1Cademy - Simplifying $$\sqrt{63u^3v^5}$$

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Simplifying $\sqrt{x^3}$

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Simplifying $\sqrt{63u^3v^5}$

Simplify a square root whose radicand contains a non-perfect-square coefficient and two variables, each raised to an odd power. $\sqrt{63u^3v^5}$

Step 1 — Find the largest perfect square factor. The largest perfect square dividing $63$ is $9$ (since $3^2 = 9$ and $63 = 9 \cdot 7$ ). For $u^3$ , the largest even power is $u^2$ ; for $v^5$ , it is $v^4$ . Together, the largest perfect square factor is $9u^2v^4$ :

$\sqrt{63u^3v^5} = \sqrt{9u^2v^4 \cdot 7uv}$

Step 2 — Apply the Product Property. Split into two radicals:

$\sqrt{9u^2v^4 \cdot 7uv} = \sqrt{9u^2v^4} \cdot \sqrt{7uv}$

Step 3 — Simplify. Since $\sqrt{9} = 3$ , $\sqrt{u^2} = |u|$ , and $\sqrt{v^4} = v^2$ :

$\sqrt{9u^2v^4} \cdot \sqrt{7uv} = 3|u|v^2\sqrt{7uv}$

The simplified form is $3|u|v^2\sqrt{7uv}$ . With two variables, each variable is treated independently: extract the largest even power of each ( $u^2$ from $u^3$ and $v^4$ from $v^5$ ), leaving one factor of each variable under the radical. Because the index is even, taking the principal square root of an even power requires absolute value signs if the resulting exponent is odd (e.g., $|u|$ ).

Updated 2026-05-01

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