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Solving radical equations: isolate, square, then check

A square root comes off when you square both sides. The points get lost in three places: squaring before the root is alone, squaring a binomial as if it were two separate terms, and keeping an answer the check should have thrown out.

Short answer

Get the radical by itself on one side. Square both whole sides: if the other side is x + 1, it becomes (x + 1)2 = x2 + 2x + 1, not x2 + 1. Solve what's left, then check every answer in the original equation, because squaring can produce answers that don't work. If there are two square roots, isolate one, square, isolate the other, and square again. A cube root comes off by cubing both sides, and that step never creates a false answer.

Why it works

If two things are equal, their squares are equal. That's why squaring both sides is allowed. But it doesn't run backwards: 3 ≠ −3, yet 32 = (−3)2. So after squaring, the new equation is satisfied by every solution of the original, and by every solution of the version where one side has the opposite sign. Those extra answers are called extraneous. They come from a step that is allowed but can't be undone, not from an arithmetic slip, and the only way to find them is to substitute into the original equation.

The other fact you need: √u means the non-negative root. √9 = 3, never −3. So a square root can't equal a negative number, and any answer that makes the other side negative is extraneous.

Cube roots don't have this problem. Every real number has exactly one cube root, and cubing keeps the sign: if a3 = b3 then a = b. Cubing both sides can be undone, so nothing extra gets in.

Example 1: isolate the radical first

Solve √2x + 3 − 1 = 4.

  1. √2x + 3 = 5Add 1 to both sides. Now the root is alone, so squaring will remove it cleanly.
  2. 2x + 3 = 25Square both sides: (√(2x + 3))² = 2x + 3 and 5² = 25.
  3. 2x = 22Subtract 3.
  4. x = 11Divide by 2. Check: √(22 + 3) − 1 = √25 − 1 = 5 − 1 = 4. It works.

Example 2: squaring a binomial, and an answer that fails

Solve √x + 7 = x + 1.

  1. x + 7 = (x + 1)2The root is already alone. Square both sides, keeping the right side in parentheses.
  2. x + 7 = x2 + 2x + 1(x + 1)² = (x + 1)(x + 1). The middle term 2x comes from the two cross products.
  3. 0 = x2 + x − 6Subtract x and 7 from both sides.
  4. 0 = (x + 3)(x − 2)Factor: 3 and −2 multiply to −6 and add to 1.
  5. x = −3 or x = 2Two candidates. Both still have to be checked.
  6. x = 2 onlyx = 2: √9 = 3 and 2 + 1 = 3. It works. x = −3: √4 = 2, but −3 + 1 = −2. 2 ≠ −2, so −3 is rejected.

Where did x = −3 come from? It solves √x + 7 = −(x + 1): √4 = 2 = −(−2). Squaring that equation gives the same line 1, so its answer got mixed in. At x = −3 the right side is negative, and a square root can't be negative.

Example 3: two radicals, so square twice

Solve √2x + 3 − √x + 1 = 1.

  1. √2x + 3 = 1 + √x + 1Isolate one radical. Moving the other root to the right keeps both sides simpler than squaring a difference of two roots.
  2. 2x + 3 = 1 + 2√x + 1 + (x + 1)Square both sides. The right side is a binomial: (1 + √(x + 1))² = 1² + 2 · 1 · √(x + 1) + (√(x + 1))².
  3. x + 1 = 2√x + 1Subtract 1, x, and 1 from both sides to get the remaining root alone.
  4. x2 + 2x + 1 = 4(x + 1)Square again: (x + 1)² on the left, 2² · (x + 1) on the right.
  5. x2 − 2x − 3 = 0Distribute the 4, then subtract 4x and 4.
  6. (x − 3)(x + 1) = 0Factor: −3 and 1 multiply to −3 and add to −2.
  7. x = 3 or x = −1Check both. x = 3: √9 − √4 = 3 − 2 = 1. x = −1: √1 − √0 = 1 − 0 = 1. Both work.

On line 3 it's tempting to divide both sides by √x + 1. Don't: that throws away x = −1, the value that makes √x + 1 = 0. Squaring keeps every answer, and the check removes any extras. Dividing by something that can be zero loses answers, and no check will bring them back.

