I got $\displaystyle 4,7,x^8,y^18,y$ --All in radicals..that is a y to the 18th power Then my answer: $\displaystyle 2x^4y^9$ radical 7y
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$\displaystyle \sqrt {448x^8 y^{19} } = \sqrt {2^6 \cdot 7x^8 y^{19} } = 2^3 x^4 y^9 \sqrt {7y} $ Do you see how it works?
Originally Posted by Plato $\displaystyle \sqrt {448x^8 y^{19} } = \sqrt {2^6 \cdot 7x^8 y^{19} } = 2^3 x^4 y^9 \sqrt {7y} $ Do you see how it works? How did you get for the first part: $\displaystyle 2^3$ I got everything except for the $\displaystyle 2^3$ I did get the 2 but no 3rd root.
Here is how it works. $\displaystyle \begin{array}{l} \sqrt {2^6 } = 2^3 \\ \sqrt {2^6 } = \left( {2^6 } \right)^{\frac{1}{2}} = 2^3 \\ \end{array} $ $\displaystyle \left( {\frac{1}{2}} \right)\left( 6 \right) = 3$
Ok. Seems I messed up because as I tried to find out how I got 4,7 I played around with the numbers again and got 4, 28. Is this the correct answer?
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