JEE Mains Maths · Quadratic Equations

Common Roots and New Equations

Two equations that share a root, and new equations whose roots are built from the roots of an old one.

All Quadratic Equations notes JEE Mains Maths strategy

Why this matters

Thirteen PYQs, eight of them multiple choice, and two from 2026. Eight have two equations sharing a root — eliminate the square term to find it, or, when the roots are not real, match the coefficients; five build an equation — four from new roots made out of old ones, one from a work problem. Two ideas cover the page.

Concept 1 of 2: A root shared by two equations

\alpha

satisfies both equations, scale them so the

x^2

terms match and subtract. What is left is linear in

\alpha

, so the common root comes out directly; the sum or product in each equation then gives its other root. If one equation has non-real roots and both have real coefficients, sharing one root means sharing both, because non-real roots come in conjugate pairs — so the coefficients are proportional.

Definition

Eliminate $x^2$ (or the constant) between the two equations; the linear equation left gives $\alpha$ .
Then Vieta on each equation gives the other roots.
Non-real roots, real coefficients: both roots are shared, so $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ .

Condition for one common root

(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)

Worked example

For which

k\ne0

x^2-3x+k=0

and

x^2-5x+2k=0

share a root?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q76Moderate

Example 1 · Quadratic Equations · Common Roots and New Equations

Let

\lambda \neq 0

be a real number. Let

\alpha,\beta

be the roots of the equation

14x^{2}- 31x + 3\lambda = 0

and

\alpha,\gamma

be the roots of the equation

35x^{2}- 53x + 4\lambda = 0

. Then

\frac{3\alpha}{\beta}

and

\frac{4\alpha}{\gamma}

are the roots of the equation

Proportional only for non-real roots

Two equations with real roots can share one root and not the other. Making the coefficients proportional then gives a wrong answer. Use proportionality only when one equation's roots are non-real.

Drill 7 more on a root shared by two equations

Concept 2 of 2: Building an equation from its roots

A quadratic with leading coefficient 1 is fixed by its roots:

x^2-(\text{sum})x+\text{product}=0

. When the new roots are made from old ones, find their sum and product from

\alpha+\beta

and

\alpha\beta

— never from

\alpha

and

\beta

separately, unless they are easy to find. Some changes have shortcuts: reciprocal roots reverse the coefficients, and scaling or shifting the roots changes

x

in the old equation.

Definition

Roots $r_1,r_2$ : $x^2-(r_1+r_2)x+r_1r_2=0$ .
Roots $\frac1\alpha,\frac1\beta$ : reverse the coefficients, $cx^2+bx+a=0$ .
Roots $k\alpha,k\beta$ : replace $x$ by $\frac xk$ .
Roots $\alpha+h,\beta+h$ : replace $x$ by $x-h$ .

Equation from its roots

x^2-(r_1+r_2)x+r_1r_2=0

Worked example

The roots of

x^2-5x+2=0

are

\alpha,\beta

. Find the equation with roots

\alpha+\frac1\beta

and

\beta+\frac1\alpha

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q51Moderate

Example 2 · Quadratic Equations · Common Roots and New Equations

Let

\alpha,\beta

be the roots of the equation

x^{2}- 3x+r= 0

, and

\frac{\alpha}{2},2\beta

be the roots of the equation

x^{2}+ 3x + r = 0

. If the roots of the equation

x^{2}+ 6x=m

are

2\alpha+\beta+ 2r

and

\alpha- 2\beta-\frac{r}{2}

, then m is equal to :-

Divide by the leading coefficient

For

ax^2+bx+c=0

the sum of the roots is

-\frac ba

, not

-b

. Read the sum and product only after dividing by

a

, and clear fractions at the end so the answer matches the options.

Drill 4 more on building an equation from its roots

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

A root shared by two equations
Condition for one common root
$(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$
Building an equation from its roots
Equation from its roots
$x^2-(r_1+r_2)x+r_1r_2=0$

Watch out for (2)

Proportional only for non-real roots
→ A root shared by two equations
Divide by the leading coefficient
→ Building an equation from its roots

Test yourself on Quadratic Equations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Quadratic Equations chapter test Drill all 13 questions