rs
Lets you find how far apart the roots are without finding either root individually — just expand (r−s)² in terms of the sum and product you already have from Vieta's.
Sum of reciprocals
1/r + 1/s = (r+s)/(rs)
A direct consequence of combining fractions over a common denominator — requires rs ≠ 0, i.e. c ≠ 0.
Worked example 1 — Sum and product directly
Problem
The roots of 3x² − 12x + 7 = 0 are r and s. Find r + s and rs.
Key insight
Read the coefficients directly into Vieta's formulas — no need to solve the equation.
Solution
Here a = 3, b = −12, c = 7. So r + s = −(−12)/3 = 4 and rs = 7/3.
Takeaway
Every Vieta's problem starts with this step — write down r+s and rs before doing anything else.
Worked example 2 — Sum of squares
Problem
The roots of x² − 9x + 8 = 0 are r and s. Find r² + s².
Key insight
r² + s² isn't one of Vieta's formulas directly, but it's built from the two that are: (r+s)² − 2rs.
Solution
r + s = 9, rs = 8. So r² + s² = 9² − 2(8) = 81 − 16 = 65.
Takeaway
Almost every "symmetric expression in the roots" question is really asking you to rewrite that expression in terms of r+s and rs, then substitute.
Worked example 3 — Sum of reciprocals
Problem
The roots of 2x² − 5x + 3 = 0 are r and s. Find 1/r + 1/s.
Key insight
1/r + 1/s = (r+s)/(rs) — both pieces come straight from Vieta's.
Solution
r + s = 5/2, rs = 3/2. So 1/r + 1/s = (5/2)/(3/2) = 5/3.
Takeaway
Resist the urge to solve the quadratic and plug in actual root values — it works, but takes far longer and risks arithmetic errors with the resulting radicals.
Worked example 4 — Extending to a cubic
Problem
The roots of x³ − 6x² + 11x − 6 = 0 are p, q, r. Find p+q+r, pq+qr+rp, and pqr.
Key insight
The cubic version of Vieta's works exactly like the quadratic version, with one more quantity and an extra sign to track.
Solution
Here a=1, b=−6, c=11, d=−6. So p+q+r = 6, pq+qr+rp = 11, and pqr = 6.
Alternate method
This cubic actually factors as (x−1)(x−2)(x−3), so the roots are 1, 2, 3 — check: 1+2+3=6, (1)(2)+(2)(3)+(3)(1)=11, (1)(2)(3)=6. Confirms Vieta's, but finding the factorization takes real work; Vieta's gets the same answer in one line without it.
Takeaway
Vieta's formulas scale to any degree — the pattern of alternating signs and elementary symmetric sums continues predictably.
Worked example 5 — Disguised application
Problem
The equation x² + 9x + k = 0 has one root equal to three times the other. Find k.
Key insight
This problem never says "Vieta's formulas" — the signal is a stated relationship between the two roots, which turns the sum equation into a single-variable equation.
Solution
Let the roots be r and 3r. The sum is r + 3r = 4r = −9, so r = −9/4. The product is k = 3r² = 3(−9/4)² = 243/16.
Takeaway
Whenever a problem states a ratio or relationship between the roots ("one is twice the other," "they differ by 5"), express both roots in terms of a single new variable, then apply Vieta's to that variable.
Strategy notes
Write down r+s and rs before anything else
For any quadratic Vieta's problem, the first line of your work should always be identifying the sum and product from the coefficients — even before you know what the question is really asking. Everything else is built from those two numbers.
Look for the algebraic identity that connects the target to r+s and rs
r² + s², r³ + s³, 1/r + 1/s, and (r − s)² are the four you'll see most often. Recognizing which identity applies is the entire skill — the arithmetic afterward is simple.
Common mistakes
Forgetting to divide by the leading coefficient
Vieta's formulas for ax²+bx+c=0 are r+s = −b/a and rs = c/a — not −b and c. This error only shows up when a ≠ 1, which is exactly when it's easy to forget.
Sign errors in the cubic pairwise-sum and product formulas
The signs alternate: sum is −b/a, pairwise sum is +c/a, product is −d/a. Writing all three with the same sign is the most common mistake when extending from quadratics to cubics.
Solving the equation anyway "to check"
If the discriminant isn't a perfect square, solving directly produces messy radicals that are easy to mishandle. Trust the symmetric-expression identity instead of verifying by brute force.
Practice
18 questions across four difficulty tiers. Type your answer and check it — up to three tries before the solution is shown. If your first-try accuracy is below 70%, you'll get one reinforcement pass through the questions you missed.