7.2 Solving Radical Equations and Extraneous Roots

7.2 Solving Radical Equations and Extraneous Roots

Simplifying a radical is not the same as solving a radical equation. Table 11’s Radical and rational equations slice includes creating and solving those equations. On AAF, a typical item isolates a square root, squares both sides, and then — the step that separates the 237–249 radical-equations band from guessing — checks every candidate in the original equation. Squaring is not reversible. It can manufacture extraneous roots.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Domain before algebra

For √(g(x)) to be a real number, g(x) ≥ 0. For ⁿ√(g(x)) with even n, the same restriction holds. Cube roots have no such restriction: ∛(x − 1) is defined for every real x.

Write the domain before you square. Candidates that solve the squared equation but miss the domain are discarded. Candidates that sit in the domain still have to satisfy the original equation, because squaring can turn a false statement into a true one.

The principal (AAF) square root never returns a negative. If, after isolating, you have √(something) = a negative number, stop. There is no real solution. Squaring anyway wastes time and produces an extraneous number you will have to throw away.

Isolate, then square

A reliable sequence:

1. State the domain of every even-root radicand.
2. Isolate one radical on one side (add or subtract first; then divide if a coefficient sits outside the radical).
3. Square both sides (or cube, if the isolated root is a cube root).
4. Solve the resulting linear or quadratic equation.
5. Check every candidate in the original equation. Keep only the values that work.

If a coefficient is outside, isolate that too: 2√(x + 1) = 10 becomes √(x + 1) = 5 before squaring. Squaring too early on 2√(x + 1) = 10 happens to survive because both sides are nonnegative, but 2√(x + 1) + 3 = 7 squared too early expands into a mess that hides the simple √(x + 1) = 2.

Worked example: √(2x + 3) + 1 = 6

Isolate: √(2x + 3) = 5.

Domain: 2x + 3 ≥ 0 ⇒ x ≥ −3/2. The isolated right side 5 is positive, so a real solution is possible.

Square: 2x + 3 = 25.

2x = 22.

x = 11.

Check in the original: √(2·11 + 3) + 1 = √25 + 1 = 5 + 1 = 6. True. And 11 ≥ −3/2. Solution: x = 11.

A nearby trap is forgetting to isolate and squaring √(2x + 3) + 1 as if it were √(2x + 3) + 1^2, or squaring termwise into 2x + 3 + 1 = 36. Another trap is √(2x + 3) + 1 = 0, which isolates to √(2x + 3) = −1. No real solution — do not square.

The isolation trap: a square root never equals a negative

Consider √(3x + 1) + 7 = 2.

Isolate: √(3x + 1) = 2 − 7 = −5.

The left side is a principal square root, so it is ≥ 0. It cannot equal −5. No real solution.

If you square blindly: 3x + 1 = 25, x = 8. Plug back: √(24 + 1) + 7 = 5 + 7 = 12, which is not 2. The candidate x = 8 is extraneous. The faster path is to notice the isolated radical equaled a negative and stop.

This is the same structural trap as an item that isolates to a negative after moving a large constant, but the numbers here are original. Train the reflex: look at the sign after isolation before you square.

Extraneous roots from squaring

Squaring is not a one-to-one operation. a = b implies a^2 = b^2, but a^2 = b^2 also holds when a = −b. That is why a candidate can satisfy the squared equation and fail the original.

Worked example: √(x + 6) = x

Domain: x + 6 ≥ 0 ⇒ x ≥ −6. Also the right side must be ≥ 0 because it equals a principal square root, so x ≥ 0.

Square: x + 6 = x^2.

x^2 − x − 6 = 0.

(x − 3)(x + 2) = 0.

Candidates: x = 3 and x = −2.

Check x = 3: √9 = 3. True. Keep.

Check x = −2: √4 = −2 would require 2 = −2. False. Discard. (It also fails x ≥ 0.)

Only x = 3. Reporting both factors of the quadratic is the most common error on this item type. Factoring is still correct algebra for the squared equation; the original equation is stricter.

Two radicals

When two square roots appear, isolate one of them, square, then isolate the remaining radical and square again. Check in the original at the end — twice squaring can add even more extras.

Worked example: √(2x + 9) = √x + 3

Domain: 2x + 9 ≥ 0 and x ≥ 0, so x ≥ 0. The right side √x + 3 is automatically ≥ 3, hence nonnegative, so isolation is already legal.

Square: 2x + 9 = (√x + 3)^2 = x + 6√x + 9.

Simplify: 2x + 9 = x + 6√x + 9.

x = 6√x.

At this stage either x = 0 or, moving everything left, x − 6√x = 0. Let u = √x (so u ≥ 0 and x = u^2): u^2 − 6u = 0 ⇒ u(u − 6) = 0 ⇒ u = 0 or u = 6. Then x = 0 or x = 36.

Check x = 0: √9 = 0 + 3 ⇒ 3 = 3. Keep.

Check x = 36: √(72 + 9) = 6 + 3 ⇒ √81 = 9 ⇒ 9 = 9. Keep.

Both x = 0 and x = 36 work. Two-radical items are not automatically “one extra, one keeper.” Checking decides. A related trap expands (√x + 3)^2 as x + 9, dropping the middle term 6√x, and then never recovers a linear-in-√x equation.

Nested radicals

A nested (composition) equation still follows isolate-then-raise-to-a-power, once per layer.

Worked example: √(5 + √(x − 2)) = 3

Domain: inner x − 2 ≥ 0 ⇒ x ≥ 2, and outer 5 + √(x − 2) ≥ 0 (always true once the inner root exists).

Square once: 5 + √(x − 2) = 9.

√(x − 2) = 4.

Square again: x − 2 = 16.

x = 18.

Check: √(5 + √16) = √(5 + 4) = √9 = 3. True. x = 18, and 18 ≥ 2.

If the first isolation had produced √(5 + √(x − 2)) = −3, you would stop with no real solution, just as in the monomial case. Nested does not change the sign rule.

Creating a radical equation

Table 11 also wants creating the equation from a sentence. “The square root of three more than twice a number, increased by 1, is 6” is exactly √(2x + 3) + 1 = 6. “A side length is the square root of an area of 50” is s = √50, which you would simplify to 5√2 using the previous section. Creating and solving are sequential skills: write the equation with the radicand grouped in parentheses, then isolate.

Cube roots

Cube both sides, not square. ∛(2x − 1) = 3 ⇒ 2x − 1 = 27 ⇒ x = 14. Check: ∛27 = 3. Cube roots can equal negatives: ∛(2x − 1) = −2 ⇒ 2x − 1 = −8 ⇒ x = −7/2, which is valid. The “never negative” rule is for even roots only. Cubing is one-to-one on the reals, so cube-root equations do not manufacture the same style of extraneous roots that squaring does — you still check, because of arithmetic slips, but the domain is all reals.

What the CAT is really testing

The 237–249 band’s simple radical equation is usually one isolated square root and one linear result, possibly with no real solution after isolation. Higher bands add the second radical or a quadratic after squaring. Either way, the check is not optional. The diagram below is the whole method: domain, isolate, refuse a negative even root, raise to a power, solve, and sort keepers from extras in the original equation. Graphing those keepers as points on y = √(g(x)) is Graphs, Domain, and Range of Radical and Rational Functions.