Example 4: a cube root

Solve ∛x − 2 + 5 = 2.

  1. ∛x − 2 = −3Subtract 5. A cube root can equal a negative number: ∛(−27) = −3, since (−3)³ = −27.
  2. x − 2 = −27Cube both sides: (−3)³ = −27.
  3. x = −25Add 2. Check: ∛(−27) + 5 = −3 + 5 = 2. Because cubing can be undone, this check is insurance, not a filter.

Compare √x − 2 = −3, which has no solution at all. The negative right side is fine for a cube root and fatal for a square root.

Common mistakes, and the exact line they happen on

1. Squaring before the radical is alone

Solve √2x + 3 − 1 = 4 (Example 1).

  1. (2x + 3) − 1 = 16This is the mistake. Each term was squared separately, but (a − b)² isn't a² − b², and the −1 should have been moved first. This line gives x = 7, and the check fails: √17 − 1 ≈ 3.12, not 4.

Squaring applies to a whole side. When the root shares its side with anything else, squaring produces a mess of cross terms with the root still in them. Move everything else away first, as in Example 1, so the square lands on the root alone.

2. Squaring x + 1 as x2 + 1

Solve √x + 7 = x + 1 (Example 2).

  1. x + 7 = x2 + 1This is the mistake. (x + 1)² = x² + 2x + 1. Dropping the 2x leads to x² − x − 6 = 0, so x = 3 or x = −2, and neither one works: √10 ≈ 3.16 but 3 + 1 = 4, and √5 ≈ 2.24 but −2 + 1 = −1.

When every candidate fails the check, suspect this line before concluding "no solution." Write (x + 1)2 with the parentheses first, then expand on the next line.

3. Keeping the extraneous answer

In Example 2, the algebra ends with x = −3 or x = 2. Writing both as the final answer loses the point even though every line was right. Substitute each one into the original equation, not a squared version: the squared version is exactly the line that let −3 in, so it will happily say −3 works.

4. Squaring a side with a root in it as if the cross term weren't there

Solve √x + 5 − √x = 1.

  1. √x + 5 = 1 + √xIsolate one radical. Fine so far.
  2. x + 5 = 1 + xThis is the mistake. (1 + √x)² = 1 + 2√x + x. Dropping the 2√x leads to 5 = 1, and "no solution." But x = 4 works: √9 − √4 = 3 − 2 = 1.

The correct line is x + 5 = 1 + 2√x + x, which gives √x = 2 and x = 4. With two radicals, the first squaring always leaves a root in the middle term. That's why there is a second squaring.

5. Solving when the root equals a negative number

  1. √x − 2 + 7 = 4Start.
  2. √x − 2 = −3Subtract 7. The root is alone, and it equals a negative number.
  3. x − 2 = 9This is the mistake. Squaring leads to x = 11, and the check fails: √(11 − 2) + 7 = 3 + 7 = 10, not 4. A square root is never negative, so the answer is no solution. Stop at the line above.

Practice

  1. Solve ∛2x + 1 = 3.

    Show answer

    Cube both sides: 2x + 1 = 27, so x = 13. Check: ∛27 = 3 ✓.

  2. Solve 2√x + 1 + 3 = 11.

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    Isolate: 2√x + 1 = 8, √x + 1 = 4. Square: x + 1 = 16, so x = 15. Check: 2√16 + 3 = 11 ✓.

  3. Solve √x + 10 = x − 2.

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    Square: x + 10 = x2 − 4x + 4, so x2 − 5x − 6 = 0, (x − 6)(x + 1) = 0. Check: x = 6 gives √16 = 4 = 6 − 2 ✓. x = −1 gives √9 = 3, but −1 − 2 = −3 ✗. Only x = 6.

  4. Solve √3x + 1 + 5 = 1.

    Show answer

    Isolate: √3x + 1 = −4. A square root can't be negative, so there is no solution.

  5. Solve √x + 7 − √x = 1.

    Show answer

    Isolate: √x + 7 = 1 + √x. Square: x + 7 = 1 + 2√x + x, so 2√x = 6, √x = 3, x = 9. Check: √16 − √9 = 4 − 3 = 1 ✓.

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