Equations with radicals — squaring both sides, extraneous solutions: Exercises

This page is a self-testing ladder. Every problem sits above a collapsible solution — try it first, then reveal. The levels climb from "spot the radical" to "invent your own trap".

Everything here builds on the parent: Equations with radicals. Before you start, hold onto the two truths that run through every problem below.

Reminder of tools you'll reuse: Inverse Operations (squaring undoes rooting only on the non-negative branch), Quadratic Equations (the squared equation is usually a quadratic), Domain and Range of Functions (the sign gate is a domain restriction), and Absolute Value Equations (because $x^{2} = ∣ x ∣$ , not $x$ ).

Level 1 — Recognition

L1.1 Solve $x - 2 = 3$ .

Recall Solution

Radical already isolated. Square both sides — squaring is chosen because it is the inverse operation to the square root, cancelling it cleanly: $x - 2 = 9 ⟹ x = 11.$ Sign gate: right side $3 \geq 0$ , so no automatic rejection. Check in the original: $11 - 2 = 9 = 3 ✓$ Answer: $x = 11$ .

L1.2 Solve $5 x + 1 = - 4$ .

Recall Solution

Stop before squaring. The left side is a principal root, so $5 x + 1 \geq 0$ always, but the right side is $- 4 < 0$ . A non-negative thing can never equal a negative thing. Answer: No solution. (If you squared anyway you'd get $5 x + 1 = 16$ , $x = 3$ , but $16 = 4 \neq = - 4$ — a pure extraneous candidate born from erasing the sign.)

L1.3 Which of these has a guaranteed extraneous candidate before you do any algebra: (a) $x = 7$ , (b) $x = - 1$ , (c) $x = x$ ?

Recall Solution

(b), because the right side is negative and a principal root cannot be negative — the sign gate is already shut. (a) is fine ( $7 \geq 0$ ). (c) may or may not (depends on $x$ ), so it is not guaranteed. Answer: (b).

Level 2 — Application

L2.1 Solve $x + 3 = 7$ .

Recall Solution

Isolate first (squaring $(x + 3)$ would create a cross term $6 x$ — see the L2 trap). Subtract $3$ : $x = 4 ⟹ x = 16.$ Check: $16 + 3 = 4 + 3 = 7 ✓$ . Answer: $x = 16$ .

L2.2 Solve $2 x + 3 = 5$ .

Recall Solution

Already isolated, right side $5 \geq 0$ . Square: $2 x + 3 = 25 ⟹ 2 x = 22 ⟹ x = 11.$ Check: $2 (11) + 3 = 25 = 5 ✓$ . Answer: $x = 11$ .

L2.3 Solve $3 x - 1 = 12$ .

Recall Solution

Isolate the radical by dividing by $3$ (do this before squaring, or the coefficient gets squared too): $x - 1 = 4 ⟹ x - 1 = 16 ⟹ x = 17.$ Check: $3 17 - 1 = 3 \cdot 4 = 12 ✓$ . Answer: $x = 17$ .

Level 3 — Analysis

L3.1 Solve $x = x - 6$ and state which root (if any) is extraneous.

Recall Solution

Sign gate: need $x - 6 \geq 0$ , i.e. $x \geq 6$ — remember this. Square: $x = (x - 6)^{2} = x^{2} - 12 x + 36 ⟹ x^{2} - 13 x + 36 = 0 ⟹ (x - 4) (x - 9) = 0.$ Candidates $x = 4, 9$ . The gate $x \geq 6$ already rejects $x = 4$ . Confirm both:

$x = 4$ : $4 = 2$ but $4 - 6 = - 2$ , $2 \neq = - 2$ ✗ extraneous.
$x = 9$ : $9 = 3$ and $9 - 6 = 3$ ✓.

Answer: $x = 9$ ; $x = 4$ is extraneous.

L3.2 Solve $x = x - 2$ .

Recall Solution

Gate: $x \geq 2$ . Square: $x = x^{2} - 4 x + 4 ⟹ x^{2} - 5 x + 4 = 0 ⟹ (x - 1) (x - 4) = 0$ . Candidates $1, 4$ .

$x = 1$ : $1 = 1$ but $1 - 2 = - 1$ ✗ (and violates gate $x \geq 2$ ).
$x = 4$ : $4 = 2$ and $4 - 2 = 2$ ✓.

Answer: $x = 4$ ; $x = 1$ extraneous.

L3.3 Solve $2 x - 1 = x - 2$ .

Recall Solution

Gate: $x \geq 2$ . Square: $2 x - 1 = (x - 2)^{2} = x^{2} - 4 x + 4 ⟹ x^{2} - 6 x + 5 = 0 ⟹ (x - 1) (x - 5) = 0$ . Candidates $1, 5$ .

$x = 1$ : fails gate $x \geq 2$ and $1 = 1 \neq = - 1$ ✗.
$x = 5$ : $9 = 3$ , $5 - 2 = 3$ ✓.

Answer: $x = 5$ ; $x = 1$ extraneous.

The picture below shows why $x = 4$ survives and $x = 1$ dies in L3.2 — the parabola $y = x - 2$ from squaring crosses $y = x$ at one real point, but the reflected candidate lands on the discarded branch.

Level 4 — Synthesis

L4.1 Solve $x + 7 = 1 + x$ .

