Quadratic Equations With a Common Root: Elimination Made Easy

Two quadratic equations appear in a question, followed by the words “have a common root”. Should you solve both equations with the quadratic formula? Compare their discriminants? Start memorising a condition involving six coefficients?

Usually, you can do something much simpler: cancel the squared term and see what remains.

The important part is what happens next. The value you obtain is a possible shared root. You still need to check it in an original equation. That small check separates a complete solution from a convincing-looking wrong answer.

Think of two trees with different branches but one shared underground foundation. You do not need to trace every branch to find the connection. You need to isolate what both trees must share.

What does a common root actually mean?

A root of an equation is a value that makes its left-hand side zero. For example, 2 is a root of x² - 7x + 10 = 0 because 4 - 14 + 10 = 0. This substitution-based definition is explained in NCERT’s chapter on quadratic equations.

A common root makes both equations zero at the same time.

Consider:

• x² - 7x + 10 = 0, with roots 2 and 5.
• x² - 9x + 14 = 0, with roots 2 and 7.

Their common root is 2. The other roots need not match.

Throughout this lesson, coefficients are real and the coefficient of x² in each quadratic is nonzero. When a question asks for a common real root, we must also respect that word “real”.

What should you eliminate first?

Before calculating, spend a few seconds looking at the coefficients.

What you notice	Best first move	What to watch
Equal coefficients of x²	Subtract the equations	Change every sign in the subtracted expression
Unequal coefficients of x²	Multiply to make them equal, then subtract	Multiply the entire equation
One quadratic factorises immediately	Find its roots and test them in the other	Each root may produce a different parameter value
A parameter appears in the same pattern	Consider eliminating the parameter	Keep every factor-zero case
One equation is a multiple of the other	Solve the single underlying quadratic	Check whether its roots are real if required

Eliminating x² is the reliable default because it normally leaves a linear equation. Eliminating the constant or the x term can work, but often leaves another squared expression to solve.

If factorisation is obvious, use it. Good algebra is about choosing a short valid route, not forcing every question into the same formula.

Example 1: equal leading coefficients

Find the common root of:

x² - 7x + 10 = 0 and x² - 9x + 14 = 0.

Let the common root be r. Then:

r² - 7r + 10 = 0

r² - 9r + 14 = 0

Subtract the second equation from the first:

2r - 4 = 0

So r = 2.

Now check:

• First equation: 2² - 7(2) + 10 = 0.
• Second equation: 2² - 9(2) + 14 = 0.

Both checks work, so 2 is the common root.

Writing r rather than x is helpful in your working: it reminds you that you are considering one value shared by both equations, not claiming that their full solution sets are identical.

Example 2: unequal leading coefficients

Find the common root of:

2x² - 9x + 7 = 0 and 3x² - 8x + 5 = 0.

Multiply the first equation by 3 and the second by 2:

6r² - 27r + 21 = 0

6r² - 16r + 10 = 0

Subtract:

-11r + 11 = 0

Therefore r = 1.

Check in the originals:

• 2 - 9 + 7 = 0.
• 3 - 8 + 5 = 0.

The common root is 1. The remaining roots are 7/2 and 5/3, so there is exactly one shared root value.

Why subtraction can produce a false candidate

Suppose the equations are:

x² - 7x + 10 = 0 and x² - 9x + 15 = 0.

Subtracting gives:

2r - 5 = 0, so r = 5/2.

But substituting in the first expression gives:

25/4 - 35/2 + 10 = -5/4, not zero.

So these equations have no common root.

What went wrong? Nothing went wrong with the subtraction. It found a value where the two polynomial expressions are equal. At r = 5/2, they are both -5/4. Equal values are not necessarily zero values.

This also explains the graph interpretation. A common real root is an x-coordinate where both parabolas meet the x-axis. The parabolas merely intersecting somewhere above or below the axis does not give a common root.

Example 3: a parameter question where factorisation wins

Find all values of k for which:

x² - 7x + 10 = 0 and x² + kx + 6 = 0

have a common root.

The first equation factorises as:

(x - 2)(x - 5) = 0.

Any common root must therefore be 2 or 5. Test both possibilities in the second equation.

If r = 2:

4 + 2k + 6 = 0, so k = -5.

The second equation becomes x² - 5x + 6 = (x - 2)(x - 3) = 0. The common root is 2.

If r = 5:

25 + 5k + 6 = 0, so k = -31/5.

The second equation becomes (x - 5)(x - 6/5) = 0. The common root is 5.

Thus k = -5 or -31/5. Neither value makes both roots common.

Notice why testing only the smaller root would lose an answer. “A common root” does not tell you in advance which root is shared.

Example 4: eliminate the parameter instead

Find all k for which:

x² + kx + 3 = 0 and 2x² + kx + 2 = 0

have a common root.

Here, subtracting the first equation from the second eliminates k immediately:

r² - 1 = 0.

