Equations with radicals — squaring both sides, extraneous solutions: Worked examples

This page is the case zoo for the radical-equations topic. The parent note taught you why squaring can invent fake ("extraneous") solutions. Here we hunt down every kind of radical equation you can meet and solve one of each, so no exam scenario surprises you.

Before we start, three plain-word reminders (every symbol earned):

See Inverse Operations for why squaring "undoes" a square root, and Domain and Range of Functions for the " output is never negative" fact we keep using.

The scenario matrix

Every radical equation falls into one of these case classes. Our worked examples below cover all of them — the label after each example title says which cell it lands in.

#	Case class	What's special	Example
A	Clean solve, one root extraneous	squaring adds a stowaway	Ex 1
B	Both quadratic roots valid	danger zone ( $x < - 1$ ) stays empty	Ex 2
C	Right side always negative	no solution at all	Ex 3
D	Two radicals (square twice)	must isolate, then square again	Ex 4
E	Zero / degenerate input	$0 = 0$ boundary case	Ex 5
F	Cube root (index 3)	odd index — no extraneous!	Ex 6
G	Real-world word problem	units + reject negative	Ex 7
H	Exam twist: radical = radical, all extraneous	every candidate fails	Ex 8

Intuition The single picture behind all of it

Every case is the same drawing, shown below. The magenta curve is $y = x + 6$ : notice it starts at $x = - 6$ and never dips below the horizontal axis — that is the " is never negative" rule made visible. The violet line is $y = x$ (this is Example 1's equation, $x + 6 = x$ ). A real solution is where the magenta curve genuinely meets the violet line — the navy dot at $(3, 3)$ . Now watch the trap: squaring both sides also lets the curve match the orange dashed line $y = - x$ (the mirror of $y = x$ below the axis), and that produces the orange phantom crossing at $x = - 2$ — a value that satisfies the squared equation but not the original, because there $4 = + 2$ while $y = x$ demanded $- 2$ . So: magenta ∩ violet = real; magenta ∩ orange = extraneous. Keep this one picture in mind for every example.

Worked Examples

Worked example Example 1 — Clean solve, one root extraneous ·

Cell A Solve: $x + 6 = x$

Forecast: Guess now — will both candidates survive, or is one a stowaway?

Isolate. The radical is already alone on the left. ✓ Why this step? If a radical is not by itself, squaring makes a mess (the isolate-first rule above).
Square both sides. $(x + 6)^{2} = x^{2} \Rightarrow x + 6 = x^{2}$ . Why this step? Squaring cancels the root because $(a)^{2} = a$ for $a \geq 0$ .
Standard form. $0 = x^{2} - x - 6 = (x - 3) (x + 2)$ , so $x = 3$ or $x = - 2$ . Why this step? A product is zero only when a factor is zero — this hands us the candidates.
Check $x = 3$ : $3 + 6 = 9 = 3 = x$ ✓. Check $x = - 2$ : $- 2 + 6 = 4 = 2$ , but the right side is $x = - 2$ . $2 \neq = - 2$ ✗.

Verify: $x = 3$ is the only solution. $x = - 2$ is extraneous because the left side is $+ 2$ (non-negative) while the right side wanted $- 2$ . This is exactly the orange phantom crossing in the figure — a "clean" quadratic still hid one stowaway, so always check.

Worked example Example 2 — Both quadratic roots valid ·

Cell B Solve: $5 x + 1 = x + 1$

Forecast: The right side $x + 1$ is negative when $x < - 1$ — that's the danger zone. Bet on trouble there — or does it stay empty?

Isolate. Already isolated. ✓ Why this step? With the root alone as $5 x + 1 = x + 1$ , we can read off the danger zone ( $x + 1 < 0$ , i.e. $x < - 1$ ) and square cleanly; a root tangled with other terms would hide both.
Square. $5 x + 1 = (x + 1)^{2} = x^{2} + 2 x + 1$ . Why this step? Expand the binomial with $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ — the middle term $2 x$ matters.
Standard form. $0 = x^{2} + 2 x + 1 - 5 x - 1 = x^{2} - 3 x = x (x - 3)$ , so $x = 0$ or $x = 3$ . Why this step? Factoring exposes both candidates.
Check $x = 0$ : $5 (0) + 1 = 1 = 1$ and $0 + 1 = 1$ . $1 = 1$ ✓. Check $x = 3$ : $15 + 1 = 16 = 4$ and $3 + 1 = 4$ . $4 = 4$ ✓.

Verify: Both survive here! A stowaway would appear only if a candidate landed in the danger zone $x < - 1$ (where $x + 1 < 0$ ); neither $0$ nor $3$ did. Solutions: $x = 0$ and $x = 3$ . This is the mirror-image lesson of Cell A: not every squaring produces a stowaway — but you only know that after checking.

Worked example Example 4 — Two radicals, square twice ·

Cell D Solve: $2 x + 3 - x + 1 = 1$

Forecast: Two roots means we square, tidy, then square again. Guess how many candidates.

