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\( \begin{array}{ll}\text { 10 } \sqrt{25 a^{6} p^{10}} & \text { 2. } \sqrt[3]{8 x^{12} y^{15}} \\ \text { 3. } \sqrt[5]{32 x^{5} y^{10}} & \text { 4. } \sqrt{2 x^{3}} \times \sqrt{8 x^{5}} \\ \)\begin{tabular}{ll}\text { (5. }\( \sqrt{2} \times \sqrt{8} \) & \text { 6. }\( \sqrt[6]{64 a^{5} b^{12} c^{18}} \) \\ \text { (7) }\( \sqrt{9 x^{10}+16 x^{10}} \) & \text { 8. }\( \sqrt{\frac{27 x^{3} y^{5}}{12 x^{3} y^{3}}} \) \\ \text { 9. }\( \sqrt{3 m^{3}} \times \sqrt{27 m} \) & \text { 10. }\( \sqrt{2 m^{3}} \times \sqrt{50 m x^{2}} \) \\ \text { 11. }\( \frac{\sqrt{9 x^{10}}+\sqrt{4 x^{10}}}{5 x^{3}} \) & \text { 12. }\( \sqrt[3]{3 x} \times \sqrt[3]{72 x^{5}} \) \\ \hline\end{tabular}\end{array}

Ask by Ramos Chen. in South Africa
Mar 11,2025

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Answer

1. \( 50a^3p^5 \) 2. \( 2x^4y^5 \) 3. \( 2xy^2 \) 4. \( 4x^4 \) 5. \( 4 \) 6. \( 2b^2c^3\sqrt[6]{a^5} \) 7. \( 5x^5 \) 8. \( \frac{3y}{2} \) 9. \( 9m^2 \) 10. \( 10m^2x \) 11. \( x^2 \) 12. \( 6x^2 \)

Solution

1. Simplify \[ 10 \sqrt{25a^6 p^{10}}. \] Inside the square root, factor: \[ 25a^6p^{10} = 25\,(a^6)(p^{10}). \] Taking the square root gives: \[ \sqrt{25}=5,\quad \sqrt{a^6}=a^3,\quad \sqrt{p^{10}}=p^5. \] Thus, \[ 10 \sqrt{25a^6p^{10}} = 10\,(5a^3p^5)=50a^3p^5. \] 2. Simplify \[ \sqrt[3]{8x^{12}y^{15}}. \] Factor each term: \[ 8=2^3,\quad x^{12}=(x^4)^3,\quad y^{15}=(y^5)^3. \] Taking the cube root yields: \[ \sqrt[3]{2^3}=2,\quad \sqrt[3]{(x^4)^3}=x^4,\quad \sqrt[3]{(y^5)^3}=y^5. \] Therefore, \[ \sqrt[3]{8x^{12}y^{15}}=2x^4y^5. \] 3. Simplify \[ \sqrt[5]{32x^5y^{10}}. \] Write each component as a fifth power: \[ 32=2^5,\quad x^5=x^5,\quad y^{10}=(y^2)^5. \] Then, \[ \sqrt[5]{2^5}=2,\quad \sqrt[5]{x^5}=x,\quad \sqrt[5]{(y^2)^5}=y^2. \] Thus, \[ \sqrt[5]{32x^5y^{10}}=2xy^2. \] 4. Simplify \[ \sqrt{2x^3}\times\sqrt{8x^5}. \] Combine under a single square root: \[ \sqrt{2x^3}\times\sqrt{8x^5}=\sqrt{(2x^3)(8x^5)}=\sqrt{16x^8}. \] Since \[ \sqrt{16}=4\quad \text{and}\quad \sqrt{x^8}=x^4, \] we obtain \[ \sqrt{16x^8}=4x^4. \] 5. Simplify \[ \sqrt{2}\times\sqrt{8}. \] Combine the radicals: \[ \sqrt{2}\times\sqrt{8}=\sqrt{16}=4. \] 6. Simplify \[ \sqrt[6]{64a^5b^{12}c^{18}}. \] Factor each term: \[ 64=2^6,\quad b^{12}=(b^2)^6,\quad c^{18}=(c^3)^6. \] The term \(a^5\) does not form a perfect sixth power, so it remains under the radical. Thus, \[ \sqrt[6]{64a^5b^{12}c^{18}}=\sqrt[6]{2^6}\,\sqrt[6]{a^5}\,\sqrt[6]{(b^2)^6}\,\sqrt[6]{(c^3)^6}=2\,b^2\,c^3\,\sqrt[6]{a^5}. \] 7. Simplify \[ \sqrt{9x^{10}+16x^{10}}. \] First, combine like terms: \[ 9x^{10}+16x^{10}=(9+16)x^{10}=25x^{10}. \] Then, \[ \sqrt{25x^{10}}=\sqrt{25}\,\sqrt{x^{10}}=5x^5. \] 8. Simplify \[ \sqrt{\frac{27x^3y^5}{12x^3y^3}}. \] Cancel the common factors \(x^3\) and \(y^3\): \[ \frac{27x^3y^5}{12x^3y^3}=\frac{27y^2}{12}. \] Simplify the fraction: \[ \frac{27}{12}=\frac{9}{4}. \] Thus, \[ \sqrt{\frac{9y^2}{4}}=\frac{\sqrt{9y^2}}{\sqrt{4}}=\frac{3y}{2}. \] 9. Simplify \[ \sqrt{3m^3}\times\sqrt{27m}. \] Multiply under the radical: \[ \sqrt{3m^3}\times\sqrt{27m}=\sqrt{(3m^3)(27m)}=\sqrt{81m^4}. \] Since \[ \sqrt{81}=9,\quad \sqrt{m^4}=m^2, \] we have \[ \sqrt{81m^4}=9m^2. \] 10. Simplify \[ \sqrt{2m^3}\times\sqrt{50mx^2}. \] Multiply under the square root: \[ (2m^3)(50mx^2)=100m^4x^2. \] So, \[ \sqrt{100m^4x^2}=\sqrt{100}\,\sqrt{m^4}\,\sqrt{x^2}=10m^2x. \] 11. Simplify \[ \frac{\sqrt{9x^{10}}+\sqrt{4x^{10}}}{5x^3}. \] Evaluate each square root: \[ \sqrt{9x^{10}}=3x^5,\quad \sqrt{4x^{10}}=2x^5. \] Then, the numerator becomes: \[ 3x^5+2x^5=5x^5. \] Thus, \[ \frac{5x^5}{5x^3}=x^{5-3}=x^2. \] 12. Simplify \[ \sqrt[3]{3x}\times\sqrt[3]{72x^5}. \] Combine under a single cube root: \[ \sqrt[3]{3x}\times\sqrt[3]{72x^5}=\sqrt[3]{(3x)(72x^5)}=\sqrt[3]{216x^6}. \] Since \[ 216=6^3\quad \text{and}\quad x^6=(x^2)^3, \] we have: \[ \sqrt[3]{216x^6}=6x^2. \]

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Beyond the Answer

Did you know that square roots were being used as far back as ancient Babylon? They had a surprising knack for calculations and often relied on geometric methods to find roots long before modern algebra came into play! They would manipulate numbers and shapes, laying the groundwork for the mathematics we know today. In real-world applications, understanding roots is crucial for various fields! For example, in engineering, calculating the area or volume of materials often involves square and cube roots. Similarly, in finance, understanding the rates of return may require knowledge of roots to compute compound interest or growth rates. So next time you encounter a root, remember its real-world significance!

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