A radical sign names one number, not two, and the choice of which one is a convention worth stating plainly. Whether that number exists at all depends on the sign of the radicand and on whether the index is even or odd.
Core ideas
Most of the difficulty here is notation rather than arithmetic: what the symbol denotes, and when it denotes nothing.
The equation x² = 9 has two solutions, 3 and −3. The expression √9 is a single number, and by convention it is the non-negative one. Confusing the equation with the symbol is the most common error in the topic, and it is not an algebra error.
An even power of a real number is never negative, so nothing real squares to −4 and √(−4) has no value on this course. Odd powers keep the sign of the base, so every real has exactly one real cube root, negatives included.
Factor the radicand into primes. For a square root, each pair of equal primes leaves one factor outside; for an n-th root, each group of n does. What is left under the sign is whatever could not form a full group.
The material
The principal square root, the general n-th root, why the even case needs absolute value, and how a radical is reduced.
For x ≥ 0, √x is the unique number r with r ≥ 0 and r² = x.
√0 = 0. For x < 0 there is no real r with r² = x, because the square of any real number is ≥ 0. On this course such a radical has no value and is said to have none; Quadratics and Complex Numbers introduces i and gives it one.
So √9 = 3, and √9 = ±3 is wrong as written. The ± belongs to solving an equation: x² = 9 gives x = ±3, and the ± is put there explicitly because √ alone would only have supplied the positive root. That is also why the quadratic formula in Quadratics and Complex Numbers carries a ± in front of its radical.
For an integer n ≥ 2, an n-th root of x is a number r with rⁿ = x. The n is called the index.
When n is odd, every real x has exactly one real n-th root and it has the same sign as x. When n is even and x > 0 there are two, one positive and one negative, and the radical denotes the positive one; when n is even and x < 0 there are none.
√(x²) is required to be non-negative, by the definition. At x = −5, x² is 25 and √25 is 5, which is −x rather than x. So the cancellation people write as √(x²) = x is false for every negative x:
The correct statement is √(x²) = |x|, which is “Absolute Value”'s two-case definition arriving from a different direction. For an odd index no absolute value is needed: the cube root of x³ is x for every real x, because a cube keeps the sign.
For a ≥ 0 and b ≥ 0: √(ab) = √a · √b. For a ≥ 0 and b > 0: √(a/b) = √a / √b.
Both rules hold for an n-th root as well. For odd n no sign restriction is needed, since every real has exactly one real odd root; for even n the radicands must be non-negative, for the same reason they must be here.
There is no corresponding rule for a sum. √(9 + 16) = √25 = 5, while √9 + √16 = 7. The radical sign brackets everything under it, exactly as “Order of Operations” said.
To simplify, factor the radicand into primes and take out each prime in complete groups of the index — pairs for a square root, triples for a cube root:
If prime factorisation is not already automatic, use repeated division: divide by the smallest prime that works until the quotient is 1. For 72 the quotients are 36, 18, 9, 3, 1, so the divisors were 2, 2, 2, 3, 3 and 72 = 2³ · 3². This is a procedure, not a factor you are expected to spot.
Fractions come apart the same way: √(18/25) = √18 / √25 = 3√2 / 5. When the denominator is not a perfect square, √(1/2) = 1/√2; the numerical standard form clears that radical by multiplying top and bottom by √2, giving √2/2. The lab performs this numerical step. Rational and Radical Expressions teaches the general rationalising method for variable and multi-term denominators.
√16 = 4. √(−16) has no real value. The 3rd root of −27 is −3, because (−3)³ = −27. The 4th root of 81 is 3, because 3⁴ = 81; −3 is also a fourth root of 81, and the radical denotes 3.
Interactive
The factorisation decides what comes out
Method
Check that the thing exists before working on it. That order saves a page of algebra now and again.
Even index and a negative radicand means there is no real value, and that is the finished answer rather than a dead end. Odd index means there is always exactly one, and it carries the sign of the radicand.
72 is 2³ · 3²; 54 is 3³ · 2. Guessing at a large square factor works sometimes and fails silently when it is not the largest, which leaves a half-simplified answer. The prime factorisation cannot leave one.
A pair of equal primes under a square root sends one copy outside; a triple under a cube root sends one copy outside. Anything that cannot fill a group stays in.
No prime under the sign should appear to the index or higher. A denominator that is already a perfect power comes out directly; otherwise the numerical standard form may clear its remaining radical as in √(1/2) = √2/2. Then raise your answer to the index and confirm you recover the radicand.
Each one is factorised first, and each answer is raised back to the index at the end.
Each check raises the answer to the index and recovers the radicand. That establishes the value is a root of the radicand. For the two square roots it takes one more observation to know it is the principal root: 6√2 and 3√2 / 5 are both positive, and positivity is the second half of the definition.
Line 2 keeps the minus sign outside the radical throughout, which is the tidy way to handle an odd index over a negative radicand. Line 3 leaves no root below the bar because 25 is a perfect square. A non-square numerical denominator uses the clearing step shown by the lab's sqrt(1/2) preset; Rational and Radical Expressions generalises it to variable and multi-term denominators.
Now leave one line for yourself: √108. Repeated division gives 108 = 2² · 3³; take one 2 and one 3 out, then stop because a single 3 remains. The result is 6√3. 6√9 failed to remove a complete pair, while 18√3 took the square itself outside instead of its root.
Decide first whether the 4th root of −81 is real; then simplify √180 and the cube root of −250 from prime factorisations, raising each answer back as a check. Also state why √9 is 3 while x² = 9 has two solutions.
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