Polynomial Roots and Coefficients
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Direct answer
Vieta's formulas tie a polynomial's roots to its coefficients without solving anything: for ax^2 + bx + c = 0 with roots α and β, α + β = −b/a and αβ = c/a; for the cubic ax^3 + bx^2 + cx + d = 0 with roots α, β, γ, the sum is −b/a, the pairwise sum αβ + βγ + γα = c/a and the product αβγ = −d/a. Every symmetric expression in the roots — squares, cubes, reciprocals — collapses to these numbers, and the remainder theorem polices individual roots: P(a) is the remainder on dividing P(x) by x − a, so (x − α) is a factor exactly when P(α) = 0.
What you must remember
- Quadratic Vieta: α + β = −b/a and αβ = c/a; then α^2 + β^2 = (α + β)^2 − 2αβ and α^3 + β^3 = (α + β)^3 − 3αβ(α + β).
- Cubic Vieta: α + β + γ = −b/a, αβ + βγ + γα = c/a, αβγ = −d/a; α^2 + β^2 + γ^2 = (α + β + γ)^2 − 2(αβ + βγ + γα).
- Roots in AP (cubic): one root equals the mean −b/(3a); the packaged condition is 2b^3 − 9abc + 27a^2 d = 0.
- Roots in GP (cubic): b^3 d = c^3 — check with roots 1, 2, 4, which give x^3 − 7x^2 + 14x − 8 and (−7)^3 × (−8) = 14^3 = 2744.
- Reciprocal-root equation: for a quadratic, interchange the coefficients end to end: roots 1/α, 1/β satisfy cx^2 + bx + a = 0.
- Common roots: a root shared by two polynomials is a root of their difference; for two quadratics the full condition is (c1b2 − c2b1)(a1b2 − a2b1) = (a1c2 − a2c1)^2.
- Complex roots: with real coefficients, non-real roots arrive in conjugate pairs, so a real cubic always owns at least one real root.
A roots-in-AP cubic, worked
Find k so that x^3 − 6x^2 + 9x + k = 0 has roots in arithmetic progression. Apply the condition 2b^3 − 9abc + 27a^2 d = 0 with a = 1, b = −6, c = 9, d = k: 2(−216) − 9(1)(−6)(9) + 27k = −432 + 486 + 27k = 0, giving k = −2. The sum of roots is 6, so the middle root is 6/3 = 2; write the roots 2 − t, 2, 2 + t. Their product equals −d = 2, so 2(4 − t^2) = 2 and t^2 = 3: the roots are 2 − √3, 2, 2 + √3. The pairwise sum checks out: (2 − √3)·2 + 2(2 + √3) + (2 − √3)(2 + √3) = 4 + 4 + 1 = 9 = c/a. Two independent verifications, both fast — that habit is what separates a guessed answer from a banked one in this chapter.
Where marks leak
The most common error is forgetting the leading coefficient: writing the sum of roots of 2x^2 − 5x + 1 as 5 instead of 5/2. JEE Main asks direct Vieta evaluations — sums of squares, reciprocals, differences squared. JEE Advanced dresses the same algebra in parameter clothing: "the roots of x^3 − 3x + k = 0 are ..." or common-root counting between two quadratics, where subtracting the equations to manufacture a linear relation is the intended move. Keep in mind the conjugate-pair rule as well: a question stating a cubic has root 2 + i has, without asking, also given you root 2 − i and therefore the third root from the sum.
Frequently asked questions
What is the sum of squares of the roots of ax^2 + bx + c = 0?
(b/a)^2 − 2(c/a), which simplifies to (b^2 − 2ac)/a^2.
What is the condition for a cubic's roots to be in AP?
2b^3 − 9abc + 27a^2 d = 0, with the middle root then equal to −b/(3a).
How do you form the quadratic whose roots are the reciprocals?
Reverse the coefficients: cx^2 + bx + a = 0, since 1/α + 1/β = (α + β)/αβ.
If 2 is a root of P(x), what follows?
P(2) = 0 and (x − 2) divides P(x) exactly — the factor theorem in both directions.
Can a real cubic have exactly two non-real roots?
No — complex roots of real-coefficient polynomials come in conjugate pairs, so the third root must be real.