So r = 1 or r = -1.

Using r² + kr + 3 = 0:

• If r = 1, then 1 + k + 3 = 0, giving k = -4.
• If r = -1, then 1 - k + 3 = 0, giving k = 4.

Check by factorisation:

k	First quadratic	Second quadratic	Shared root
-4	(x - 1)(x - 3)	2(x - 1)²	1
4	(x + 1)(x + 3)	2(x + 1)²	-1

Each case has exactly one distinct common root. The repeated root in the second quadratic does not create a second distinct shared value.

This is a useful exception to the default method: eliminate whatever makes the remaining equation easiest, provided you preserve all cases.

The general common-root formula, derived carefully

Let:

f(x) = ax² + bx + c

g(x) = dx² + ex + h

with a and d nonzero. Suppose r is common, so f(r) = g(r) = 0.

Multiply the first equation by d and the second by a, then subtract:

(db - ae)r + (dc - ah) = 0.

For shorter working, define:

L = db - ae, and M = dc - ah.

We now have Lr + M = 0.

If L is nonzero:

r = -M/L = (ah - dc)/(db - ae).

This gives the only possible common root. Substitute it into f(r) = 0:

a(M²/L²) - b(M/L) + c = 0.

Multiplying by L² gives the condition:

aM² - bML + cL² = 0.

For L nonzero, this condition is both necessary and sufficient. Why sufficient? It makes f(r) zero, while the elimination relation gives d f(r) - a g(r) = 0. Since a is nonzero, g(r) must then be zero too.

You may also meet the compact version:

(ah - dc)² = (ae - db)(bh - ce).

The common-root condition is included in SATHEE’s quadratic-equation study notes. Our elimination working above shows how to obtain and check the candidate instead of relying on a memorised coefficient pattern.

For everyday numerical questions, doing the subtraction directly is usually easier than assigning six letters. The general formula is useful when the question asks for a condition on coefficients.

What if the denominator is zero?

Never put L = 0 into r = -M/L and continue. Return to the equation Lr + M = 0.

There are three possibilities:

L	M	Meaning
Nonzero	Any value	One candidate, r = -M/L; substitute to check
Zero	Nonzero	Impossible equation M = 0; no common root
Zero	Zero	The quadratics are proportional; solve the underlying equation

Example 5: zero denominator, no common root

Take:

2x² + 6x + 1 = 0 and x² + 3x + 4 = 0.

Subtract twice the second expression from the first:

-7 = 0.

That contradiction proves there is no common root. You do not need the quadratic formula.

Example 6: zero denominator, proportional equations

Take:

2x² - 10x + 12 = 0 and x² - 5x + 6 = 0.

The first is twice the second. Subtracting gives 0 = 0, which supplies no new restriction.

Solve the second equation:

(x - 2)(x - 3) = 0.

Both 2 and 3 are common roots.

For genuine quadratics, proportionality means there is a nonzero constant t such that:

a = td, b = te, c = th.

This wording is safer than writing a/d = b/e = c/h, because some middle or constant coefficients may be zero.

Proportional does not always mean common real roots

The equations x² + 4 = 0 and 3x² + 12 = 0 are proportional. They share the complex roots 2i and -2i, where i² = -1, but have no real roots.

If a question specifies real roots, check the discriminant of the underlying quadratic. Proportional quadratics have two distinct common real roots when the discriminant is positive, one repeated common real value when it is zero, and none when it is negative.

Example 7: the factor you must not cancel

Suppose:

x² + px + q = 0 and x² + qx + p = 0

have a common root r. Subtraction gives:

(p - q)r + (q - p) = 0,

or (p - q)(r - 1) = 0.

There are two branches.

If p differs from q, then r = 1. Substitution gives:

1 + p + q = 0.

This condition also makes 1 a root of both equations, so it works in this branch.

If p = q, the equations are identical. You cannot conclude r = 1. Instead, solve x² + px + p = 0.

For a common real root in this branch:

p² - 4p must be nonnegative, giving p ≤ 0 or p ≥ 4.

For instance, p = q = 0 gives x² = 0 in both equations. Their common root is 0, even though p + q + 1 is not zero.

That one example shows exactly why cancelling p - q without checking it can remove valid answers.

A zero common root needs special care too

If both constant terms are zero, x = 0 is automatically a common root.

For example:

x² - 4x = x(x - 4) = 0

2x² + 3x = x(2x + 3) = 0.

The common root is 0. Dividing both equations by x would erase it and leave only x = 4 and x = -3/2, incorrectly suggesting there is no common root.

The safe move is to factor x out, record x = 0, and then examine the other factors.

Similarly, if a parameter can make a leading coefficient zero, treat that value separately. A formula derived for two quadratics assumes both leading coefficients remain nonzero. If the question explicitly requires two quadratic equations, parameter values that destroy that requirement must be excluded.

