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Simplify the following expressions without ever appealing to radical notation.
$\displaystyle{\left[ \frac{-5x^{-3}}{(16y^2)^{-1/2}} \right]^{-1}}$
$\displaystyle{(3x^2y^3)(-2x^{-3}y^2)(xy^2z^{1/3})^0}$
$\displaystyle{(a+b)^{2/3}(a+b)}$
$\displaystyle{\frac{(3x^{-1/2}y^{2/3})^{-2}}{(xy^{-1})^{-3}}}$
$\displaystyle{\left( \frac{25a^6 b^{-2}}{16x^4 a^4} \right)^{-\frac{1}{2}}}$
$\displaystyle{\left( \frac{16xy^{-2}}{81a^4b^6} \right)^{-\frac{1}{4}}}$
$\displaystyle{\left( \frac{27x^2}{w^6} \right)^{-\frac{1}{3}}}$
$\displaystyle{\left( \frac{16^{-1/4} + (-4)^{-2} - 2^{-3}}{8^{1/3}} \right)^{-1}}$
${\displaystyle{\left[ \frac{-5x^{-3}}{(16y^2)^{-1/2}} \right]^{-1} = \frac{(-5)^{-1} x^3}{(16y^2)^{1/2}} = \frac{(-5)^{-1} x^3}{4 |y|} = \frac{x^3}{-20 |y|} = \boxed{\frac{-x^3}{20|y|}}}}$
${\displaystyle{(3x^2y^3)(-2x^{-3}y^2)(xy^2z^{1/3})^0 = (3x^2y^3)(-2x^{-3}y^2) = -6x^{-1} y^5 = \boxed{\frac{-6y^5}{x}}}}$
${\displaystyle{(a+b)^{2/3}(a+b) = (a+b)^{2/3 + 1} = \boxed{(a+b)^{5/3}}}}$
${\displaystyle{\frac{(3x^{-1/2}y^{2/3})^{-2}}{(xy^{-1})^{-3}} = \frac{9^{-1} x y^{-4/3}}{x^{-3} y^3} = 9^{-1} x^{4} y^{-13/3} = \boxed{\frac{x^4}{9y^{13/3}}}}}$
${\displaystyle{\left( \frac{25a^6 b^{-2}}{16x^4 a^4} \right)^{-\frac{1}{2}} = \left( \frac{5^2 a^6 b^{-2}}{2^4 x^4 a^4} \right)^{-1/2} = \left(\frac{5^2 a^2 b^{-2}}{2^4 x^4}\right)^{-1/2} = \frac{5^{-1} |a|^{-1} b}{2^{-2} x^{-2}} = \boxed{\frac{4|b|x^2}{5|a|}}}}$
${\displaystyle{\left( \frac{16xy^{-2}}{81a^4b^6} \right)^{-\frac{1}{4}} = \left(\frac{2^4 x y^{-2}}{3^4 a^4 b^6} \right)^{-\frac{1}{4}} = \frac{2^{-1} x^{-1/4} |y|^{1/2}}{3^{-1} |a|^{-1} |b|^{-3/2}} = \boxed{\frac{3|a| |b|^{3/2} |y|^{1/2}}{2x^{1/4}}}}}$
${\displaystyle{\left( \frac{27x^2}{w^6} \right)^{-\frac{1}{3}} = \left( \frac{3^3 x^2}{w^6} \right)^{-\frac{1}{3}} = \frac{3^{-1} x^{-2/3}}{w^{-2}} = \boxed{\frac{w^2}{3x^{2/3}}}}}$
${\displaystyle{\left( \frac{16^{-1/4} + (-4)^{-2} - 2^{-3}}{8^{1/3}} \right)^{-1} = \left(\frac{\frac{1}{2} + \frac{1}{16} - \frac{1}{8}}{2} \right)^{-1} = \left(\frac{\frac{8}{16} + \frac{1}{16} - \frac{2}{16}}{2}\right)^{-1} = \boxed{\frac{32}{7}}}}$
Simplify by converting to rational exponent notation first -- converting to radical notation (if needed) only in your last step.
