# Midterm Three Review

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