When fnding a square root of 64x4y16 do you minus the powers by two or divide them by two
+26375 When fnding a square root of 64x4y16 do you minus the powers by two or divide them by two
\(\begin{array}{rcll} 64x^4y^{16} &=& 8^2x^4y^{16} \\\\ \sqrt{ 64x^4y^{16} } &=& \sqrt{ 8^2x^4y^{16} }\\ \sqrt{ 64x^4y^{16} } &=& \left( 8^2x^4y^{16} \right)^{\frac12}\\ &=& \left( 8^2\right)^{\frac12} \cdot \left( x^4 \right)^{\frac12} \cdot \left( y^{16} \right)^{\frac12}\\ &=& \left( 8 \right)^{2\cdot \frac12} \cdot \left( x \right)^{4\cdot \frac12} \cdot \left( y \right)^{16\cdot \frac12}\\ &=& \left( 8 \right)^{\frac22} \cdot \left( x \right)^{\frac42} \cdot \left( y \right)^{\frac{16}{2}}\\ &=& \left( 8 \right)^{1} \cdot \left( x \right)^{2} \cdot \left( y \right)^{8}\\ &=& 8^1 \cdot x^2 \cdot y^8 \\ &=& 8 \cdot x^2 \cdot y^8 \\ \end{array}\)
You divide the powers by two.