$\displaystyle{\sqrt[4]{81} + \sqrt[5]{-32}}$
$\displaystyle{(\sqrt{3} - \sqrt{7})(\sqrt{3} + \sqrt{7})}$
$\displaystyle{(5x^2 - \sqrt{2})^2}$
$\displaystyle{8\sqrt{5} + \frac{25}{\sqrt{5}}}$
$\displaystyle{3\sqrt{32} - 2\sqrt{18} - \sqrt{8}}$
$\displaystyle{\frac{\sqrt[3]{(x+1)^4} \sqrt{(x+1)^3}} {\sqrt[6]{(x+1)^5}}}$
$\displaystyle{\frac{\sqrt[3]{(a+b)^5} \cdot \sqrt[4]{(a+b)^2}}{\sqrt{(a+b)^3}}}$
$\displaystyle{\sqrt[4]{\frac{x^8 y^4 z^{16}}{w^8}}}$
$\displaystyle{\sqrt[3]{a^4} - \frac{\sqrt[3]{a^2}}{\sqrt[3]{a^{-1}}}+\sqrt[3]{\frac{27}{a^{-1}}}}$
${\displaystyle{\sqrt[4]{81} + \sqrt[5]{-32} = 3 + (-2) = \boxed{1}}}$
${\displaystyle{ \begin{array}[t]{rcl} (\sqrt{3} - \sqrt{7})(\sqrt{3} + \sqrt{7}) &=& (3^{1/2} - 7^{1/2})(3^{1/2} + 7^{1/2})\\ &=& (3^{1/2} - 7^{1/2})3^{1/2} + (3^{1/2} - 7^{1/2})7^{1/2}\\ &=& 3 - 3^{1/2}7^{1/2} + 3^{1/2}7^{1/2} - 7\\ &=& \boxed{-4} \end{array}}}$
${\displaystyle{ \begin{array}[t]{rcl} (5x^2 - \sqrt{2})^2 &=& (5x^2 - 2^{1/2})^2\\ &=& (5x^2 - 2^{1/2})(5x^2 - 2^{1/2})\\ &=& (5x^2 - 2^{1/2})5x^2 - (5x^2 - 2^{1/2})2^{1/2}\\ &=& 25x^4 - 5x^2 2^{1/2} - (5x^2 2^{1/2} - 2)\\ &=& 25x^4 - 5x^2 2^{1/2} - 5x^2 2^{1/2} + 2\\ &=& 25x^4 - 10x^2 2^{1/2} + 2\\ &=& \boxed{25x^4 - 10x^2 \sqrt{2} + 2} \end{array}}}$
$\displaystyle{8\sqrt{5} + \frac{25}{\sqrt{5}} = 8 \cdot 5^{1/2} + \frac{5^2}{5^{1/2}} = 8 \cdot 5^{1/2} + 5^{3/2} = 8 \cdot 5^{1/2} + 5 \cdot 5^{1/2} = 13 \cdot 5^{1/2} = \boxed{13\sqrt{5}}}$
$\displaystyle{ \begin{array}[t]{rcl} 3\sqrt{32} - 2\sqrt{18} - \sqrt{8} &=& 3 \cdot (2^5)^{1/2} - 2 \cdot (2 \cdot 3^2)^{1/2} - (2^3)^{1/2}\\ &=& 3 \cdot 2^{5/2} - 2 \cdot 2^{1/2} \cdot 3 - 2^{3/2}\\ &=& 3 \cdot 2^2 \cdot 2^{1/2} - 6 \cdot 2^{1/2} - 2 \cdot 2^{1/2}\\ &=& 12 \cdot 2^{1/2} - 6 \cdot 2^{1/2} - 2 \cdot 2^{1/2}\\ &=& 4 \cdot 2^{1/2}\\ &=& \boxed{4\sqrt{2}} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \cfrac{\sqrt[3]{(x+1)^4} \sqrt{(x+1)^3}} {\sqrt[6]{(x+1)^5}} &=& \cfrac{(x+1)^{4/3} (x+1)^{3/2}}{(x+1)^{5/6}}\\ &=& (x+1)^{\frac{4}{3} + \frac{3}{2} - \frac{5}{6}}\\ &=& \boxed{(x+1)^2} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \cfrac{\sqrt[3]{(a+b)^5} \cdot \sqrt[4]{(a+b)^2}}{\sqrt{(a+b)^3}} &=& \cfrac{(a+b)^{5/3} \cdot (a+b)^{1/2}}{(a+b)^{3/2}}\\ &=& (a+b)^{\frac{5}{3} + \frac{1}{2} - \frac{3}{2}}\\ &=& (a+b)^{\frac{2}{3}}\\ &=& \boxed{\sqrt[3]{(a+b)^2}} \end{array}}$