Recall Solution

One radical is already alone on the left; the right side is a sum $\geq 0$ , gate satisfied for all $x \geq 0$ . Square both sides, expanding the right with $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ : $x + 7 = 1 + 2 x + x .$ Subtract $x$ : $7 = 1 + 2 x ⟹ 6 = 2 x ⟹ x = 3 ⟹ x = 9.$ Check: $16 = 1 + 9 \Rightarrow 4 = 4$ ✓. Answer: $x = 9$ .

L4.2 Solve $x + 5 - x = 1$ .

Recall Solution

Isolate one radical first: $x + 5 = 1 + x$ . Square: $x + 5 = 1 + 2 x + x ⟹ 4 = 2 x ⟹ x = 2 ⟹ x = 4.$ Check: $9 - 4 = 3 - 2 = 1$ ✓. Answer: $x = 4$ .

L4.3 Solve $3 x + 1 - x - 1 = 2$ .

Recall Solution

Isolate: $3 x + 1 = 2 + x - 1$ . Square: $3 x + 1 = 4 + 4 x - 1 + (x - 1) ⟹ 3 x + 1 = x + 3 + 4 x - 1 .$ Collect the surviving radical: $2 x - 2 = 4 x - 1 ⟹ x - 1 = 2 x - 1$ . Let $u = x - 1 (u \geq 0)$ , so $u^{2} = 2 u ⟹ u (u - 2) = 0 ⟹ u = 0$ or $u = 2$ .

$u = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1$ : check $4 - 0 = 2 - 0 = 2$ ✓.
$u = 2 \Rightarrow x - 1 = 4 \Rightarrow x = 5$ : check $16 - 4 = 4 - 2 = 2$ ✓.

Answer: both $x = 1$ and $x = 5$ are valid (a two-solution radical equation).

Level 5 — Mastery

L5.1 Find all $x$ with $x^{2} - 3 x = x - 3$ , using $t^{2} = ∣ t ∣$ where needed.

Recall Solution

Gate: $x - 3 \geq 0 \Rightarrow x \geq 3$ . Square: $x^{2} - 3 x = (x - 3)^{2} = x^{2} - 6 x + 9$ . The $x^{2}$ cancels: $- 3 x = - 6 x + 9 ⟹ 3 x = 9 ⟹ x = 3.$ Gate $x \geq 3$ allows it. Check: $9 - 9 = 0 = 0$ and $3 - 3 = 0$ ✓. Answer: $x = 3$ (a boundary root — the gate is equality here, so keep it).

L5.2 For which real values of $k$ does $x = x - k$ have exactly one valid solution? (Treat $k$ as a parameter.)

Recall Solution

Square: $x = (x - k)^{2} \Rightarrow x^{2} - (2 k + 1) x + k^{2} = 0$ . Discriminant of this quadratic: $Δ = (2 k + 1)^{2} - 4 k^{2} = 4 k + 1.$ Real candidates exist only when $4 k + 1 \geq 0 \Rightarrow k \geq - \frac{1}{4}$ . Roots: $x = \frac{( 2 k + 1 ) \pm 4 k + 1}{2}$ . Now apply the sign gate $x - k \geq 0 \Rightarrow x \geq k$ , plus $x \geq 0$ .

When $Δ = 0$ (i.e. $k = - \frac{1}{4}$ ): one double root $x = \frac{2 k + 1}{2} = \frac{1}{4}$ . Check: $1/4 = 1/2$ , $x - k = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ ✓. One solution.
When $k \geq 0$ : the larger root passes the gate and the smaller one fails it (it lands below $k$ ), giving exactly one valid solution. (E.g. $k = 6$ gave the single answer $x = 9$ in L3.1.)
When $- \frac{1}{4} < k < 0$ : both roots can pass, giving two solutions.

Answer: exactly one valid solution when $k = - \frac{1}{4}$ or $k \geq 0$ .

L5.3 (Design your own trap.) Construct a radical equation of the form $x + a = x + b$ that produces two algebraic candidates but only one valid solution, and prove it does.

Recall Solution

Choose $a = 2, b = - 4$ : equation $x + 2 = x - 4$ . Gate: $x \geq 4$ . Square: $x + 2 = (x - 4)^{2} = x^{2} - 8 x + 16 \Rightarrow x^{2} - 9 x + 14 = 0 \Rightarrow (x - 2) (x - 7) = 0$ . Two candidates $2, 7$ .

$x = 2$ : fails gate $x \geq 4$ ; check $4 = 2$ but $2 - 4 = - 2$ ✗ extraneous.
$x = 7$ : $9 = 3$ , $7 - 4 = 3$ ✓.

Exactly one valid solution ( $x = 7$ ), one extraneous ( $x = 2$ ). Trap constructed and verified.

Recall Rapid self-check (cloze)

The reverse of " $a = b \Rightarrow a^{2} = b^{2}$ " only gives ::: $a = \pm b$ Before substituting, the sign gate for $f (x) = g (x)$ is ::: $g (x) \geq 0$ $t^{2}$ equals ::: $∣ t ∣$ , not $t$ In $x + 2 = x - 4$ the extraneous root is ::: $x = 2$ The rule for two-radical equations is ::: isolate one radical, square, repeat