1
The radical sign names the principal root
√9 is 3, not ±3. Both 3 and −3 square to 9, but the symbol denotes the non-negative one so that √ is a function. The ± in the quadratic formula is written explicitly because the symbol does not supply it.
Lesson · Rational and Radical Expressions · Radicals
Simplifying Radical Expressions
The second half of the course opens the way the first did: with the values that are not allowed. An even root demands a non-negative radicand, the principal root is a single number rather than two, and both facts show up in the identity √(x²) = |x|.
√(a²) = |a| not a
ⁿ√a · ⁿ√b = ⁿ√(ab) n even requires a ≥ 0 and b ≥ 0
√200 = √(100 · 2) = 10√2
n even → radicand ≥ 0; n odd → any real
Four facts about roots
Core ideas
What this lesson establishes
A radical sign is not an instruction to find every number that works. It names one specific number, and which one depends on whether the index is even or odd.
2
Even and odd indices behave differently
An even power is never negative, so √(−4) names nothing real and −4 is outside the domain. An odd power keeps its sign, so ∛(−8) is −2 with no difficulty at all.
3
Simplifying means extracting perfect powers
√200 is not made simpler by a decimal. It is made simpler by writing 200 = 100 · 2 and taking the perfect square out, leaving 10√2 — an exact value in a form that can be compared and combined.
The material
Roots, principal values and the product rule
The definitions first, then the two rules that do the simplifying and the condition attached to them.
Definition · nth root
For an integer n ≥ 2, b is an nth root of a if bⁿ = a. In ⁿ√a the number n is the index and a is the radicand. With no index written, the index is 2.
Definition · Principal root
When n is even and a ≥ 0, ⁿ√a denotes the non-negative root. When n is odd, ⁿ√a denotes the unique real root, which has the same sign as a. When n is even and a < 0, ⁿ√a is not a real number.
That last clause is the radical half's version of "the denominator is not zero". It is a restriction on the domain, it is decided before any algebra happens, and no manipulation later can be allowed to lose it. √(−4) is treated here as undefined; Quadratics and Complex Numbers introduces i and gives it a value.
Theorem · The identity that catches everyone
For every real x, √(x²) = |x|.
The left side is non-negative by definition of the principal root, so the right side must be too. Writing √(x²) = x asserts something false for every negative x: at x = −5 the left side is 5 and the right side is −5.
More generally ⁿ√(xⁿ) is |x| when n is even and x when n is odd. The absolute value is not decoration. It is the price of an even power destroying sign information that the root then cannot restore.
Theorem · Product and quotient rules
ⁿ√a · ⁿ√b = ⁿ√(ab) and ⁿ√a / ⁿ√b = ⁿ√(a/b), provided every radical named is defined — for even n that means a ≥ 0 and b ≥ 0, and b ≠ 0 in the quotient.
The condition is not fine print. Without it one could write √(−4) · √(−9) = √36 = 6, and the left side is not a real number at all.
Notice what is absent: there is no rule for ⁿ√(a + b). Roots distribute over multiplication and division and over nothing else. √(9 + 16) is √25 = 5, while √9 + √16 is 7. This is the same shape of error as cancelling a term rather than a factor in “Simplifying Rational Expressions”.
Simplified form
- No perfect nth power left under the radical. √72 becomes 6√2 because 72 = 36 · 2.
- No fraction under the radical and no radical in a denominator — “Rationalizing Denominators” handles the second half of that.
Example · Variables under an even root
√(50x³) simplifies to 5x√(2x). No absolute value is needed: for the expression to be defined at all, 50x³ ≥ 0 forces x ≥ 0, and a non-negative x is its own absolute value. The domain did the work that | | would otherwise have to do.
Interactive
Simplifying a radical
The factorisation decides what comes out
Method
Simplifying a radical
Four steps. The first decides whether there is anything to do at all.
1
Check the index and the sign of the radicand
Even index with a negative radicand: stop, it is undefined here. Odd index: any real number is fine, and a negative sign may be brought out front.
2
Factor the radicand into perfect nth powers times the rest
For a square root look for the largest perfect square that divides it; 200 = 100 · 2 in one step beats 200 = 4 · 50 in three.
3
Take each perfect power out, one factor at a time
ⁿ√(cⁿ · d) = c · ⁿ√d by the product rule. For an even index with a variable base, this is where |x| appears unless the domain already forces the base non-negative.
4
Confirm nothing extractable is left
Look at what remains under the radical. If any factor is still a perfect nth power, another complete group can come out and the answer is not yet simplified.
Roots and their conditions
Score 0 / 0Choose the answer you can justify.
Three radicals, three different obstacles
One numeric square root, one odd index with a negative radicand, and one with variables.
(a) √200
200 = 100 · 2, and 100 = 10²
√200 = √100 · √2 = 10√2 check: 10 × 1.41421… = 14.1421…
(b) ∛(−108)
index 3 is odd, so a negative radicand is fine
−108 = (−27) · 4, and −27 = (−3)³
∛(−108) = ∛(−27) · ∛4 = −3∛4 check: −3 × 1.5874… = −4.7622…
(c) √(48x⁵) defined only where 48x⁵ ≥ 0, so x ≥ 0
48x⁵ = (16x⁴) · (3x)
√(48x⁵) = 4x²√(3x)
no absolute value is needed on x²: it is non-negative for every real x
In (b) the sign came out with the root because the index is odd. Had the index been even, there would have been nothing to do — the expression would not name a real number and the work would stop at the first line.
In (c) the domain restriction was written before simplifying, exactly as in “Rational Expressions and Their Domains”, and it makes √(3x) defined. The extracted factor is x², already non-negative, so √(x⁴) = x² needs no bars. The bars are genuinely at stake in the earlier √(50x³) = 5x√(2x): extracting √(x²) first gives |x|, and only the domain condition x ≥ 0 lets that become x.
For a faded variable pass, simplify √(18t²). The supplied factorisation is 18t² = 9 · 2 · t²; decide the domain, extract both square factors and preserve the principal-root sign. Check against 3|t|√2, defined for every real t. Writing 3t√2 reveals that the absolute value was dropped even though negative t is allowed.
Common mistakes
1
Writing √(x²) = xTrue only for x ≥ 0. It is the same class of error as cancelling without recording the domain: a step that gives the right answer on the examples you tried is not thereby a valid rule.
2
Distributing a root over a sum√(a + b) is not √a + √b. Test it once with 9 and 16 and the habit dies: 5 against 7.
3
Reading √9 as ±3The symbol names one number. The two solutions of x² = 9 come from the square root property, which supplies the ± explicitly — the radical sign never does.
Check your understanding
✓
Finish when you can state the condition alongside every rule you use.
Given any radical you should be able to say whether it is defined, simplify it exactly, and justify each extraction by the product or quotient rule with its hypothesis checked. Knowing that √200 = 10√2 matters less than knowing why √(−4) · √(−9) is not 6.
Note: Everything in “Adding, Subtracting and Multiplying Radicals”, “Rationalizing Denominators”, “Solving Radical Equations” and “Radical Functions and Their Graphs” rests on this one. Adding radicals needs simplified form to see which are alike, rationalizing needs the product rule, and radical equations need the domain restriction that the even index imposes here.