${\displaystyle{\sqrt[4]{\frac{x^8 y^4 z^{16}}{w^8}} = \left(\frac{x^8 y^4 z^{16}}{w^8} \right)^{\frac{1}{4}} = \boxed{\frac{x^2 |y| \,z^4}{w^2}}}}$
$\displaystyle{ \begin{array}[t]{rcl} \sqrt[3]{a^4} - \cfrac{\sqrt[3]{a^2}}{\sqrt[3]{a^{-1}}}+\sqrt[3]{\cfrac{27}{a^{-1}}} &=& a^{4/3} - \cfrac{a^{\frac{2}{3}}}{a^{\frac{-1}{3}}} + \left( \frac{3^3}{a^{-1}} \right)^{1/3}\\ &=& (a \cdot a^{1/3}) - a + 3a^{1/3}\\ &=& \boxed{a \sqrt[3]{a} - a + 3\sqrt[3]{a}} \end{array}}$
Rework problem #1 (all parts), but by converting to radical notation first -- converting back to rational exponents only in your last step.
All the answers should be the same as they were before.
Rework problem #2 (all parts), but without ever appealing to rational exponents.
All the answers should be the same as they were before.
Simplify each, making sure to leave no radicals in any denominator.
$\displaystyle{\frac{\sqrt[3]{12}}{\sqrt[3]{9}}}$
$\displaystyle{\frac{5}{\sqrt[3]{2xy^2}}}$
$\displaystyle{\frac{\sqrt[3]{xy}}{\sqrt[3]{9ab^2}}}$
${\displaystyle{\frac{\sqrt[3]{12}}{\sqrt[3]{9}} = \frac{(2^2 \cdot 3)^{1/3}}{(3^2)^{1/3}} = 2^{\frac{2}{3}} \cdot 3^{\frac{1}{3}-\frac{2}{3}} = 2^{2/3} \cdot 3^{-1/3} = 2^{2/3} \cdot 3^{-1} \cdot 3^{2/3} = \frac{\sqrt[3]{2^2 \cdot 3^2}}{3} = \boxed{\frac{\sqrt[3]{36}}{3}}}}$
Although, this is probably quicker using radicals: $\displaystyle{\frac{\sqrt[3]{12}}{\sqrt[3]{9}} = \frac{\sqrt[3]{12}}{\sqrt[3]{3^2}} \cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}} = \frac{\sqrt[3]{36}}{\sqrt[3]{3^3}} = \frac{\sqrt[3]{36}}{3}}$
$\displaystyle{\frac{5}{\sqrt[3]{2xy^2}} = \frac{5}{\sqrt[3]{2xy^2}} \cdot \frac{\sqrt[3]{2^2 x^2 y}}{\sqrt[3]{2^2 x^2 y}} = \boxed{\frac{5\sqrt[3]{4x^2y}}{2xy}}}$
$\displaystyle{\frac{\sqrt[3]{xy}}{\sqrt[3]{9ab^2}} = \frac{\sqrt[3]{xy}}{\sqrt[3]{9ab^2}} \cdot \frac{\sqrt[3]{3 a^2 b}}{\sqrt[3]{3 a^2 b}} = \boxed{\frac{\sqrt[3]{3a^2bxy}}{3ab}}}$
Write each expression below using only a single radical.
