Lesson 10: Radical Functions and Equations

A radical function involves a root: f(x) = √x, g(x) = ∛x, and their transformations. Radical equations — where the variable lives under the root — introduce the single most-tested caution in Algebra II: squaring both sides can manufacture extraneous solutions, answers that satisfy the squared equation but not the original. This lesson graphs the functions, solves the equations, and builds the checking habit that makes extraneous roots harmless.

The square root and cube root functions

f(x) = √x starts at (0, 0) and rises with decreasing slope; its domain and range are both x ≥ 0, y ≥ 0. It is the inverse of x^2 restricted to x ≥ 0 (Lesson 2): reflect y = x^2 (right half) in the line y = x and you get y = √x. Transformations shift the starting point: y = √(x − 3) + 1 starts at (3, 1), with domain x ≥ 3 and range y ≥ 1.

(Notice the pattern: perfect squares 0, 1, 4, 9, 16 under the root occur at x = 3, 4, 7, 12, 19 — graph key points by choosing inputs that make the radicand a perfect square.) The cube root function y = ∛x is altogether more relaxed: defined for all real x, with range all reals, passing through (−8, −2), (0, 0), (8, 2), and rotationally symmetric about the origin. Odd roots accept negative inputs; even roots do not. That one distinction drives everything below.

Solving radical equations — and the mandatory check

The method: isolate the radical, raise both sides to the matching power, solve the resulting polynomial equation, then check every candidate in the original equation.

Worked Example 1 — the classic extraneous root. Solve √(x + 7) = x − 5. Before squaring, note the right side must be non-negative (it equals a square root), so any solution needs x ≥ 5. Squaring: x + 7 = x^2 − 10x + 25, so x^2 − 11x + 18 = 0 = (x − 9)(x − 2), candidates x = 9 and x = 2. Check: x = 9 gives √16 = 4 and 9 − 5 = 4 ✓. x = 2 gives √9 = 3 but 2 − 5 = −3 ✗. Squaring turned the equation into one that also admits the "solution" of √(x + 7) = −(x − 5) — the sign information was destroyed. Only x = 9 solves the equation.

Worked Example 2 — two radicals. Solve √(x + 5) − √(x − 3) = 2. Isolate one radical: √(x + 5) = 2 + √(x − 3). Square: x + 5 = 4 + 4√(x − 3) + (x − 3), which simplifies to 4 = 4√(x − 3), so √(x − 3) = 1 and x − 3 = 1, giving x = 4. Check in the original: √9 − √1 = 3 − 1 = 2 ✓. Domain note: the equation only makes sense for x ≥ 3, and x = 4 complies.

Worked Example 3 — a cube root equation. Solve ∛(2x + 1) = 3. Cube both sides: 2x + 1 = 27, so x = 13. No extraneous root is possible: cubing is a one-to-one operation on the reals (unlike squaring, it preserves sign), and the check confirms ∛27 = 3 ✓. Extraneous solutions are a disease of even powers and even roots.

Modelling: pendulums, skid marks and horizons

A pendulum's period is T = 2π√(L/g), with L the length in metres and g = 9.8 m/s^2. Solving for the length that gives the stately 2-second swing of a grandfather clock: L = g(T/(2π))^2 = 9.8 · (1/π)^2 ≈ 0.993 m — almost exactly a metre, which is no coincidence: the metre was nearly defined this way. Note the algebra: isolating L required squaring T/(2π), an inverse operation, not a source of extraneous roots here because lengths and periods are positive by context.

Police investigators estimate a car's speed from its skid marks with v = √(30·d·f), where d is the skid length in feet and f the road's friction coefficient. For d = 100 ft on dry asphalt (f = 0.7): v = √(30 · 100 · 0.7) = √2100 ≈ 45.8 mph. Inverting the model answers the courtroom question — how long would the skid be at the 30 mph limit? d = v^2/(30f) = 900/21 ≈ 42.9 ft. A 100 ft skid is mathematically incompatible with a 30 mph claim, and the radical function is the witness.

Common mistakes

• Skipping the check. The single most penalised habit in this topic. Worked Example 1's x = 2 looks algebraic and is simply wrong.
• Splitting roots over sums. √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7. Roots distribute over products and quotients, never over sums or differences.
• Writing √(x^2) = x. √(x^2) = |x|: at x = −3, √9 = 3, not −3. The principal root is never negative.
• Squaring a sum incorrectly. (√(x − 3) + 2)^2 is not (x − 3) + 4; it is (x − 3) + 4√(x − 3) + 4 — the cross term is where the remaining radical hides (Worked Example 2).
• Imposing even-root fears on odd roots. ∛(x − 5) needs no domain restriction and cubing creates no extraneous roots; conversely √(x) = −4 has no solution at all, immediately, since a principal root is never negative.

