JEE Mains Maths · Quadratic Equations

Equations with Modulus and Greatest Integer

Equations with |x| or [x] in them: split the line where each part changes form, solve an ordinary quadratic on each piece, and keep only the roots that lie in their piece.

All Quadratic Equations notes JEE Mains Maths strategy

Why this matters

Twenty-four PYQs, fifteen of them multiple choice, and five from 2026. Thirteen split the number line where each modulus changes sign and solve one quadratic per piece; eight take one modulus such as |x| or |x − 1| as the unknown, or use |A + B| = |A| + |B|; three mix [x] or {x} into a quadratic. Three ideas cover the page.

Concept 1 of 3: Splitting at the critical points

|x-a|

equals

x-a

to the right of

a

and

a-x

to the left. Mark every such point on the line. On each piece every modulus has a fixed sign, so the equation is an ordinary quadratic. Solve it, then keep a root only if it lies in its own piece. Most questions ask how many roots there are, so this last check decides the answer.

Definition

Critical points: where each expression inside a modulus is $0$ .
On each piece, drop the bars with the right sign and solve.
Keep a root only if it lies in its piece.
$x|x-a|$ is $x(x-a)$ for $x\ge a$ and $-x(x-a)$ for $x<a$ .

Definition of modulus

|x-a|=\begin{cases}x-a,& x\ge a\\ a-x,& x<a\end{cases}

Worked example

Solve

x^2-3|x-1|-1=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q70Moderate

Example 1 · Quadratic Equations · Equations with Modulus and Greatest Integer

The number of the real solutions of the equation :

x|x + 3| + |x - 1| - 2 = 0

A root on the boundary is counted once

A critical point can solve the equation in both neighbouring pieces. Count it once. And reject a root from a piece it does not lie in, even though it solves that piece's quadratic.

Drill 12 more on splitting at the critical points

Concept 2 of 3: One modulus as the unknown

x^2=|x|^2

, so

x^2-5|x|+6=0

is a quadratic in

t=|x|

: solve for

t\ge0

, then each positive

t

gives two values of

x

. The same works for

(x-1)^2

with

|x-1|

. A second shortcut:

|A+B|=|A|+|B|

holds exactly when

A

and

B

have the same sign, so such an equation becomes the inequality

AB\ge0

Definition

$x^2=|x|^2$ : put $t=|x|$ and keep roots $t\ge0$ .
$t>0$ gives $x=\pm t$ ; $t=0$ gives $x=0$ only.
$|A+B|=|A|+|B|$ exactly when $AB\ge0$ .
$|A|<k$ means $-k<A<k$ .

Equality in the triangle inequality

|A+B|=|A|+|B|\quad\text{exactly when}\quad AB\ge0

Worked example

Find the sum of the squares of the roots of

x^2-5|x|+4=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q52Moderate

Example 2 · Quadratic Equations · Equations with Modulus and Greatest Integer

If the set of all solutions of

\left| x^{2}+x- 9 \right|= |x| +\left| x^{2}- 9 \right|

\lbrack\alpha,\beta\rbrack \cup \lbrack\gamma,\infty)

, then

\left( \alpha^{2}+\beta^{2}+\gamma^{2} \right)

is equal to :

Reject negative values of |x|

The quadratic in

t=|x|

can have a negative root, and it gives no

x

at all. Count roots only from

t\ge0

, and remember that

t=0

gives one root, not two.

Drill 7 more on one modulus as the unknown

Concept 3 of 3: Greatest integer and fractional part

Write

x=[x]+\{x\}

, with

[x]

an integer and

0\le\{x\}<1

. Two routes. Factor the equation and see what

[x]

\{x\}

must equal — a factor that forces

\{x\}=3

gives nothing. Or put

[x]=n

, a constant: then

x

lies in

[n,n+1)

, the equation is a quadratic in

x

, and only roots inside that interval count.

Definition

$x=[x]+\{x\}$ , $[x]\in\mathbb Z$ , $0\le\{x\}<1$ .
Put $[x]=n$ : then $n\le x<n+1$ , and the equation is a quadratic in $x$ .
An equation forcing $\{x\}\ge1$ or $\{x\}<0$ has no solution.

Greatest integer

[x]=n\quad\text{exactly when}\quad n\le x<n+1

Worked example

Solve

x^2-3[x]-4=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q72Moderate

Example 3 · Quadratic Equations · Equations with Modulus and Greatest Integer

The equation

x^{2}- 4x + \lbrack x\rbrack + 3 = x\lbrack x\rbrack

, where

\lbrack x\rbrack

denotes the greatest integer function, has :

The fractional part is below 1

\{x\}

is never

1

or more and never negative. A factor that forces

\{x\}=3

, or any value outside

[0,1)

, gives no root.

Drill 2 more on greatest integer and fractional part

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 (3)

Splitting at the critical points
Definition of modulus
$|x-a|=\begin{cases}x-a,& x\ge a\\ a-x,& x<a\end{cases}$
One modulus as the unknown
Equality in the triangle inequality
$|A+B|=|A|+|B|\quad\text{exactly when}\quad AB\ge0$
Greatest integer and fractional part
Greatest integer
$[x]=n\quad\text{exactly when}\quad n\le x<n+1$

Watch out for (3)

A root on the boundary is counted once
→ Splitting at the critical points
Reject negative values of |x|
→ One modulus as the unknown
The fractional part is below 1
→ Greatest integer and fractional part

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 24 questions