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"Find k if the two equations have a common root" is a standard JEE question, often solved slowly by finding the roots of one equation and substituting them into the other. This post gives you the subtraction method, the general condition and when to use each, then works three examples, including a case where sharing one root forces sharing both.
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" in Unit 2. The JEE (Advanced) 2026 brochure lists quadratic equations with real coefficients and the relations between roots and coefficients. Neither names common roots; they are an application of these lines.
In NCERT, the Class 11 chapter Complex Numbers and Quadratic Equations (reprint 2026-27) keeps its title but has no separate section on solving quadratics; its sections end at the Argand plane. The groundwork is in Class 10: Polynomials gives the sum and product of zeroes, and Quadratic Equations gives the discriminant. Check the current year's documents.
The fast method: subtract
If α is a root of both equations, it satisfies both, so it satisfies their difference. When the x² coefficients are equal, subtracting removes x² and leaves a linear equation for α. When they are not equal, multiply first so they are.
Then put α back into either equation to find the unknown.
The general condition
Write both equations with x = α and treat α² and α as two unknowns:
a₁α² + b₁α + c₁ = 0 a₂α² + b₂α + c₂ = 0
Solving by cross-multiplication gives
α² ÷ (b₁c₂ − b₂c₁) = α ÷ (c₁a₂ − c₂a₁) = 1 ÷ (a₁b₂ − a₂b₁).
So α = (c₁a₂ − c₂a₁) ÷ (a₁b₂ − a₂b₁), and since α² = α × α, the condition for a common root is
(c₁a₂ − c₂a₁)² = (a₁b₂ − a₂b₁)(b₁c₂ − b₂c₁).
Check it on x² − 5x + 6 = 0 (roots 2 and 3) and x² − 7x + 12 = 0 (roots 3 and 4). Here c₁a₂ − c₂a₁ = 6 − 12 = −6, a₁b₂ − a₂b₁ = −7 + 5 = −2 and b₁c₂ − b₂c₁ = −60 + 42 = −18. The left side is 36, the right side is (−2)(−18) = 36, and α = −6 ÷ −2 = 3. Correct.
Worked example 1: find k
Find k so that x² − 5x + k = 0 and x² − 7x + 2k = 0 have a common root.
Subtract the second from the first: 2x − k = 0, so α = k/2.
Put α into the first equation: k²/4 − 5k/2 + k = 0, which is k²/4 − 3k/2 = 0, or k(k − 6) = 0.
So k = 0 or k = 6.
Check both. For k = 6: x² − 5x + 6 has roots 2 and 3; x² − 7x + 12 has roots 3 and 4. The common root is 3 = k/2. For k = 0: x² − 5x has roots 0 and 5; x² − 7x has roots 0 and 7. The common root is 0. Both values are valid, and dropping k = 0 is the most common lost mark on this type.
Worked example 2: a relation between p and q
If x² + px + q = 0 and x² + qx + p = 0 have a common root and p ≠ q, show that p + q + 1 = 0.
Subtract: (p − q)x + (q − p) = 0. Since p ≠ q, divide by p − q to get x = 1.
Put x = 1 into the first equation: 1 + p + q = 0.
You can also find the other roots. In the first equation the product of roots is q, so the other root is q; in the second it is p. Check with p = 2, q = −3: x² + 2x − 3 has roots 1 and −3, and x² − 3x + 2 has roots 1 and 2. Correct.
Worked example 3: when one shared root means both
If x² + x + 1 = 0 and ax² + bx + c = 0 have a common root, with a, b, c real, find a : b : c.
The discriminant of x² + x + 1 is 1 − 4 = −3, so its roots are a non-real complex pair. A quadratic with real coefficients has non-real roots only in conjugate pairs. If ax² + bx + c = 0 shares one of them, it must share the other too. Two quadratics with the same two roots are multiples of each other, so a : b : c = 1 : 1 : 1.
The same idea works for irrational roots: if both equations have rational coefficients and the common root is irrational, its conjugate surd is common too. In general, both roots are common when a₁/a₂ = b₁/b₂ = c₁/c₂.
Which mistakes cost marks?
Losing a solution such as k = 0 by dividing by k.
Dividing by p − q without stating p ≠ q.
Mixing up the order of terms in the condition. Write α first, then square it, rather than memorising the final line.
The sum and product of roots, which you need for the other roots, are covered in our post on nature and location of roots, and the next step is symmetric functions of roots.
If you get stuck on which equation to subtract or scale, 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 condition by cross-multiplication once, then verify it on a pair built from known roots.
Make five pairs yourself by choosing roots, such as (1, 4) and (4, −2), and solve each with the subtraction method.
Do two "find the relation" problems like Example 2.
Do one conjugate-root problem like Example 3.
Today, solve Example 1 again with x² − 4x + k and x² − 6x + 3k, then check each k by factorising.
