Common Roots

Take two quadratic equations:

The first has roots 2 and 3. The second has roots 1 and 2. They share the root x = 2.

That was easy to spot because both equations factored into clean integers. But what if the coefficients are messier? What if someone hands you 3x^2 + 7x - 6 = 0 and 2x^2 + 5x - 3 = 0 and asks, do these two equations share a root? You could solve both equations fully using the quadratic formula, compare all four roots, and check for matches. That works, but it is slow — four roots, six possible comparisons, and the roots themselves might be ugly irrational numbers that are hard to compare.

There is a much faster way. Instead of solving both equations and comparing, you can test a single algebraic condition on the coefficients alone. If the condition holds, the equations share a root. If it doesn't, they don't. You never have to solve either equation.

The subtraction trick

Here is the key idea. Suppose \alpha is a root common to both equations. Then both of these are true at the same time:

Subtract equation (2) from equation (1):

This is still a quadratic in \alpha — unless the leading coefficient happens to be zero. And here is the useful case: when a_1 = a_2 (or more generally, when you can arrange the subtraction to kill the \alpha^2 term), the result is a linear equation in \alpha, which you can solve immediately.

Even when a_1 \neq a_2, you can multiply equation (1) by a_2 and equation (2) by a_1 before subtracting, to eliminate the \alpha^2 term on purpose:

Expanding: multiply (1) by a_2 to get a_1a_2\alpha^2 + a_2b_1\alpha + a_2c_1 = 0, and multiply (2) by a_1 to get a_1a_2\alpha^2 + a_1b_2\alpha + a_1c_2 = 0. Subtract:

So the common root is:

provided the denominator is not zero. This is a direct formula for the common root — you read it off the coefficients without ever touching the quadratic formula.

But there is still a question: how do you know a common root exists in the first place? The formula above gives you a candidate for \alpha. You still need to verify that this \alpha actually satisfies both original equations. That verification gives rise to the formal condition.

Condition for one common root

Take the candidate \alpha = \frac{a_1c_2 - a_2c_1}{a_2b_1 - a_1b_2} and substitute it back into one of the original equations. After algebraic simplification (which is lengthy but entirely mechanical), the condition that \alpha satisfies both equations reduces to a single equation in the coefficients.

The condition looks imposing, but it has a clean structure. Notice the three factors involved: (a_1c_2 - a_2c_1), (a_1b_2 - a_2b_1), and (b_1c_2 - b_2c_1). Each one is a "2 \times 2 cross-product" of two columns from the coefficient matrix. If you have met determinants, each factor is a 2 \times 2 determinant built from two of the three columns \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}, \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}, \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}.

A quick mnemonic: write the six coefficients in a 2 \times 3 grid:

Take the three 2 \times 2 minors by dropping one column at a time. Call them P = b_1c_2 - b_2c_1 (drop column a), Q = a_1c_2 - a_2c_1 (drop column b), and R = a_1b_2 - a_2b_1 (drop column c). The condition is Q^2 = PR.

Where the condition comes from

Here is the derivation in full. Suppose \alpha is the common root. From the subtraction trick:

Now use the same cross-multiplication idea but eliminate \alpha instead of \alpha^2. Multiply (1) by b_2 and (2) by b_1:

Subtract:

So \alpha^2 = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}.

Now you have both \alpha and \alpha^2 in terms of the coefficients. For consistency, \alpha^2 must equal (\alpha)^2. Squaring the expression for \alpha and setting it equal to the expression for \alpha^2:

Since a_2b_1 - a_1b_2 = -(a_1b_2 - a_2b_1), the left side is \frac{(a_1c_2 - a_2c_1)^2}{(a_1b_2 - a_2b_1)^2}. Cross-multiplying:

That is the condition, derived cleanly in five lines.

Condition for both common roots

If both roots of the first equation are also roots of the second, the two equations describe the same pair of roots. Since a quadratic is entirely determined by its leading coefficient and its roots, the two equations must be proportional — one is just a scalar multiple of the other.

This makes geometric sense. Two parabolas y = a_1x^2 + b_1x + c_1 and y = a_2x^2 + b_2x + c_2 cross the x-axis at the same two points only if they are scalar multiples of each other. They might have different steepness (different leading coefficients), but their roots — where they touch zero — are identical.

Finding the common root in practice

Here is the practical procedure. Given two quadratics, the fastest route to the common root is almost always the subtraction trick, not the formula.

Step 1. Write both equations in standard form: a_1x^2 + b_1x + c_1 = 0 and a_2x^2 + b_2x + c_2 = 0.

Step 2. Eliminate the x^2 term. If a_1 = a_2, simply subtract. Otherwise, multiply the first equation by a_2 and the second by a_1, then subtract. You get a linear equation in x.

Step 3. Solve the linear equation. The solution is your candidate common root \alpha.

Step 4. Verify by substituting \alpha back into both original equations. If both give zero, \alpha is indeed a common root. If one of them doesn't, the equations have no common root.

