# Equations With Transformed Roots

> Root transformations in JEE Mathematics: forming equations with roots 1/alpha, alpha + h, k alpha or alpha^2 using substitution and Vieta arithmetic with worked checks.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/sum-roots-transformations
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Equations With Transformed Roots", PrepElephant, https://prepelephant.com/topics/jee/mathematics/sum-roots-transformations

## Direct answer

Given an equation with roots α and β, examiners love asking for the equation whose roots are 1/α, α + h, kα or α^2 — and two routes produce it. The substitution method runs f(x) = 0 through the change of variable backwards: for roots α + h, set x = y - h (since y = x + h) and expand f(y - h) = 0; for roots kα, use f(y/k) = 0 and clear denominators; for reciprocal roots 1/α, replace x by 1/y and multiply by y^2. The Vieta method computes the new sum and product directly from S = α + β and P = αβ — roots α + h and β + h carry sum S + 2h and product P + Sh + h^2, while α^2 and β^2 carry sum S^2 - 2P and product P^2. Both routes must agree, and checking that agreement is the fastest self-verification available in algebra.

## What you must remember

- **Substitution direction:** new roots α + h mean y = x + h, so substitute x = y - h into f(x) = 0; writing x = y + h is the classic reversal error.
- **Reciprocal roots:** replace x by 1/y in f(x) = 0 and multiply through by y^n — the coefficients of the original run backwards.
- **Scaled roots kα:** f(y/k) = 0, then clear the powers of k from denominators; for k = -1 (negated roots) alternate the coefficient signs.
- **Vieta building blocks:** S = α + β, P = αβ; for roots α + h, β + h: sum S + 2h, product P + Sh + h^2; for α^2, β^2: sum S^2 - 2P, product P^2.
- **Reciprocal by Vieta:** for roots 1/α and 1/β, the new sum is 1/α + 1/β = S/P and the new product is 1/P, so the equation is Py^2 - Sy + 1 = 0.
- **Diminishing by turns:** roots one less than the originals (h = -1) give sum S - 2 and product P - S + 1 — used repeatedly in root-shifting descent arguments for cubics.
- **Cubic extensions:** for ax^3 + bx^2 + cx + d with roots α, β, γ, transformed Vieta sums follow from S1 = -b/a, S2 = c/a, S3 = -d/a with the same substitution machinery.

## Two routes to the same polynomial

Let α, β be the roots of x^2 - 5x + 6 = 0 (so S = 5, P = 6) and build the equation with roots 2α + 1 and 2β + 1. Vieta route: the new sum is 2S + 2 = 12, and the new product is (2α + 1)(2β + 1) = 4P + 2S + 1 = 24 + 10 + 1 = 35, so the equation is y^2 - 12y + 35 = 0, which factors as (y - 5)(y - 7) — consistent, since the original roots are 2 and 3 and their images 5 and 7. Substitution route: y = 2x + 1 means x = (y - 1)/2; substituting into x^2 - 5x + 6 = 0 gives (y - 1)^2/4 - 5(y - 1)/2 + 6 = 0, and multiplying by 4: y^2 - 2y + 1 - 10y + 10 + 24 = y^2 - 12y + 35. Identical output, and when the two routes disagree you have made an algebra slip — the check costs thirty seconds and saves four marks. For practice, run the same pair on roots α^2, β^2: sum S^2 - 2P = 25 - 12 = 13, product 36, equation y^2 - 13y + 36 = 0, roots 4 and 9, exactly the squares of 2 and 3.

## Where students slip

JEE Main asks this as a direct construction — "form the equation whose roots are the reciprocals of..." — answerable in under a minute by either route, with 3-5-6 and 2-3-5 style integers keeping the arithmetic transparent. Advanced wraps transformations inside larger arguments: showing a polynomial exists whose roots are all one less than another's (then iterating toward a root count), or combining transformation with symmetric-function identities like α^3 + β^3. The most frequent error is the shift's direction: for roots α + h the substitution is x = y - h, because the new variable y must equal the old root plus h — doing x = y + h produces roots α - h, a fully worked wrong answer that looks plausible. The second is clearing denominators only from some terms (for f(y/k), every power of k must clear, degree by degree). A quieter trap: for equations with non-real roots, transformed roots inherit conjugacy — the new polynomial still has real coefficients, and candidates who compute numeric roots first waste the structure. Quadratic theory of equations is core syllabus; transformations extend naturally to cubics in Advanced.

## Frequently asked questions

### What substitution gives the equation with roots α + h?

Replace x by y - h in f(x) = 0, since the new variable y = x + h; expanding and simplifying yields the transformed polynomial.

### How is the equation with reciprocal roots formed?

Substitute x = 1/y and multiply by y^n; equivalently reverse the coefficient sequence — for a quadratic with sum S and product P, the new equation is Py^2 - Sy + 1 = 0.

### What are the sum and product of the roots α^2 and β^2?

Sum S^2 - 2P and product P^2, where S and P are the original sum and product — no root extraction needed.

### How do you square-check a transformed equation?

Solve the original if it factors nicely, transform the numeric roots, and verify they satisfy your constructed polynomial; better still, derive it by both Vieta and substitution.

### What happens to the roots when each is multiplied by k?

They become kα and kβ, obtained from f(y/k) = 0; the transformed sum is kS and the product k^2 P, so constants scale the coefficients degree by degree.