Isolate one radical. $2 x + 3 = 1 + x + 1$ . Why this step? Squaring works cleanest with a single radical alone on one side (isolate-first rule).
Square (first time). Left: $2 x + 3$ . Right: $(1 + x + 1)^{2} = 1 + 2 x + 1 + (x + 1)$ . So $2 x + 3 = x + 2 + 2 x + 1$ . Why this step? One root vanishes; one remains — that's expected.
Isolate the surviving radical. $x + 1 = 2 x + 1$ . Why this step? Set up for the second squaring.
Square (second time). $(x + 1)^{2} = 4 (x + 1) \Rightarrow (x + 1)^{2} - 4 (x + 1) = 0 \Rightarrow (x + 1) (x + 1 - 4) = 0$ . So $x = - 1$ or $x = 3$ . Why this step? A product equals zero only when a factor does.
Check $x = - 1$ : $2 (- 1) + 3 - - 1 + 1 = 1 - 0 = 1 - 0 = 1$ ✓. Check $x = 3$ : $6 + 3 - 3 + 1 = 9 - 4 = 3 - 2 = 1$ ✓.

Verify: Both $x = - 1$ and $x = 3$ are valid. Squaring twice doubled the extraneous risk, yet here both survived — proof that checking is not optional even when you square many times.

Worked example Example 6 — Cube root, index 3 ·

Cell F Solve: $3 2 x - 3 = - 3$

Forecast: The index is 3 (an odd number). Does the "never negative" rule apply?

Recognise the index. $3 a$ asks "which number cubed gives $a$ ?" Cubing keeps sign ( $(- 3)^{3} = - 27$ ), so a cube root can be negative. See Rational Exponents. Why this step? This tells us the sign trap of square roots does not exist here.
Cube both sides. $(3 2 x - 3)^{3} = (- 3)^{3} \Rightarrow 2 x - 3 = - 27$ . Why this step? Cubing is the exact inverse of the cube root — and it's reversible, so no stowaways.
Solve. $2 x = - 24 \Rightarrow x = - 12$ . Why this step? Add $3$ to both sides, then divide by $2$ — plain inverse operations to peel $x$ out of $2 x - 3$ .
Check $x = - 12$ : $3 2 (- 12) - 3 = 3 - 27 = - 3$ ✓.

Verify: $x = - 12$ is valid. Odd-index radicals never produce extraneous solutions because raising to an odd power is one-to-one — a crucial contrast with square roots.

Worked example Example 7 — Real-world word problem ·

Cell G Problem: A pendulum's period (seconds for one swing) is $T = 2 π \frac{L}{g}$ , with length $L$ in metres and $g = 9.8 m/s^{2}$ . For what length $L$ is the period exactly $T = 2 s$ ?

Forecast: Length must be positive — watch for a rejected negative.

Isolate the radical. Divide by $2 π$ : $\frac{T}{2 π} = \frac{L}{g}$ , i.e. $\frac{2}{2 π} = \frac{1}{π} = \frac{L}{9.8}$ . Why this step? Squaring is clean only with the root alone (isolate-first rule).
Square both sides. $(\frac{1}{π})^{2} = (\frac{L}{9.8})^{2}$ , giving $\frac{1}{π ^{2}} = \frac{L}{9.8}$ . Why this step? Removes the root; both sides are positive so no sign trap.
Where would a negative candidate come from? Squaring $L /9.8 = \frac{1}{π}$ forgot that the root could also equal $- \frac{1}{π}$ : the equation $\frac{1}{π ^{2}} = \frac{L}{9.8}$ is what you also get from $L /9.8 = - \frac{1}{π}$ . That branch corresponds to a negative period, which is physically impossible, so we discard it. The surviving branch keeps the root non-negative. Why this step? To show explicitly that the sign-erasing trap is present even in a word problem — we just reject the unphysical branch on sight.
Solve for $L$ . $L = \frac{9.8}{π ^{2}} \approx 0.9929 m$ . Why this step? Rearranging gives the physical length; $L > 0$ ✓ (a length must be positive).

Verify: Plug back: $T = 2 π 0.9929/9.8 = 2 π 0.10132 = 2 π (0.31831) \approx 2.00 s$ ✓. Units: metres in, seconds out — consistent. Answer: $L \approx 0.993 m$ .

Worked example Example 8 — Exam twist: two roots, ALL candidates extraneous ·

Cell H Solve: $x - 3 = 2 x + 1$

Forecast: Two square roots set equal. Guess: does any $x$ work?

Both already isolated. ✓ Why this step? Each side is a single radical standing alone, so squaring will strip both roots in one move — no expansion or cross terms to worry about (this is the isolate-first rule already satisfied on both sides).
Square both sides. $x - 3 = 2 x + 1$ . Why this step? Squaring removes both roots at once since each side is a single radical.
Solve. $- 3 - 1 = 2 x - x \Rightarrow x = - 4$ . Why this step? One linear equation, one candidate.
Check $x = - 4$ : inside left $= - 4 - 3 = - 7 < 0$ — $- 7$ is not a real number. ✗

Verify: No (real) solution. The single candidate $x = - 4$ violates the domain (you cannot take the real square root of a negative). This is the classic exam trap: even root-equals-root can leave you empty-handed. Compare with Absolute Value Equations, where $∣ a ∣ = ∣ b ∣$ likewise splits into cases you must test.

Recall Quick self-test

Q: Why must you always re-check every candidate in the original equation? A: Because squaring is a one-way street — the original being true forces the squared equation to be true, but not the reverse, so squaring can add stowaways.

Q: Why does an odd-index radical like a cube root never create extraneous solutions? A: Because cubing keeps sign and is reversible (one-to-one), so no information is lost and no fake candidates appear.

Q: How many solutions does $2 x + 5 = - 4$ have? A: Zero — a principal square root is never negative, so it can never equal $- 4$ .

Related vault topics: Quadratic Equations (the polynomial you land on after squaring) and Function Transformations (how $y = x$ shifts to $y = x + 6$ ).