DiagnosticTestAlgebra < MSC

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---+ Algebra Diagnostic Test

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---++ Algebra

__Combine and simplify expressions as much as possible.%BR%Solve equations for x.__


---+++ !!1.  <latex>\mbox{\small $\frac{1}{a+b} - \frac{2a}{a^2-b^2}$}</latex>

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<latex>=\mbox{ \small $\frac{1}{b-a}$ }</latex>

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<latex>\mbox{ \small $\frac{1}{a+b} - \frac{2a}{a^2-b^2}$ }</latex>

<latex>=\mbox{ \small $\frac{1}{a+b} \cdot \frac{a-b}{a-b} - \frac{2a}{(a+b)(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{a-b}{(a+b)(a-b)} - \frac{2a}{(a+b)(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{a-b-2a}{(a+b)(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{-b-a}{(a+b)(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{-(b+a)}{(a+b)(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{-(a+b)}{(a+b)(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{-1}{(a-b)}$ }</latex>

<latex>=\mbox{ \small $\frac{1}{b-a}$ }</latex>

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---+++ !! 2. <latex>\mbox{ \small $\frac{x^2 + 2x + 1}{2x^2} \div \frac{x+1}{x+2}$ }</latex> 
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<latex>=\mbox{ \small $\frac{ x^2+3x+2 }{2x^2}$ }</latex>

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<latex>\mbox{ \small $\frac{x^2 + 2x + 1}{2x^2} \div \frac{x+1}{x+2}$ }</latex>

<latex>=\mbox{ \small $\frac{(x+1)(x+1)}{2x^2}  \cdot \frac{x+2}{x+1}$ }</latex>

<latex>=\mbox{ \small $\frac{(x+1)}{2x^2} \cdot \frac{(x+2)}{1}$ }</latex>

<latex>=\mbox{ \small $\frac{(x+1)(x+2)}{2x^2}$ }</latex>

<latex>=\mbox{ \small $\frac{ x^2+3x+2 }{2x^2}$ }</latex>

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---+++ !!3.  <latex>\mbox{ \small $-\frac{a+b}{ac+bd}$ }</latex>
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<latex>=\mbox{ \small $-\frac{a+b}{ac+bd}$ }</latex>

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   * Cannot be simplified.
   * _however,_ if we are told further that <latex>\mbox{ \small $d = c$}</latex> then we have a common factor in the denominator:
      * <latex>\mbox{ \small $-\frac{a+b}{ac+bd}$ }</latex>
      * <latex>=\mbox{ \small $-\frac{a+b}{ac+bc}$ }</latex>
      * <latex>=\mbox{ \small $ -\frac{(a+b)}{(a+b)c}$ }</latex>
      * <latex>=\mbox{ \small $ -\frac{1}{c}$ }</latex>, provided <latex>\mbox{ \small $ (a+b) \ne 0$ }</latex>

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---+++ !!4. <latex>\mbox{ \small $\frac{(2a)^3}{a^5}$ }</latex>
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<latex>=\mbox{ \small $\frac{8}{a^2}$ }</latex>

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<latex>=\mbox{ \small $\frac{2^3a^3}{a^2a^3}$ }</latex>

<latex>=\mbox{ \small $\frac{8}{a^2}$ }</latex>

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---+++ !!5.  <latex>\mbox{ \small $(0.2a^2)^4$ }</latex>
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<latex>=\mbox{ \small $=.0016 a^{8}$ }</latex>

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<latex>=\mbox{ \small $(0.2)^4 \cdot (a^2)^4$ }</latex>

<latex>=\mbox{ \small $(2 \cdot .1)^4 \cdot a^{2*4}$ }</latex>

<latex>=\mbox{ \small $2^4 \cdot 10^{-4} \cdot a^{8}$ }</latex>

<latex>=\mbox{ \small $16 \cdot 10^{-4} a^{8}$ }</latex>

<latex>=\mbox{ \small $=.0016 a^{8}$ }</latex>

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---+++ !!6. <latex>\mbox{ \small $\frac{8y^n}{-2y^{n-1}}$ }</latex>
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<latex>=\mbox{ \small $-\frac{8 \cdot y}{2} $ }</latex>

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<latex>=\mbox{ \small $\frac{8y^n}{-2y^{n-1}} \cdot \frac{y}{y}$ }</latex>

<latex>=\mbox{ \small $\frac{8(y^n) \cdot y}{-2(y^{n})} $ }</latex>

<latex>=\mbox{ \small $\frac{8 \cdot y}{-2} $ }</latex>

<latex>=\mbox{ \small $-\frac{8 \cdot y}{2} $ }</latex>

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---+++ !!7. <latex>\mbox{ \small $\sqrt[3]{-64y^{27}}$ }</latex>
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<latex>=\mbox{ \small $ -4 y^{9}$ }</latex>

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<latex>=\mbox{ \small $ \sqrt[3]{-1} \cdot \sqrt[3]{64} \cdot \sqrt[3]{y^{27}} $ }</latex>

<latex>=\mbox{ \small $ \sqrt[3]{-1} \cdot \sqrt[3]{64} \cdot y^{\frac{27}{3}}$ }</latex>

<latex>=\mbox{ \small $ -1 \cdot 4 \cdot y^{9}$ }</latex>

<latex>=\mbox{ \small $ -4 y^{9}$ }</latex>

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---+++ !!8. <latex>\mbox{ \small $\sqrt{a^2 + b^2}$ }</latex>
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<latex>=\mbox{ \small $\sqrt{a^2 + b^2}$ }</latex>