$\displaystyle{\frac{\sqrt{(2x-3)^3} \cdot \sqrt[4]{2x-3}}{\sqrt[3]{(2x-3)^4}}}$
$\displaystyle{\frac{\left(\sqrt{x}\right)^2}{x \, \sqrt[17]{\sqrt[3]{x} \cdot \sqrt{x^5}}}}$
$\displaystyle{\sqrt{x^3 \sqrt[3]{x^{10} y^4 \sqrt[5]{z}}}}$
${\displaystyle{\frac{\sqrt{(2x-3)^3} \cdot \sqrt[4]{2x-3}}{\sqrt[3]{(2x-3)^4}} = (2x-3)^{\frac{3}{2} + \frac{1}{4} - \frac{4}{3}} = (2x-3)^{\frac{5}{12}} = \boxed{\sqrt[12]{(2x-3)^5}}}}$
$\displaystyle{\frac{\left(\sqrt{x}\right)^2}{x \, \sqrt[17]{\sqrt[3]{x} \cdot \sqrt{x^5}}} = \frac{x}{x \cdot (x^{\frac{1}{3}} \cdot x^{\frac{5}{2}})^{\frac{1}{17}}} = \frac{1}{(x^{\frac{17}{6}})^{\frac{1}{17}}} = \boxed{\frac{1}{\sqrt[6]{x}}}}$
$\sqrt{x^3 \sqrt[3]{x^{10} y^4 \sqrt[5]{z}}} = x^{\frac{3}{2}} x^{\frac{10}{6}} y^{\frac{4}{6}} z^{\frac{1}{30}} = x^{\frac{3}{2} + \frac{10}{6}} y^{\frac{20}{30}} z^{\frac{1}{30}} = x^{\frac{95}{30}} y^{\frac{20}{30}} z^{\frac{1}{30}} =\\ x^{3 + \frac{5}{30}} y^{\frac{20}{30}} z^{\frac{1}{30}} = \boxed{x^3 \sqrt[30]{x^5 y^{20} z}}$
Explain the difference between the two expressions $\displaystyle{\sqrt{{}^* 25} \textrm{ and } \sqrt{25}}$
$\sqrt{{}^* 25}$ is the set of all square roots of $25$, and is given by $\{-5,5\}$, while $\sqrt{25}$ is the principle square root of $25$ (i.e., the positive one), which is simply $5$.
Approximate each square root below using the Babylonian method and the given initial guess $x_1$. In finding this approximation, only calculate three "iterates" (i.e., find $x_2$, $x_3$, and then stop after finding $x_4$). Compare the result with what the square root button on your calculator finds. How close are you in each case?
$\displaystyle{\sqrt{5}, \quad x_1 = 2}$
$\displaystyle{\sqrt{113}, \quad x_1 = 10}$ (an initial underestimate)
$\displaystyle{\sqrt{300}, \quad x_1 = 20}$ (an initial overestimate)
$x_4 = 2.236067978$ (versus $2.236067977$)
$x_4 = 10.63014581$ (versus $10.63014581$)
$x_4 = 17.3205081$ (versus $17.32050808$)
What happens in the Babylonian Method for approximating the square root of $N$ if your initial guess $x_1$ actually equals $\sqrt{N}$ exactly? What is the value of the next iterate, $x_2$?
The following "proof without words" is meant to justify why the arithmetic mean $(a+b)/2$ of two positive values $a$ and $b$ is always greater than or equal to their geometric mean, $\sqrt{ab}$, but less than or equal to the larger of $a$ and $b$. Explain in detail how it does this.
Can you use similar triangles to establish the length of segment $GQ$ in the image? Also, how does the length of $AO$ compare to the length of $PO$? ..and to the length of the larger of the two segments $PQ$ and $QR$?
Prove the following are irrational values.
Construct an argument similar to that used earlier to prove $\sqrt{2}$ is irrational.