Simplifying Radical Expressions

The method: find the largest perfect square factor

Perfect squares are the squares of whole numbers: $4, 9, 16, 25, 36, 49, 64, 81, 100, \ldots$ To simplify $\sqrt{n}$, find the largest perfect square that divides $n$. For $\sqrt{48}$: does $16$ divide $48$? Yes, $48 = 16 \cdot 3$. So $\sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}$.

If you don't spot the largest one, you can simplify in stages. $\sqrt{48} = \sqrt{4 \cdot 12} = 2\sqrt{12}$, and $\sqrt{12} = 2\sqrt{3}$, so you still land on $4\sqrt{3}$ — it just takes two passes. Grabbing the largest perfect square first finishes it in one.

A radical is fully simplified when no perfect square bigger than $1$ divides what's left under the root. $4\sqrt{3}$ is done because $3$ has no perfect-square factor; $3\sqrt{8}$ is not done because $8 = 4 \cdot 2$ still has one.

Checking your answer by squaring

Every simplification can be verified in five seconds: square your answer and see if you get the original number back. Claim $\sqrt{48} = 4\sqrt{3}$? Then $(4\sqrt{3})^2 = 16 \cdot 3 = 48$ ✓. Square the coefficient, multiply by the number under the root, compare. If the check fails, you either picked a wrong factor or took a wrong root.

Variables under the root

Variable powers follow the same idea, and even exponents are the perfect squares: $\sqrt{x^2} = x$, $\sqrt{x^4} = x^2$, $\sqrt{x^6} = x^3$ — the square root cuts the exponent in half. (In Algebra 1 we assume the variables represent nonnegative numbers, so no absolute-value bars are needed.)

For a mixed radicand like $\sqrt{75x^4}$, handle the number and the variable separately: $75 = 25 \cdot 3$ gives $5\sqrt{3}$, and $\sqrt{x^4} = x^2$. Put the pieces together: $5x^2\sqrt{3}$. An odd exponent splits into an even power times one extra factor: $x^5 = x^4 \cdot x$, so $\sqrt{x^5} = x^2\sqrt{x}$.

Worked examples

Try one yourself

Common questions