How to present a complete exam solution

A clear solution usually follows this sequence:

1. Write, “Let r be the common root.”
2. Substitute r into both equations and label them.
3. State the operation used, such as “3 × (i) - 2 × (ii)”.
4. Solve the simpler relation, keeping every case before division.
5. Substitute each candidate into an original equation to find any parameter.
6. Verify the resulting root and parameter in both originals.
7. State whether there is exactly one shared value or both roots are shared, if asked.

Checking both equations is quick and catches arithmetic slips. Algebraically, once the elimination relation is correct and the candidate satisfies one original, the other follows. In written practice, showing the two short numerical checks is often the clearest finish.

Do not use equal discriminants as a common-root test. For example, x² - 5x + 6 = 0 and x² - 9x + 20 = 0 both have discriminant 1. Their roots are 2, 3 and 4, 5 respectively, so they share nothing.

Try these before looking at the answers

For questions 1 to 3, find the common root or show that none exists.

1. x² - 8x + 15 = 0 and x² - 10x + 21 = 0.
2. 2x² - 5x - 3 = 0 and 3x² - 8x - 3 = 0.
3. x² - 6x + 8 = 0 and x² - 8x + 15 = 0.
4. Find all k for which x² - 5x + 4 = 0 and x² + kx + 2 = 0 have a common root.
5. How many distinct common real roots do x² - 6x + 9 = 0 and 4x² - 24x + 36 = 0 have?
6. Find the common root of x² + 2x = 0 and 3x² - x = 0. Explain why dividing by x first is unsafe.
7. Do x² + 9 = 0 and 2x² + 18 = 0 have a common real root? What if complex roots are allowed?
8. For x² + px + q = 0 and x² + qx + p = 0 with p unequal to q, find the condition for a common root.

Answers with the key working

1. Common root 3. Subtraction gives 2r - 6 = 0. Both originals are zero at r = 3.

2. Common root 3. Three times the first minus twice the second gives r - 3 = 0. The checks are 18 - 15 - 3 = 0 and 27 - 24 - 3 = 0.

3. No common root. Subtraction gives r = 7/2, but the first expression then equals -3/4.

4. k = -3 or -9/2. The fixed quadratic has roots 1 and 4. Substitution gives 3 + k = 0 or 18 + 4k = 0. The second quadratic’s roots are 1, 2 in the first case and 4, 1/2 in the second.

5. One distinct common real root, 3. Both quadratics are multiples of (x - 3)². The shared root is repeated, but its numerical value is still just 3.

6. Common root 0. The remaining roots are -2 and 1/3. Dividing by x would discard the shared zero root.

7. No common real root. Over the complex numbers, both 3i and -3i are shared.

8. p + q + 1 = 0. Since p differs from q, subtraction forces r = 1. Substitution produces the condition, and the same condition verifies 1 in both equations.

Frequently asked questions

What is the fastest way to find a common root of two quadratic equations?

Usually, eliminate x² by subtraction, multiplying the equations first if necessary. Solve the resulting linear equation and check the candidate. If one quadratic already factorises easily, testing its roots may be faster.

Is the value obtained after subtraction always the common root?

No. Subtraction gives a necessary condition. The candidate might make both polynomial expressions equal to the same nonzero number. Substitute it into an original equation to establish that the value is actually zero.

Can two different quadratic equations have both roots common?

Yes. One equation can be a nonzero constant multiple of the other. They then have the same roots, including multiplicities. For example, x² - 5x + 6 = 0 and 2x² - 10x + 12 = 0 share 2 and 3.

Does a zero denominator in the common-root formula mean no common root?

Not automatically. Return to the linear elimination equation. A nonzero constant equated to zero means no common root. If the entire equation becomes 0 = 0, the quadratics are proportional and must be examined directly.

Can zero be a common root?

Yes. For two genuine quadratics, zero is common exactly when both constant terms are zero. Record that solution before dividing by x.

Does having a common root mean the discriminants are equal?

No. Discriminants describe the nature of each equation’s roots separately. Equations can share a root while having different discriminants, and equal discriminants do not guarantee any shared root.

Can two quadratics with real coefficients share exactly one non-real root?

No. Non-real roots occur in conjugate pairs for real-coefficient quadratics. If they share a non-real root, they also share its conjugate. Consequently, non-proportional real-coefficient quadratics cannot share a non-real root.

What is the difference between a repeated root and a common root?

A repeated root occurs twice within one quadratic, as in (x - 3)² = 0. A common root is shared between two equations. A value can be repeated in one equation and still be common to another.

Should I always use the general common-root condition?

No. Direct elimination or factorisation usually gives shorter working for numerical questions. Use the general condition when asked for a coefficient relation, and still check proportional cases and any requirement for real roots.

How do I avoid losing parameter values?

Before cancelling a factor involving a parameter, check when that factor is zero. Solve that branch separately, then continue with the nonzero branch. Finally, verify every proposed parameter value in the original equations.