CDS Mathematics · Quadratic Equations

Equations Reducible to Quadratics

Equations in x², in x − 1/x, with square roots or with fractions become quadratics after a substitution or after clearing — and every root must then be checked in the original.

All Quadratic Equations notes CDS Mathematics strategy

Why this matters

Six PYQs, three of them HARD. The substitution is always one of a few — u = x², t = x ± 1/x — and the check at the end is not optional: squaring can add a root that the original equation does not have.

Concept 1 of 2: Substitute to get a quadratic

An equation that uses only

x^2

and

x^4

is a quadratic in

u = x^2

. One whose terms pair as

x^2 + \dfrac{1}{x^2}

and

x \pm \dfrac1x

is a quadratic in

t = x \pm \dfrac1x

Definition

Biquadratic: $ax^4 + bx^2 + c = 0$ with $u = x^2$ ; keep only $u \ge 0$ .
Reciprocal pairs: group as $A\left(x^2 + \dfrac{1}{x^2}\right) + B\left(x - \dfrac1x\right) + C = 0$ and put $t = x - \dfrac1x$ , so $x^2 + \dfrac{1}{x^2} = t^2 + 2$ .
With $t = x + \dfrac1x$ , use $x^2 + \dfrac{1}{x^2} = t^2 - 2$ .
If $u^2 = u + 1$ , higher powers reduce: $u^2 = u + 1$ , $u^3 = 2u + 1$ .

Reciprocal substitution

t = x - \tfrac1x \;\Rightarrow\; x^2 + \tfrac{1}{x^2} = t^2 + 2

Worked example

Solve

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (II) 2025 — Elementary Mathematics · Q13Moderate

Example 1 · Quadratic Equations · Equations Reducible to Quadratics

x^4=x^2+1

, where

x>0

, then what is

2x^4

equal to?

Some roots of the new quadratic give no real x

u = x^2

must be non-negative, and

t = x + \dfrac1x

must satisfy

|t| \ge 2

. A root of the

u

t

equation outside that range is dropped.

Drill 2 more on substitute to get a quadratic

Concept 2 of 2: Clear roots and fractions, then check

Squaring both sides removes a square root but also accepts solutions of the equation with the opposite sign. Clearing fractions is safe, but a root that makes a denominator zero must be thrown out. So check every candidate in the original.

Definition

$\sqrt{f(x)} = g(x)$ : square to get $f(x) = g(x)^2$ , then keep only roots with $g(x) \ge 0$ .
For fractions, multiply by the common denominator, solve, and reject roots that make a denominator zero.
$x = 0$ may satisfy a rearranged form trivially; the question usually wants the non-zero roots.

Squaring a root equation

\sqrt{f(x)} = g(x) \iff f(x) = g(x)^2 \text{ and } g(x) \ge 0

Worked example

Solve

\sqrt{x + 7} = x - 5

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (II) 2023 — Elementary Mathematics · Q11Moderate

Example 2 · Quadratic Equations · Equations Reducible to Quadratics

How many real roots does the equation

\sqrt{x + 9} = x - 3

have ?

Squaring adds roots

Both

\sqrt{f} = g

and

\sqrt{f} = -g

square to the same equation. Every root of the squared equation must be tried in the original; count only those that pass.

Drill 2 more on clear roots and fractions, then check

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)

Substitute to get a quadratic
Reciprocal substitution
$t = x - \tfrac1x \;\Rightarrow\; x^2 + \tfrac{1}{x^2} = t^2 + 2$
Clear roots and fractions, then check
Squaring a root equation
$\sqrt{f(x)} = g(x) \iff f(x) = g(x)^2 \text{ and } g(x) \ge 0$

Watch out for (2)

Some roots of the new quadratic give no real x
→ Substitute to get a quadratic
Squaring adds roots
→ Clear roots and fractions, then check

Test yourself on Quadratic Equations

20 past CDS questions from this chapter, timed at 24 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 6 questions