Simplify $\sqrt{-81x^4y^2}$

Q: What is $\sqrt{-81x^4y^2}$ in the real numbers?

It is NOT REAL because the radicand is negative.

Q: Why is a negative radicand not real?

A square root of a negative number is not defined in the real-number system.

Question

Simplify. (If the solution is not a real number, enter NOT REAL.)

$\sqrt{-81x^4y^2}$

$-9x^2y$

Acemy AI · Accepted Answer

**Key takeaway**: Before simplifying a square root, check the sign of the radicand. A square root of a negative number is not real in the real-number system.

We look at the radicand:

$$\sqrt{-81x^4y^2}$$

Since $-81x^4y^2$ is negative whenever $x$ and $y$ are real numbers, the square root is not a real number.

Therefore the answer is

**NOT REAL**

---
**Pitfalls the pros know** 👇
Do not try to simplify the factors first and ignore the negative sign. The presence of a negative radicand means the expression has no real value.

**What if the problem changes?**
If the expression were $\sqrt{81x^4y^2}$, then it would simplify to $9x^2|y|$. The negative sign is what makes the original problem not real.

`Tags`: radicand, real number, square root

Question

Simplify $\sqrt{-81x^4y^2}$

Expert Verified Solution

FAQ

What is $\sqrt{-81x^4y^2}$ in the real numbers?

Why is a negative radicand not real?