The decision and its conditions
An exponent with a denominator describes a root, while its numerator describes a power. A negative exponent additionally takes a reciprocal; keep those operations separate. Even roots use the nonnegative principal value, so a square root of a square produces an absolute value. Rationalizing a denominator multiplies by a form of one, not by a number that changes the expression. For a binomial radical denominator, its conjugate converts the denominator into a difference of squares. Verify the original domain before simplifying.
Worked example
Evaluate the real principal value √(√(81)). The inner principal square root is 9. The outer principal square root of 9 is 3. Both roots are nonnegative; do not retain a negative branch.
A contrasting decision
Find the exact real value of 27^(-2/3). The cube root of 27 is 3. The numerator two squares that root to 9. The negative exponent takes its reciprocal, giving 1/9.
Diagnose a competing method
The first example’s tempting answer 81 follows this mistake: Leave both root operations unevaluated. Apply each root in sequence; two square roots together take a fourth root.
Transfer before revealing
Write 1/(5+√2) as a+b√2 with rational a and b. What is a? Pause and identify the original restrictions. Solution: Multiply numerator and denominator by 5−√2. The denominator becomes 5²−2=23. The rational component is 5/(23)=5/23.
Check the boundary of the method
For a binomial radical denominator, its conjugate converts the denominator into a difference of squares. Read root, power and reciprocal as three separate operations. Check the solution in the original statement. A transformed equation can generate candidates rather than equivalent solutions.
Evaluate the real principal value √(√(81)).
- The inner principal square root is 9.
- The outer principal square root of 9 is 3.
- Both roots are nonnegative; do not retain a negative branch.
Compare the other answer choices
Leave both root operations unevaluated.
Divide by four instead of taking a fourth root.
Treat a principal root as a signed solution pair.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
CLEP College Algebra exam overview