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<latex>=\mbox{ \small $\sqrt{a^2 + b^2}$ }</latex>
   * No _algebraic_ simplification.
   *  _however..._ A beautiful _geometric_ simplification is possible:
      * <latex>\mbox{ \small $\sqrt{a^2 + b^2}$ }</latex> is the hypotenuse of the right triangle with sides <latex>a</latex> and <latex>b</latex>

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---+++ !!9. <latex>\mbox{ \small $(  a + b  )^3 $ }</latex>
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<latex>=\mbox{ \small $   a^3 + 3 a^2  b + 3 a b^2  + b^3  $ }</latex>

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<latex>=\mbox{ \small $( a + b ) \cdot (  a + b  )^2 $ }</latex>

<latex>=\mbox{ \small $( a + b ) \cdot (  a^2 + 2 a b + b^2 ) $ }</latex>

<latex>=\mbox{ \small $ a (  a^2 + 2 a b + b^2 ) + b (  a^2 + 2 a b + b^2 ) $ }</latex>

<latex>=\mbox{ \small $   a^3 + 2 a^2  b + a b^2  +  b a^2 + 2 a b^2 + b^3  $ }</latex>

<latex>=\mbox{ \small $   a^3 + 3 a^2  b + 3 a b^2  + b^3  $ }</latex>

---++++ !!Alternatively:
Recall Pascal's Triangle:

<pre>
                1    1
             1    2    1
          1    3    3    1
       1    4     6    4     1
    1    5   10   10    5    1
</pre>

...in which elements in a row are the sum of the two elements in the row above it. These numbers are the _Binomial Coefficients._
<latex>\mbox{ \small $(  a + b  )^3 $ }</latex> is the cube of the binomial <latex>\mbox{ \small $(  a + b  ) $ }</latex> so the coefficients can be taken from the third row above:
=1  3  3  1= and the exponents on <latex>\mbox{ \small $  a $ }</latex> start with <latex>\mbox{ \small $  3 $ }</latex> and count down, while the exponents on <latex>\mbox{ \small $  b $ }</latex> 
start with <latex>\mbox{ \small $  0 $ }</latex> and count up:

<latex>=\mbox{ \small $  1\cdot a^3b^0 + 3 \cdot a^2b^1 + 3 \cdot a^1b^2 + 1\cdot a^0b^3  $ }</latex> which simplifies to:

<latex>=\mbox{ \small $   a^3 + 3 a^2  b + 3 a b^2  + b^3  $ }</latex>

as previously.

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---+++ !!10. <latex>\mbox{ \small $( \sqrt{x} + 3\sqrt{y} ) ( \sqrt{x} - \sqrt{y} )$ }</latex>
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<latex>=\mbox{ \small $ x + 2 \sqrt{x \cdot y}  - 3 y $ }</latex>

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<latex>=\mbox{ \small $ \sqrt{x}   ( \sqrt{x} - \sqrt{y} ) + 3\sqrt{y}   ( \sqrt{x} - \sqrt{y} )$ }</latex>

<latex>=\mbox{ \small $ ( \sqrt{x}   \sqrt{x}  -  \sqrt{x}  \sqrt{y} ) + (3\sqrt{y} \sqrt{x} - 3  \sqrt{y} \sqrt{y}  )$ }</latex>

<latex>=\mbox{ \small $ x -  \sqrt{x}  \sqrt{y}  + 3\sqrt{y}  \sqrt{x} - 3  y $ }</latex>

<latex>=\mbox{ \small $ x + 2 \sqrt{x \cdot y}  - 3 y $ }</latex>

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---+++ !!11. <latex>\mbox{ \small $x^3 - x^2 - 6x = 0$ }</latex>
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either <latex>\mbox{ \small $x=3$ }</latex> or <latex>\mbox{ \small $x=-2$ }</latex> or <latex>\mbox{ \small $x=0$ }</latex>

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<latex>0 = \mbox{ \small $x^3 - x^2 - 6x $ }</latex>

<latex> = \mbox{ \small $(x^2 - x - 6+)(x)$ }</latex>

<latex> = \mbox{ \small $(x-3) \cdot (x+2) \cdot (x) $ }</latex>

if a product is _zero_ one of its factors must be _zero_ so:

either <latex>\mbox{ \small $ (x-3)  = 0$ }</latex> or <latex>\mbox{ \small $ (x+2)  = 0$ }</latex> or <latex>\mbox{ \small $ (x) = 0$ }</latex> 

and either <latex>\mbox{ \small $x=3$ }</latex> or <latex>\mbox{ \small $x=-2$ }</latex> or <latex>\mbox{ \small $x=0$ }</latex>

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---+++ !!12. <latex>\mbox{ \small $x^2 + 7x = -3$ }</latex>
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<latex>  = \mbox{ \small $\frac{ -7\pm\sqrt{37} }{ 2 }$ } </latex>

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<latex>-3 = \mbox{ \small $x^2 + 7x $ }</latex>

<latex> 0 = \mbox{ \small $x^2 + 7x + 3 $ }</latex>

<latex> x = \mbox{ \small $\frac{ -7\pm\sqrt{7^2-4 \cdot 1 \cdot 3} }{ 2 \cdot 1 }$ } </latex>

<latex>  = \mbox{ \small $\frac{ -7\pm\sqrt{49-12} }{ 2 }$ } </latex>

<latex>  = \mbox{ \small $\frac{ -7\pm\sqrt{37} }{ 2 }$ } </latex>

   * Rearrange so the equation is of the form:
   * <latex>\mbox{ \small $a x^2 + bx + c = 0$ }</latex>
   * and use the _Quadratic Formula:_
   * <latex> x = \mbox{ \small $\frac{ -b\pm\sqrt{b^2-4ac} }{ 2a }$ } </latex>

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-- Main.DickFurnas - 21 Apr 2007
Topic revision: r4 - 2009-01-26 - DickFurnas