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Many root questions never ask for the roots. They ask for α⁴ + β⁴, or 1/α² + 1/β², or a ratio of high powers that would take a page to expand. Solving the quadratic first is slow and invites arithmetic slips with surds. This post shows why sum and product are always enough for symmetric expressions, gives the identities you need, and works one equation all the way to fourth powers, then a high-power ratio.
Where is this in the syllabus?
The JEE (Advanced) 2026 brochure lists "symmetric functions of roots" alongside relations between roots and coefficients. The JEE (Main) 2026 syllabus lists "relations between roots and coefficients", which is where these questions sit.
In NCERT, the relations α + β = −b/a and αβ = c/a are in section 2.3 of the Class 10 chapter Polynomials (reprint 2026-27), together with the cubic versions. The Class 11 chapter Complex Numbers and Quadratic Equations (reprint 2026-27) no longer has a section on solving quadratics, so the symmetric-function technique comes from JEE preparation rather than the current NCERT book. Check the current year's documents.
What makes an expression symmetric?
An expression in α and β is symmetric if swapping α and β leaves it unchanged. α² + β², αβ and 1/α + 1/β are symmetric. α − β is not, since swapping gives β − α.
The useful fact is that every symmetric polynomial in α and β can be written using only s = α + β and p = αβ. You never need the roots, only s and p, which you read off the coefficients.
Which identities do you need?
With s = α + β and p = αβ:
α² + β² = s² − 2p
α³ + β³ = s³ − 3ps
α⁴ + β⁴ = (α² + β²)² − 2p²
α²β + αβ² = ps
1/α² + 1/β² = (α² + β²) ÷ p²
(α − β)² = s² − 4p, so |α − β| = √(s² − 4p)
The last one shows how to handle a non-symmetric expression: square it, use s and p, then decide the sign from the question.
Worked example 1: one equation, many expressions
Take 3x² − 5x + 1 = 0. Then s = 5/3 and p = 1/3.
α² + β² = 25/9 − 2/3 = 25/9 − 6/9 = 19/9.
α³ + β³ = 125/27 − 3 × (1/3) × (5/3) = 125/27 − 45/27 = 80/27.
α⁴ + β⁴ = (19/9)² − 2/9 = 361/81 − 18/81 = 343/81.
1/α² + 1/β² = (19/9) ÷ (1/9) = 19.
|α − β| = √(25/9 − 4/3) = √(13/9) = √13/3.
Check. The roots are (5 ± √13)/6, about 1.4343 and 0.2324. Their squares add to about 2.1111, which is 19/9, and their cubes to about 2.9630, which is 80/27. The difference is about 1.2019, which is √13/3.
How do you get higher powers quickly?
Let Sₙ = αⁿ + βⁿ. Since α is a root, 3α² = 5α − 1. Multiply by αⁿ⁻² to get 3αⁿ = 5αⁿ⁻¹ − αⁿ⁻², do the same for β, and add:
3Sₙ = 5Sₙ₋₁ − Sₙ₋₂.
Start with S₀ = 2 and S₁ = 5/3:
S₂ = (5 × 5/3 − 2) ÷ 3 = (25/3 − 6/3) ÷ 3 = 19/9
S₃ = (5 × 19/9 − 5/3) ÷ 3 = (95/9 − 15/9) ÷ 3 = 80/27
S₄ = (5 × 80/27 − 19/9) ÷ 3 = (400/27 − 57/27) ÷ 3 = 343/81
All three agree with the identities. In general, for ax² + bx + c = 0 the recurrence is aSₙ + bSₙ₋₁ + cSₙ₋₂ = 0.
Worked example 2: a ratio of high powers
α and β are the roots of x² − 6x − 2 = 0, and aₙ = αⁿ − βⁿ. Find (a₁₀ − 2a₈) ÷ (2a₉).
Here aₙ is not symmetric, but the recurrence still works because it holds for each root separately. From α² = 6α + 2, multiplying by α⁸ gives α¹⁰ = 6α⁹ + 2α⁸, and the same for β. Subtracting the β line from the α line gives
a₁₀ = 6a₉ + 2a₈, so a₁₀ − 2a₈ = 6a₉,
and the ratio is 6a₉ ÷ 2a₉ = 3. No powers were computed. A numerical check with the roots 3 ± √11 gives 3.000.
Which mistakes cost marks?
Sign of s. For ax² + bx + c = 0, s = −b/a. For 3x² − 5x + 1, s = +5/3.
α³ + β³ = s³ − 3ps, not s³ − 3p.
Treating α − β as symmetric and taking only the positive root without checking the question.
Forgetting S₀ = 2, not 1, when starting a recurrence.
To build a new equation from these values, see our post on quadratics with transformed roots. For when two equations share a root, see the common root condition.
If a step in the algebra keeps slipping, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Derive the six identities above from (α + β)² and (α + β)³ without notes.
Pick one quadratic a day and compute S₂ to S₅ both by identity and by recurrence.
Do two ratio problems like Example 2 a week.
Check one answer per session numerically by solving the quadratic.
Today, for 2x² + 3x − 4 = 0, find α³ + β³ two ways and confirm they match.
