[latexpage]

Midterm Three Review

1. 1. \(\dfrac{6\cancel{(a-b)}}{(a+b)\cancel{(a^2-ab+b^2)}}\cdot \dfrac{\cancel{a^2-ab+b^2}}{(a+b)\cancel{(a-b)}}\Rightarrow \dfrac{b}{(a+b)^2}\)
   2. \(\begin{array}{l}
      \\ \\ \\ \\ \\ \\ \\ \\ \\
      \dfrac{x}{(x+5)(x-5)}-\dfrac{2}{(x-5)(x-1)} \\ \\
      \text{LCD}=(x+5)(x-5)(x-1) \\ \\
      \therefore \dfrac{x(x-1)-2(x+5)}{(x+5)(x-5)(x-1)}\Rightarrow \dfrac{x^2-x-2x-10}{(x+5)(x-5)(x-1)}\Rightarrow \dfrac{x^2-3x-10}{(x+5)(x-5)(x-1)} \\ \\
      \Rightarrow \dfrac{\cancel{(x-5)}(x+2)}{(x+5)\cancel{(x-5)}(x-1)}\Rightarrow \dfrac{x+2}{(x+5)(x-1)}
      \end{array}\)
   3. \(\dfrac{\left(1-\dfrac{6}{x}\right)x^2}{\left(\dfrac{4}{x}-\dfrac{24}{x^2}\right)x^2}\Rightarrow \dfrac{x^2-6x}{4x-24}\Rightarrow \dfrac{x\cancel{(x-6)}}{4\cancel{(x-6)}}\Rightarrow \dfrac{x}{4}\)
   4. \(\phantom{1}\)
      \(\left(\dfrac{4}{x+4}-\dfrac{5}{x-2}=5\right)(x+4)(x-2) \\ \)
      \(\begin{array}{rrrrrrrrrrcrr}
      4(x&-&2)&-&5(x&+&4)&=&5(x&+&4)(x&-&2) \\
      4x&-&8&-&5x&-&20&=&5(x^2&+&2x&-&8) \\
      &&&&-x&-&28&=&5x^2&+&10x&-&40 \\
      &&&&+x&+&28&&&+&x&+&28 \\
      \midrule
      &&&&&&0&=&5x^2&+&11x&-&12 \\ \\
      &&&&&&0&=&5x^2&+&15x-4x&-&12 \\
      &&&&&&0&=&5x(x&+&3)-4(x&+&3) \\
      &&&&&&0&=&(x&+&3)(5x&-&4) \\ \\
      &&&&&&x&=&-3,&\dfrac{4}{5}&&& \\
      \end{array}\)
   5. True
   6. False
   7. \(\begin{array}{l}
      \\ \\
      4\cdot 6+3\sqrt{36\cdot 2}+4 \\
      24+3\cdot 6\sqrt{2}+4 \\
      28+18\sqrt{2}
      \end{array}\)
   8. \(\dfrac{\sqrt{\cancel{300}100a^{\cancel{5}4}\cancel{b^2}}}{\sqrt{\cancel{3}\cancel{a}\cancel{b^2}}}\Rightarrow \sqrt{100a^4}\Rightarrow 10a^2\)
   9. \(\dfrac{(12)(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\Rightarrow \dfrac{36+12\sqrt{6}}{9-6}\Rightarrow \dfrac{\cancel{36}12+\cancel{12}4\sqrt{6}}{\cancel{3}1}\Rightarrow 12+4\sqrt{6}\)
   10. \(\left(\dfrac{\cancel{a^0}1b^3}{c^6d^{-12}}\right)^{\frac{1}{3}}\Rightarrow \left(\dfrac{b^3d^{12}}{c^6}\right)^{\frac{1}{3}}\Rightarrow \dfrac{bd^4}{c^2}\)
   11. \(\begin{array}{rrl}
       \\ \\ \\ \\ \\
       (\sqrt{5x-6})^2& =& (x)^2 \\
       5x-6&=&x^2 \\
       0&=&x^2-5x+6 \\
       0&=&(x-3)(x-2) \\ \\
       x&=&3,2
       \end{array}\)
   12. \(\begin{array}{rrcrrrrrr}
       \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
       \sqrt{2x+9}&+&3&=&x&&&& \\
       &-&3&&&-&3&& \\
       \midrule
       &&\sqrt{2x+9}&=&x&-&3&& \\ \\
       &&(\sqrt{2x+9})^2&=&(x&-&3)^2&& \\
       2x&+&9&=&x^2&-&6x&+&9 \\
       -2x&-&9&&&-&2x&-&9 \\
       \midrule
       &&0&=&x^2&-&8x&& \\
       &&0&=&x(x&-&8)&& \\ \\
       &&x&=&0,&8&&&
       \end{array}\)
   13. \(\phantom{1}\)
       \((\sqrt{x-3})^2=(\sqrt{2x-5})^2 \\ \)
       \(\begin{array}{rrrrrrrr}
       &x&-&3&=&2x&-&5 \\
       -&x&+&5&&-x&+&5 \\
       \midrule
       &&&2&=&x&&
       \end{array}\)
   14. \(\phantom{1}\)
       1. \(\begin{array}{l}
          \\ \\ \\ \\
          b^2-4ac \\
