JEE Mains Maths · Quadratic Equations

Equations Reducible to Quadratics

Exponential, logarithmic and other equations that turn into a quadratic after one substitution, where the range of the substitution decides which roots survive.

All Quadratic Equations notes JEE Mains Maths strategy

Why this matters

Twenty-six PYQs, sixteen of them multiple choice, and three from 2026. Ten are exponential — put t = eˣ or t = aˣ, often ending in t + 1/t; seven are logarithmic, where taking logs or changing the base gives a quadratic and the domain removes some roots; nine find a quadratic hidden elsewhere — a repeated block such as x² − 9x or x + 1/x, square roots, a continued fraction, or a polynomial that factors. Three ideas cover the page.

Concept 1 of 3: Exponential equations

Put

t=a^x

. Then

a^{2x}=t^2

, and every

t

is positive. Two powers whose bases multiply to 1 are

t

and

\frac1t

, so their sum gives

t+\frac1t=k

, a quadratic. A quartic in

t

whose coefficients read the same both ways divides by

t^2

into a quadratic in

u=t+\frac1t

— and for

t>0

u\ge2

, which removes roots.

Definition

$t=a^x>0$ : keep only the positive roots in $t$ .
Each positive $t$ gives exactly one $x=\log_a t$ .
$t+\frac1t=k$ with $t>0$ has roots only when $k\ge2$ : two if $k>2$ , one if $k=2$ .
$t^4+pt^3+qt^2+pt+1=0$ : divide by $t^2$ and put $u=t+\frac1t$ .

Reciprocal pair

t+\frac1t=k\ \Rightarrow\ t=\frac{k\pm\sqrt{k^2-4}}{2}

Worked example

How many real roots has

e^{4x}-3e^{3x}+4e^{2x}-3e^x+1=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q69Moderate

Example 1 · Quadratic Equations · Equations Reducible to Quadratics

Let

S =\left\{ x\in R:(\sqrt{3}+\sqrt{2})^{x}+ (\sqrt{3}-\sqrt{2})^{x}= 10 \right\}

. Then the number of elements in

S

is :

Every t must be positive

a^x

is never

0

or negative, so such a root in

t

gives no

x

. The sum of the

x

-roots is the log of the product of the valid

t

-roots only — not of every root of the polynomial in

t

Drill 9 more on exponential equations

Concept 2 of 3: Logarithmic equations

Two routes. If the unknown sits both in a power and in a log, as in

x^{\log_3 x}

, take logs so the log itself becomes the unknown

t

. If two logs have swapped bases,

\log_a b

and

\log_b a

, they are

y

and

\frac1y

, and the equation becomes

y+\frac cy=k

, a quadratic. Either way, finish with the domain: every argument positive, every base positive and not 1.

Definition

$\log_b a=\frac{1}{\log_a b}$ ; put $y=\log_a b$ .
$\log_{a^k}x=\frac1k\log_a x$ .
Unknown in a power and a log: take logs, put $t=\log x$ .
Domain: arguments $>0$ ; bases $>0$ and $\ne1$ .

Swapped bases

\log_b a=\frac{1}{\log_a b}

Worked example

Solve

\log_2 x+2\log_x 4=5

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q57Moderate

Example 2 · Quadratic Equations · Equations Reducible to Quadratics

The sum of all the real solutions of the equation

\log_{(x + 3)}\left( 6x^{2}+ 28x + 30 \right)= 5 - 2\log_{(6x + 10)}\left( x^{2}+ 6x + 9 \right)

is equal to :

Check every base

A root can make a base negative or equal to 1 even when every argument is positive. Test each root in each base and each argument before counting it.

Drill 6 more on logarithmic equations

Concept 3 of 3: A repeated expression as the unknown

Look for a block that appears twice:

x^2+5x

(x^2+5x+4)(x^2+5x+6)

x+\frac1x

inside

x^2+\frac1{x^2}

\sqrt x

inside

x

. Call it

t

, solve the quadratic in

t

, then undo

t

— each block has its own range. A continued fraction that repeats contains a copy of itself, so its value satisfies a quadratic.

Definition

Spot a repeated block and call it $t$ .
$t=x+\frac1x$ : $x^2+\frac1{x^2}=t^2-2$ , and real $x$ needs $|t|\ge2$ .
$t=\sqrt x$ or $t=x^2$ : keep $t\ge0$ .
A repeating continued fraction $y=a+\frac{1}{b+\frac1y}$ gives a quadratic in $y$ ; keep the positive root.

Reciprocal substitution

x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2

Worked example

Solve

(x+1)(x+2)(x+3)(x+4)=24

over the reals.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q67Moderate

Example 3 · Quadratic Equations · Equations Reducible to Quadratics

The product of all the rational roots of the equation

\left( x^{2}- 9x + 11 \right)^{2}- (x - 4)(x - 5) = 3

, is equal to :

Undo the substitution with its range

Each root in

t

must be a value its block can take.

x+\frac1x

never lies strictly between

-2

and

2

, and

\sqrt x

and

x^2

are never negative; such roots give no

x

Drill 8 more on a repeated expression as the unknown

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)

Exponential equations
Reciprocal pair
$t+\frac1t=k\ \Rightarrow\ t=\frac{k\pm\sqrt{k^2-4}}{2}$
Logarithmic equations
Swapped bases
$\log_b a=\frac{1}{\log_a b}$
A repeated expression as the unknown
Reciprocal substitution
$x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2$

Watch out for (3)

Every t must be positive
→ Exponential equations
Check every base
→ Logarithmic equations
Undo the substitution with its range
→ A repeated expression as the unknown

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 26 questions