Step 4 is not optional. The subtraction trick always produces a candidate, but the candidate is genuine only when the consistency condition (a_1c_2 - a_2c_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1) is satisfied. Substituting back is the fastest verification.

Applications

Common roots appear in several places across algebra.

Intersecting loci in coordinate geometry. When two curves are defined by quadratic equations (a circle and a parabola, or two conics), their intersection points satisfy both equations simultaneously. Finding common roots is exactly the algebra behind finding intersection points.

Parameter-dependent equations. A typical JEE problem gives you x^2 + px + q = 0 and x^2 + rx + s = 0 where p, q, r, s depend on some parameter k, and asks: for what values of k do the two equations share a root? You apply the condition Q^2 = PR and solve for k.

Consistency of systems. If a system of equations has a common solution, the equations are "consistent" at that point. The common-root condition is a special case of checking consistency for quadratic systems.

Interactive exploration

Drag the red point along the x-axis below. The two readouts show the value of each quadratic at that point. When both readouts hit zero at the same x, you have found a common root.

Common confusions

• "If two quadratics share a root, the other root must also be the same." Not at all. The examples above show equations sharing exactly one root while having different second roots. Both-roots-common is a strictly stronger condition (proportional coefficients).

• "I can always find a common root by subtracting." The subtraction trick always produces a candidate. But the candidate is a genuine common root only if it satisfies both original equations. If the condition (a_1c_2 - a_2c_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1) is not met, the candidate fails the check, and the equations have no common root.

• "The condition for one common root and the condition for both common roots are alternatives — exactly one holds." They are not mutually exclusive in that way. If the equations are proportional (both roots common), the one-common-root condition is also satisfied — it is automatically true. The one-common-root condition is necessary for any common root; the proportionality condition is the special case where both roots coincide.

• "If a_1/a_2 = b_1/b_2 but b_1/b_2 \neq c_1/c_2, the equations have one common root." No — if the first two ratios match but the third doesn't, the equations have no common root. For both roots common, all three ratios must match. For one common root, the Q^2 = PR condition must hold. Partial proportionality of coefficients is not enough for anything.

• "Common roots only matter for quadratics." The idea extends to polynomials of any degree. Two polynomials share a common root if and only if their resultant is zero — the common-root condition for quadratics is the simplest case of a much larger theory.

Going deeper

The resultant

The condition (a_1c_2 - a_2c_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1) is the vanishing of the resultant of the two quadratics. The resultant of two polynomials f(x) and g(x) is a single number computed from their coefficients — it is zero precisely when f and g share a root. For two quadratics, the resultant is a 4 \times 4 determinant (the Sylvester matrix), which simplifies to the Q^2 - PR expression above.

The resultant has a clean product formula. If f(x) = a_1(x - \alpha)(x - \beta) and g(x) = a_2(x - \gamma)(x - \delta), then

This is zero exactly when one of the factors (\alpha - \gamma), (\alpha - \delta), (\beta - \gamma), or (\beta - \delta) is zero — that is, when a root of f coincides with a root of g. The resultant packages this "do any roots match?" question into a single polynomial expression in the coefficients, with no need to actually compute the roots.

GCD of polynomials

There is another way to think about common roots. Two polynomials share a common root \alpha if and only if (x - \alpha) divides both of them — that is, (x - \alpha) is a common factor. The greatest common divisor (GCD) of the two polynomials captures all such shared factors. If \gcd(f, g) = 1 (a constant), they share no root. If \gcd(f, g) is linear, they share exactly one root. If \gcd(f, g) is quadratic (and f, g are both quadratic), they share both roots.

You can compute the GCD of two polynomials using the Euclidean algorithm for polynomials — the same idea as the Euclidean algorithm for integers, but with polynomial long division replacing integer division. The subtraction trick in this article is essentially the first step of that algorithm.

Connection to Vieta's formulas

From the sum and product of roots, if the first equation has roots \alpha, \beta and the second has roots \alpha, \gamma (sharing \alpha), then:

• \alpha + \beta = -b_1/a_1 and \alpha\beta = c_1/a_1
• \alpha + \gamma = -b_2/a_2 and \alpha\gamma = c_2/a_2

Subtracting: \beta - \gamma = b_2/a_2 - b_1/a_1. And from the product relations, \alpha = \frac{c_1/a_1}{\beta} = \frac{c_2/a_2}{\gamma}. These two relations, combined with the sum relations, give another route to the common root and the consistency condition — entirely through Vieta's formulas without needing the subtraction trick.

Where this leads next

• Quadratic Formula — the formula that finds all roots of a single quadratic, which you need before you can talk about roots being shared.
• Discriminant and Nature of Roots — the single number b^2 - 4ac that decides whether the roots are real, repeated, or complex.
• Sum and Product of Roots — Vieta's formulas, which give another route to common-root problems.
• Quadratic Equations — Advanced — harder problems involving parameters, symmetric functions of roots, and inequalities.
• Polynomial Factorization — factoring higher-degree polynomials, where the GCD idea from the going-deeper section becomes essential.