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A JEE question gives you a quadratic with roots α and β, then asks for the equation whose roots are α², or 1/α, or (α − 1)/(α + 1). You never need the roots themselves. This post shows the two methods that work, when to choose each, and how to check an answer numerically in a minute.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "relations between roots and coefficients, nature of roots, the formation of quadratic equations with given roots". The JEE (Advanced) 2026 brochure lists the same and adds "symmetric functions of roots". Quadratic equations with complex roots are in NCERT's Class 11 chapter on complex numbers and quadratic equations (reprint 2026-27). Check the documents for your exam year.
Method 1: new sum and new product
If the new roots are p and q, the equation is x² − (p + q)x + pq = 0. Write p + q and pq in terms of α + β and αβ, which you read from the coefficients.
Use 2x² − 3x − 1 = 0, so α + β = 3/2 and αβ = −1/2.
Roots α² and β².
Sum = (α + β)² − 2αβ = 9/4 + 1 = 13/4.
Product = (αβ)² = 1/4.
Equation: x² − (13/4)x + 1/4 = 0, or 4x² − 13x + 1 = 0.
Roots α/β and β/α.
Sum = (α² + β²)/αβ = (13/4) ÷ (−1/2) = −13/2.
Product = 1.
Equation: x² + (13/2)x + 1 = 0, or 2x² + 13x + 2 = 0.
Method 2: substitution
If each new root is y = g(x), solve for x = h(y) and put h(y) into the original equation. This works best when g is easy to invert.
Roots α² and β² again. Here y = x², so x = √y. Substituting gives 2y − 3√y − 1 = 0. Isolate the root, 2y − 1 = 3√y, then square: 4y² − 4y + 1 = 9y, so 4y² − 13y + 1 = 0. Both methods agree.
Roots (α − 1)/(α + 1) and (β − 1)/(β + 1). Here y = (x − 1)/(x + 1), so x = (1 + y)/(1 − y). Substitute and multiply through by (1 − y)²:
2(1 + y)² − 3(1 + y)(1 − y) − (1 − y)² = 0
Expanding: 2 + 4y + 2y² − 3 + 3y² − 1 + 2y − y² = 4y² + 6y − 2 = 0, so 2y² + 3y − 1 = 0. Method 1 would need several lines of algebra here.
Which quick rules should you know?
For ax² + bx + c = 0 with roots α and β:
Roots α + k: replace x by x − k. From x² − 5x + 6 = 0 (roots 2 and 3), roots 4 and 5 give (x − 2)² − 5(x − 2) + 6 = x² − 9x + 20 = 0.
Roots kα: replace x by x/k, giving ax² + bkx + ck² = 0. Doubling 2 and 3 gives x² − 10x + 24 = 0.
Roots 1/α: reverse the coefficients, cx² + bx + a = 0 (with c ≠ 0). Here 6x² − 5x + 1 = 0, with roots 1/2 and 1/3.
Roots −α: change the sign of b: x² + 5x + 6 = 0.
Each one checks against the known roots 2 and 3.
How do you check an answer numerically?
The roots of 2x² − 3x − 1 = 0 are (3 ± √17)/4, about 1.7808 and −0.2808.
Their squares are about 3.1712 and 0.0788. The sum is 3.2500 and the product 0.2500, matching 13/4 and 1/4.
(α − 1)/(α + 1) ≈ 0.7808/2.7808 ≈ 0.2808, and (β − 1)/(β + 1) ≈ −1.2808/0.7192 ≈ −1.7808. Their sum is −1.5 and product about −0.5, matching 2y² + 3y − 1 = 0.
Which mistakes cost marks?
Replacing x by x + k for roots α + k. The new variable is x = α + k, so α = x − k.
Forgetting to clear fractions, then giving an equation with fractional coefficients when integers are expected.
Squaring before isolating the root in the substitution method, which leaves √y terms behind.
Sign of the sum. The equation is x² − (sum)x + product = 0.
Using the 1/α rule when c = 0. Then one root is zero and has no reciprocal.
For the sum and product rules and the conditions on roots, see our post on nature and location of roots.
A practice routine
Seeing a method across varied examples helps you recognise it. A review by Atkinson, Derry, Renkl and Wortham (2000) concluded that effective instruction uses multiple examples for each problem type and varies their format.
One quadratic, many transforms (twice a week): take one equation and form five transformed equations.
Both methods: solve at least one transform both ways each session.
Numerical check: for one answer per session, compute the new roots on a calculator and confirm the sum and product.
Mix: add these to questions on symmetric functions of roots.
If the algebra in a substitution keeps going wrong, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, form the equation whose roots are α + 1/β and β + 1/α for 2x² − 3x − 1 = 0, then check it numerically.
