1. 2 b) -1 c) 10 d) -22 e) -3 f) -1
    1. $ x^{1/2} $ b) $ y^{1/3} $ c) $ 15^{1/5} $ d) $ z^{5/3} $ e) $ t^{3/10} $ f) $ a^{9} $ g) $ (ab)^{1/n} $
    1. $ $ b) $ $ c) $ $ d) $ $ e) $ $ f) $ 1/ $
    1. $ 2 $ b) $ 5 $ c) $ 6 $ d) 14 e) $ 4 $
    1. No b) False
    1. No b) False

\[ 41.\ a)\ 4/5\quad b)\ -10\quad c)\ 5\sqrt{2}/6\quad d)\ 3\sqrt{2}\quad e)\ 3/2\quad f)\ 216\quad g)\ 16\quad h)\ 36\quad i)\ 100,000\quad j)\ 49\quad k)\ 7a^{4}/b^{2} \]

  1. xy m) x - y n) $ a^{8/9} $ o) xy

\[ 42.a)\sqrt{2}\quad b)5\sqrt{6}\quad c)2\sqrt{7}\quad d)\frac{3x(3-\sqrt{x})}{9-x}\quad e)\sqrt{7}-\sqrt{5}\quad f)5+2\sqrt{6}\quad g)\frac{5}{3\sqrt{3}}\quad h)\frac{4-x}{10\sqrt{x}+5x} \]

Chapter 3

  1. The expression $ (-2) $ /-5 is shorthand for two numbers: $ (-2+)/(-5) $ and $ (-2-)/(-5) $ . Multiplying their tops and bottoms by -1 (to make their denominators positive), we can rewrite them as $ (2-)/5 $ and $ (2+)/5 $ . We can then describe these numbers together by the shorthand $ (2)/5 $ . Thus, $ (-2)/(-5)=(2)/5 $ as claimed.

\[ 6.a)x=16/3\qquad b)u=49/5\qquad c)x=-26/3\qquad d)x=\pm11/4\qquad e)x=-1,5\qquad f)z=-3,4 \]

\[ g)x=\frac{-15\pm\sqrt{281}}{4}\quad h)t=\frac{1\pm\sqrt{5}}{2}\qquad i)x=-1,5/2\qquad j)w=\frac{-5\pm\sqrt{13}}{6} \]

\[ \mathsf{k})x=\left(-\sqrt[3]{7}\pm\sqrt[3]{49}+8\pi\right)/2\pi\qquad\mathsf{l})x=-1\pm\sqrt{2}\qquad\mathsf{m})x=-1\pm\sqrt{7}\qquad\mathsf{n})no solutions\qquad\mathsf{o})x=1 \]

7c. i) $ x = $ ii) No solution iii) x = 5, -2

8c. i) $ x = - $ ii) $ x = $ iii) $ x = $ 9. $ x = $

  1. The true solutions: $ x = $ . 12b. i) $ x = $ ii) $ x = $ iii) $ x = $

  2. 2567, 2568 14. $ 2 $ inches $ ^{2} $ 15. $ $ 16. 6 by $ $ furlongs 17. $ 300 $ cm $ ^{2} $ 18. 5:33 pm

  3. $ $ 20. The pieces should be 1 and 4 units. 21. 41 and 35.

  4. $ 6 + $ on each side. $ 23. $ sec. 24. He is summoning the stork.

\[ x=\frac{-1\pm\sqrt{5}}{2}\quad b)x=8/5,5\quad c)x=\frac{-1\pm\sqrt{3}}{2}\quad d)t=-3\quad e)z=-1\quad f)x=\frac{6\pm\sqrt{6}}{3}\quad g)x=\pm12 \]

\[ h)x=4\quad i)x=-1\quad j)x=-28\quad k)w=2\quad l)x=2,3 \]

29a. Our original equation $ (x = 1) $ had one solution, 1. When we multiplied its sides by x, we quietly introduced an additional - and bogus - solution, 0. (Go back and read the section’s footnotes, and you’ll understand.) Thus, the variable x now secretly carried two numerical values: 1 (correctly), and 0 (incorrectly). Later, when we divided both sides by $ (x - 1) $ , we accidentally left 1 in our “blind spot”. After we dropped the 1 this way, x held only one value again – namely, the bogus solution, 0.

\[ \mathsf{f})-3(x-1/8)^{2}+(3/64)\quad\mathsf{g})(x+3/2)^{2}-5/4\quad\mathsf{h})2(x-1)^{2}+2 \]