Why can an even root of a negative number fail to be real?
Quick answer
For any real number, raising it to an even power gives a number that is zero or positive. So there is no real number whose even power equals a negative number: for example, √(−9) is not real. Odd powers can stay negative, so an odd root can be real; ∛(−8)=−2 because (−2)³=−8. This explanation is about real numbers, not complex numbers. [S1]
The question, the details, the next step.
Why can’t I take the square root of a negative number and get a real answer?
A square root asks, “What real number multiplied by itself gives this number?” A positive number squared is positive, and a negative number squared is also positive: 2²=4 and (−2)²=4. Zero squared is zero. In general, an even power of a real number cannot be negative, so no real number squared gives −9. That means √(−9) has no real value. [S1]
A useful way to check is to reverse the root: if √(−9) were a real number, call it x, then x² would have to equal −9. But a real square cannot be negative. This is a limitation of the real number system; it does not mean the question is meaningless in every larger number system.
Does the same reasoning apply to every even root, not just square roots?
Yes. For an even root such as the fourth root, the question is whether some real number raised to the fourth power equals the number inside the root. Since an even power of a real number cannot be negative, a negative radicand has no real even root. For example, there is no real x satisfying x⁴=−16. The same reasoning applies to any even index. [S1]
The sign of the number being raised to the power is the key. A negative base does not solve the problem: (−2)⁴=16, not −16. For nonnegative radicands, even roots are real; for a negative radicand, they are not real. [S1]
Why is an odd root of a negative number different?
With an odd power, a negative base stays negative: (−2)³=−8. So the cube root of −8 is the real number −2, because (−2)³ gives the number under the root. More generally, odd roots of negative numbers can be real, and odd roots are real for all real radicands. [S1]
Compare these two checks:
- ∛(−8)=−2, since (−2)³=−8.
- √(−9) has no real answer, since there is no real x with x²=−9.
To check a root, raise your proposed answer to the root’s index and see whether you get the original number. The even-versus-odd distinction matters: an even power cannot give a negative real result, while an odd power can.
Can you give me a short method I can use when solving exercises?
First identify the root’s index: a square root has index 2, a cube root has index 3, and so on. Then check whether the number under the root is negative. If it is negative and the index is even, write “not real” when the exercise asks for real-number answers. If the index is odd, look for a real number whose power matches the radicand, including its sign.
Worked check: For ⁴√(−81), the index 4 is even and the radicand −81 is negative, so there is no real answer. For ∛(−27), the index 3 is odd, and (−3)³=−27, so the answer is −3.
This method establishes whether a real answer exists; it does not describe complex-number answers. [S1] If an exercise asks specifically for complex numbers, use the conventions and methods from that topic instead.
A practical checklist
Identify the root’s index.
Check whether the radicand is negative.
If both are true and the index is even, state that there is no real root.
For an odd root, test a number by raising it to the root’s index.
Substitute your proposed answer back into the power to verify it.
A few more useful answers
Is √(−9) the same as −√9?
No. −√9 equals −3, and (−3)²=9, not −9. The negative sign outside the root does not make the radicand negative.
Does “not real” mean there is no answer in any number system?
No. It means there is no answer among the real numbers. This guide covers real-number roots; it does not explain how to calculate roots using complex numbers.
Sources
References used for this guide.
- [S1] OpenStax Elementary Algebra 2e: Higher Roots openstax.org · Checked