Exam Experience

1. The solution set of √(2x + 3) = 5 is

• (A) {11}
• (B) {14}
• (C) {11, −14}
• (D) {4}

Answer: (A). Square: 2x + 3 = 25, so x = 11; check √25 = 5 ✓. (B) forgets the +3 when isolating... precisely: 2x = 28 comes from adding 3 to the wrong side's arithmetic (25 + 3), a sign slip in rearrangement.

2. The domain of f(x) = √(5 − x) + 2 is

• (A) x ≥ 5
• (B) x ≤ 5
• (C) x ≥ −5
• (D) all real numbers

Answer: (B). The radicand needs 5 − x ≥ 0, i.e. x ≤ 5 — the inequality flips when isolating x, which is exactly the step (A) gets wrong.

3. Consider √(2x + 1) = x − 1. Squaring gives x = 0 and x = 4 as candidates. The true solution set is

• (A) {0, 4}
• (B) {0}
• (C) {4}
• (D) { } (empty)

Answer: (C). Check x = 0: √1 = 1 but 0 − 1 = −1 ✗ extraneous. Check x = 4: √9 = 3 and 4 − 1 = 3 ✓.

Constructed response. The distance to the horizon from a height h feet above the sea is approximately d = 1.22√h miles. (a) A lighthouse lantern is 36 ft above the water. How far can its light be seen? (b) A ship's lookout must spot the coast 20 miles away. At what minimum height must the lookout stand? Give your answer to the nearest foot. (c) A student claims that doubling the height doubles the horizon distance. Use the structure of the formula to correct the claim precisely.

Scoring guide. (a) 1 point: d = 1.22√36 = 1.22 · 6 = 7.32 miles. (b) 2 points: 1 for isolating the root, √h = 20/1.22 ≈ 16.39; 1 for squaring to get h ≈ 268.7, i.e. about 269 ft. (c) 2 points: 1 for identifying the relationship as proportional to √h, not h; 1 for the precise correction — doubling h multiplies d by √2 ≈ 1.41 (a 41% increase); to double the distance you must quadruple the height.

Going deeper: rational exponents — radicals in exponential clothing

Radicals and fractional powers are the same objects: x^(1/n) = ⁿ√x and x^(m/n) = (ⁿ√x)^m = ⁿ√(x^m). The exponent laws then do the work that radical rules did. Simplify 16^(3/4): 16^(1/4) = 2, then 2^3 = 8 (take the root first — the numbers stay small; raising first gives 16^3 = 4,096 and a miserable fourth root). Similarly 27^(2/3) = 3^2 = 9, and (8x^6)^(1/3) = 2x^2 by applying the power to each factor.

Equations in rational exponents solve by raising to the reciprocal power, with Lesson 10's checks still in force for even denominators. Solve x^(3/2) = 8: raise both sides to the 2/3 power: x = 8^(2/3) = (∛8)^2 = 2^2 = 4. Check: 4^(3/2) = (√4)^3 = 2^3 = 8 ✓. Solve x^(2/3) = 9: ∛x = ±3 (because the square in (∛x)^2 erases sign), so x = ±27 — both check: (∛(−27))^2 = (−3)^2 = 9 ✓. The two answers surprise students who expect the ± only from square roots; the structure (an even power applied last) is what creates the pair.

Fractional exponents also close the loop with Lesson 5: the graph of y = x^(2/3) is the cube-root graph squared — defined for all x, symmetric about the y-axis, with a cusp at the origin — and expressions like a^(1/2) · a^(1/3) = a^(5/6) obey the product law of exponents, which is precisely why the notation was invented: one rule-set for integers, fractions and (eventually) all reals.

One more extraneous root, caught in the wild

Solve √(5x − 4) = x − 2. The right side must be non-negative, so candidates need x ≥ 2. Square: 5x − 4 = x^2 − 4x + 4, hence x^2 − 9x + 8 = 0 = (x − 1)(x − 8). Candidate x = 1 fails the x ≥ 2 screen before any substitution — and indeed √1 = 1 ≠ −1. Candidate x = 8: √36 = 6 and 8 − 2 = 6 ✓. Solution: x = 8 only. Notice the efficiency: the sign screen (RHS ≥ 0) rejected the impostor faster than the check would have. Use both defences — the screen while solving, the check after.