          =(4)^2-4(2)(3) \\
          =16-24 \\
          =-8 \\
          \text{2 non-real solutions}
          \end{array}\)
       2. \(\begin{array}{l}
          \\ \\ \\ \\
          b^2-4ac \\
          =(-2)^2-4(3)(-8) \\
          =4+96 \\
          =100 \\
          \text{2 real solutions}
          \end{array}\)
   15. \(\phantom{1}\)
       1. \(\begin{array}{rrl}
          \\ \\ \\
          \dfrac{3x^2}{3}&=&\dfrac{27}{3} \\
          x^2&=&9 \\
          x&=&\pm 3
          \end{array}\)
       2. \(\begin{array}{rrl}
          \\ \\
          2x^2-16x&=&0 \\
          2x(x-8)&=&0 \\
          x&=&0,8
          \end{array}\)
   16. \(\phantom{1}\)
       1. \((x-4)(x+3)\Rightarrow x=4,-3\)
       2. \(\begin{array}{rrl}
          \\ \\
          x^2+9x+8&=&0 \\
          (x+8)(x+1)&=&0 \\
          x&=&-1,-8
          \end{array}\)
   17. \(\left(\dfrac{x-3}{2}+\dfrac{6}{x+3}=1\right)(2)(x+3) \\ \)
       \(\begin{array}{rrrrcrrrrrr}
       (x&-&3)(x&+&3)&+&6(2)&=&2(x&+&3) \\
       x^2&&&-&9&+&12&=&2x&+&6 \\
       &-&2x&&&-&6&&-2x&-&6 \\
       \midrule
       &&x^2&-&2x&-&3&=&0&& \\
       &&(x&-&3)(x&+&1)&=&0&& \\ \\
       &&&&&&x&=&3,&-1&
       \end{array}\)
   18. \(\left(\dfrac{x-2}{x}=\dfrac{x}{x+4}\right)(x)(x+4) \\ \)
       \(\begin{array}{rrrcrrrl}
       &(x&-&2)(x&+&4)&=&x^2 \\
       &x^2&+&2x&-&8&=&x^2 \\
       -&x^2&&&+&8&&-x^2+8 \\
       \midrule
       &&&&&\dfrac{2x}{2}&=&\dfrac{8}{2} \\ \\
       &&&&&x&=&4
       \end{array}\)
   19. \(\phantom{1}\)
       \(\text{width}=W\hspace{0.5in}\text{length}=L=3+2W \\ \)
       \(\begin{array}{rrl}
       A&=&L\cdot W \\
       65&=&W(3+2W) \\
       65&=&3W+2W^2 \\ \\
       0&=&2W^2+3W-65 \\
       0&=&2W^2-10W+13W-65 \\
       0&=&2W(W-5)+13(W-5) \\
       0&=&(W-5)(2W+13) \\ \\
       W&=&5, \cancel{-\dfrac{13}{2}} \\ \\
       L&=&3+2W \\
       L&=&13
       \end{array}\)
   20. \(\phantom{1}\)
       \(x, x+2, x+4 \\ \)
       \(\begin{array}{rrcrrrrrrrl}
       &&x(x&+&2)&=&68&+&x&+&4 \\
       x^2&+&2x&&&=&x&+&72&& \\
       &-&x&-&72&&-x&-&72&& \\
       \midrule
       x^2&+&x&-&72&=&0&&&& \\ \\
       (x&+&9)(x&-&8)&=&0&&&& \\
       &&&&x&=&\cancel{-9},&8&&&
       \end{array}\)
       ∴ 8, 10, 12
   21. \(\phantom{1}\)
       \(d=r\cdot t\text{ and }d_{\text{up}}=d_{\text{down}} \\ \)
       \(\begin{array}{rrrrrrcr}
       &8(r&-&4)&=&6(r&+&4) \\
       &8r&-&32&=&6r&+&24 \\
       -&6r&+&32&&-6r&+&32 \\
       \midrule
       &&&2r&=&56&& \\
       &&&r&=&28&\text{km/h}& \\
       \end{array}\)
   22. \(\begin{array}{rrl}
       \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
       A&=&\dfrac{1}{2}bh \\ \\
       (330&=&\dfrac{1}{2}(h+8)h)(2) \\ \\
       660&=&h^2+8h \\ \\
       0&=&h^2+8h-660 \\
       0&=&h^2+30h-22h-660 \\
       0&=&h(h+30)-22(h+30) \\ \\
       0&=&(h+30)(h-22) \\
       h&=&\cancel{-30}, 22 \\ \\
       \therefore b&=&h+8 \\
       &=&22+8 \\
       &=&30
       \end{array}\)