Irrational equations

In an irrational (radical) equation the unknown sits under a root: √(x + 7) = x + 1, √x + √(x + 5) = 5. The main method is to get rid of the root by squaring both sides (or raising them to the required power). But this step is not an equivalence: it produces extraneous roots. So at every step you must either write down the conditions or check at the end — this is where most exam mistakes happen. This lesson covers the main types of equations and inequalities, substitution and using the domain.

§1. Why extraneous roots appear

Squaring turns a = b into a² = b², but not conversely: a² = b² gives a = b or a = −b. For example, squaring x = 3 gives x² = 9, whose roots are 3 and −3 — the extraneous root −3 has appeared.

There are two safeguards: an equivalent transition (write down the conditions) or a check (substitute every number found into the original equation). If the roots are “nice”, checking is convenient; if they are “ugly”, conditions are safer.
Raising to an odd power (e.g. cubing) is an equivalence: ∛f = g ⇔ f = g³, and no extraneous roots arise.

§2. The basic type

Why g(x) ≥ 0? A principal root is never negative; squaring loses this information. There is no need to write f(x) ≥ 0 separately — it follows automatically from f = g².

• √f(x) = a with a < 0 — no solutions: √(x − 3) = −2.
• √f(x) = a with a ≥ 0 — f(x) = a².

Worked example

Solve √(x + 7) = x + 1.

1. Condition: x + 1 ≥ 0 → x ≥ −1.
2. Square: x + 7 = x² + 2x + 1 → x² + x − 6 = 0.
3. Roots: x = 2 or x = −3.
4. x = −3 violates the condition — extraneous. Answer: x = 2.

Graphically: y = √(x + 7) is half a parabola (the upper branch only) and y = x + 1 is a line; they meet once, at x = 2. The extraneous x = −3 is where the line meets the lower branch y = −√(x + 7) — squaring merged the two branches.

§3. Equations with two roots

• √f = √g ⇔ f = g and f ≥ 0 (or g ≥ 0 — whichever is simpler).
• √f + √g = c: isolate one root and square twice. You may square with both roots on one side, but then a product of roots appears.

Worked example

Solve √(x + 5) + √x = 5.

1. Isolate one root: √(x + 5) = 5 − √x.
2. Square: x + 5 = 25 − 10√x + x.
3. 10√x = 20 → √x = 2 → x = 4.
4. Check: √9 + √4 = 3 + 2 = 5. Answer: x = 4.

§4. Substitution and the domain

If an expression contains both √x and x (or √f and f), put t = √x ≥ 0 — the equation becomes quadratic. Negative values of t are discarded.
x − 5√x + 6 = 0 → t² − 5t + 6 = 0 → t = 2; 3 → x = 4; 9.
Sometimes the domain alone gives the answer without solving: if no x makes all the roots exist at once, the equation has no solutions.

Worked example

Solve √(x − 5) + √(3 − x) = 1.

1. First root: x − 5 ≥ 0 → x ≥ 5.
2. Second root: 3 − x ≥ 0 → x ≤ 3.
3. x ≥ 5 and x ≤ 3 cannot both hold — the domain is empty.
4. Answer: no solutions.

§5. Irrational inequalities

Checking does not work for inequalities (there are infinitely many solutions), so only equivalent transitions are used:

The second rule has two systems: if g is negative, the inequality holds for every x where the root exists (a non-negative number exceeds a negative one).

Worked example (from the official sample)

Solve √(x + 18) < 2 − x.

1. x + 18 ≥ 0 → x ≥ −18; 2 − x > 0 → x < 2.
2. x + 18 < (2 − x)² = x² − 4x + 4 → x² − 5x − 14 > 0.
3. (x − 7)(x + 2) > 0 → x < −2 or x > 7.
4. Intersection: [−18; −2).

Worked example

Solve √(x + 2) > x.

1. Case 1: x < 0 and x + 2 ≥ 0 → [−2; 0).
2. Case 2: x ≥ 0 and x + 2 > x² → x² − x − 2 < 0 → −1 < x < 2 → [0; 2).
3. Union: [−2; 2).

Check yourself

1. Solve √(2x − 1) = 3.

2. Solve √(x + 2) = x.

3. How many solutions does √(x − 3) = −2 have?

4. Solve x − 5√x + 6 = 0.

5. Solve √(x − 1) < 2.

6. Solve √(3x + 1) = √(x + 5).

7. Solve ∛(x + 6) = 2.

8. How many solutions does √(x − 4) + √(1 − x) = 3 have?

Common mistakes

• Not checking after squaring and keeping an extraneous root.
• Expanding (5 − √x)² as 25 − x — the middle term −10√x is lost.
• Forgetting the condition ≥ 0 under the root in an inequality.
• Dropping the case g < 0 in √f > g.
• Forgetting t = √x ≥ 0 in a substitution and finding x from a negative t.
• Trying to “check” an inequality — only equivalent transitions work.