Common Roots of Quadratic Equations
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Direct answer
Two quadratics a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 share exactly one root when (c₁a₂ − c₂a₁)² = (a₁b₂ − a₂b₁)(b₁c₂ − b₂c₁), and share both roots only when a₁/a₂ = b₁/b₂ = c₁/c₂ — proportional coefficients. The working engine beneath both results is elimination: if α satisfies both equations, subtract suitable multiples to kill x² or x, solve the resulting linear equation for α, and verify in an original equation, since manipulation can smuggle in extraneous values. Graphically, a common root is a shared x-intercept of the two parabolas y = f(x) and y = g(x), a reading JEE occasionally exploits in coefficient-range questions.
What you must remember
- One common root condition: (c₁a₂ − c₂a₁)² = (a₁b₂ − a₂b₁)(b₁c₂ − b₂c₁) — memorise it as a symmetric determinant-style identity.
- Both roots common: a₁/a₂ = b₁/b₂ = c₁/c₂, valid whether the roots are real or complex; the equations are then the same equation scaled.
- The common root itself: from cross-multiplication, α = (c₁a₂ − c₂a₁)/(a₁b₂ − a₂b₁) = (b₁c₂ − b₂c₁)/(c₁a₂ − c₂a₁).
- Linear-combination trick: the common root of f = 0 and g = 0 also satisfies f + λg = 0 for every λ; choosing λ to make the combination a linear equation or a perfect square is the speed move.
- Recovering the other roots: once α is known, the other root of f is c₁/(a₁α) (product c₁/a₁) or −b₁/a₁ − α (sum −b₁/a₁).
- Derivation route: treat aα² + bα + c = 0 for both equations as a linear system in (α², α, 1); non-trivial solution forces the determinant to vanish — this regenerates the one-root condition on demand.
- Reality check: a (real) common root requires each discriminant ≥ 0; coefficient conditions alone do not guarantee the shared root is real.
The symmetric-coefficient classic
Consider x² + ax + b = 0 and x² + bx + a = 0 with a ≠ b, asked to show they share exactly one root and find it. Subtract the second from the first: (a − b)x + (b − a) = 0, which is (a − b)(x − 1) = 0. Since a ≠ b, the only possible common root is x = 1. Substituting back: 1 + a + b = 0 is then the consistency condition — the equations share the root 1 precisely when a + b = −1. The other roots fall out by product: for the first equation the roots multiply to b, so the other root is b; for the second, the other root is a. The entire problem is three lines once subtraction is instinctive.
The same subtraction idea powers parameter questions. Given x² − 5x + k = 0 and x² − kx + 5 = 0, subtraction yields (k − 5)x + (k − 5) = 0, i.e. (k − 5)(x + 1) = 0. So either k = 5 (the equations become identical) or the only possible common root is x = −1, which substituted into the first equation gives 1 + 5 + k = 0, k = −6. One candidate rejected, one confirmed — the verification step is where examiners set the trap.
Traps in common-root problems
The lethal trap is declaring a common root without checking it satisfies both original equations; elimination narrows candidates but never certifies them, and JEE answer options are built on the unchecked value. The second trap is direction: a₁/a₂ = b₁/b₂ = c₁/c₂ guarantees both roots common, but one accidental equality of ratios guarantees nothing. The third is reality: the algebra of shared roots works over complex numbers too, so a question saying "real common root" demands a discriminant check on top. Advanced-level versions ask for all k making the quadratics share a root, then expect a discriminant filter on the resulting relation; skipping it leaves spurious values among the options.
Frequently asked questions
What is the condition for exactly one common root between two quadratics?
(c₁a₂ − c₂a₁)² = (a₁b₂ − a₂b₁)(b₁c₂ − b₂c₁) for a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0.
When do two quadratics have both roots in common?
When their coefficients are proportional: a₁/a₂ = b₁/b₂ = c₁/c₂, real or complex roots alike.
If x² + ax + b = 0 and x² + bx + a = 0 (a ≠ b) share a root, what is it?
Subtraction forces x = 1, and consistency requires a + b + 1 = 0.
Why verify the common root after elimination?
Because subtracting or multiplying equations can introduce extraneous candidate values; only substitution in the originals confirms a genuine common root.
Does a common root mean the discriminants are positive?
For a real common root, yes — each equation must have discriminant ≥ 0; complex common roots need no such condition.