which expression is equivalent to $sqrt{128x^{8}y^{3}z^{9}}$? assume $ygeq0$ and $zgeq0$.\n$2x^{2}z^{2}sqrt{8…

which expression is equivalent to $sqrt{128x^{8}y^{3}z^{9}}$? assume $ygeq0$ and $zgeq0$.\n$2x^{2}z^{2}sqrt{8y^{3}z}$\n$4x^{2}yz^{3}sqrt{2x^{2}}$\n$8x^{4}yz^{4}sqrt{2yz}$\n$64x^{4}yz^{4}sqrt{2yz}$

which expression is equivalent to $sqrt{128x^{8}y^{3}z^{9}}$? assume $ygeq0$ and $zgeq0$.\n$2x^{2}z^{2}sqrt{8y^{3}z}$\n$4x^{2}yz^{3}sqrt{2x^{2}}$\n$8x^{4}yz^{4}sqrt{2yz}$\n$64x^{4}yz^{4}sqrt{2yz}$

Answer

Explanation:

Step1: Prime - factorize 128

$128 = 2^7$. So, $\sqrt{128x^{8}y^{3}z^{9}}=\sqrt{2^{7}x^{8}y^{3}z^{9}}$.

Step2: Use the square - root property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$

$\sqrt{2^{7}x^{8}y^{3}z^{9}}=\sqrt{2^{7}}\cdot\sqrt{x^{8}}\cdot\sqrt{y^{3}}\cdot\sqrt{z^{9}}$.

Step3: Simplify each square - root term

  • $\sqrt{2^{7}}=\sqrt{2^{6}\times2}=2^{3}\sqrt{2}=8\sqrt{2}$ since $\sqrt{a^{m}} = a^{\frac{m}{2}}$ and $\sqrt{2^{6}}=2^{3}$.
  • $\sqrt{x^{8}}=x^{4}$ because $\sqrt{x^{8}}=x^{\frac{8}{2}} = x^{4}$.
  • $\sqrt{y^{3}}=y\sqrt{y}$ since $\sqrt{y^{3}}=\sqrt{y^{2}\times y}=y\sqrt{y}$.
  • $\sqrt{z^{9}}=z^{4}\sqrt{z}$ since $\sqrt{z^{9}}=\sqrt{z^{8}\times z}=z^{4}\sqrt{z}$.

Step4: Combine the simplified terms

$8\sqrt{2}\cdot x^{4}\cdot y\sqrt{y}\cdot z^{4}\sqrt{z}=8x^{4}yz^{4}\sqrt{2yz}$.

Answer:

C. $8x^{4}yz^{4}\sqrt{2yz}$