Solution 2.2:3d

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Current revision (10:18, 24 September 2008) (edit) (undo)
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(The exercise is taken from an actual exam in Spring Term 1945!)
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There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side
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There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side:
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<math>\begin{align}
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& \left( \frac{2}{x}-3 \right)\left( \frac{1}{4x}+\frac{1}{2} \right)=\frac{2}{x}\centerdot \frac{1}{4x}-\frac{2}{x}\centerdot \frac{1}{2}-3\centerdot \frac{1}{4x}-3\centerdot \frac{1}{2} \\
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& =\frac{1}{2x^{2}}+\frac{1}{x}-\frac{3}{4x}-\frac{3}{2}=\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}, \\
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& \\
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& \left( \frac{1}{2x}-\frac{2}{3} \right)^{2}=\frac{1}{\left( 2x \right)^{2}}-2\centerdot \frac{1}{2x}\centerdot \frac{2}{3}+\left( \frac{2}{3} \right)^{2} \\
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& =\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}, \\
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& \\
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& \\
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\end{align}</math>
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<math>\begin{align}
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& \left( \frac{1}{2x}+\frac{1}{3} \right)\left( \frac{1}{2x}-\frac{1}{3} \right)=\left\{ \text{conjugate rule} \right\}=\frac{1}{\left( 2x \right)^{2}}-\frac{1}{3^{2}} \\
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& =\frac{1}{4x^{2}}-\frac{1}{9} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\biggl(\frac{2}{x}-3\biggr)\biggl(\frac{1}{4x}+\frac{1}{2}\biggr)
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&= \frac{2}{x}\cdot\frac{1}{4x} - \frac{2}{x}\cdot\frac{1}{2} - 3\cdot\frac{1}{4x} - 3\cdot\frac{1}{2}\\[5pt]
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&= \frac{1}{2x^{2}} + \frac{1}{x} - \frac{3}{4x} - \frac{3}{2}\\[5pt]
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&= \frac{1}{2x^{2}} + \frac{1}{4x} - \frac{3}{2}\,,\\[15pt]
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\biggl(\frac{1}{2x}-\frac{2}{3}\biggr)^{2}
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&= \frac{1}{(2x)^{2}} - 2\cdot\frac{1}{2x}\cdot\frac{2}{3} + \biggl(\frac{2}{3}\biggr)^{2}\\[5pt]
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&= \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\,,\\[15pt]
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\biggl(\frac{1}{2x}+\frac{1}{3}\biggr)\biggl(\frac{1}{2x}-\frac{1}{3}\biggr)
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&= \bigl\{\,\text{difference of two squares}\,\}\\[5pt]
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&= \frac{1}{(2x)^{2}}-\frac{1}{3^{2}}\\[5pt]
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&= \frac{1}{4x^{2}}-\frac{1}{9}\,\textrm{.}
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\end{align}</math>}}
Collecting up terms, the left-hand side becomes
Collecting up terms, the left-hand side becomes
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{{Displayed math||<math>\begin{align}
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&\biggl(\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{1}{9}\biggr)\\[7pt]
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&\qquad\quad{}= \biggl(\frac{1}{2}-\frac{1}{4}-\frac{1}{4}\biggr)\frac{1}{x^{2}} + \biggl(\frac{1}{4}+\frac{2}{3}\biggr)\frac{1}{x} + \biggl(-\frac{3}{2}-\frac{4}{9}+\frac{1}{9}\biggr)\\[7pt]
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&\qquad\quad{}= \frac{2-1-1}{4}\,\frac{1}{x^{2}} + \frac{3+2\cdot 4}{3\cdot 4}\,\frac{1}{x} + \frac{-3\cdot 9 - 4\cdot 2 + 1\cdot 2}{2\cdot 9}\\[7pt]
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&\qquad\quad{}=\frac{11}{3\cdot 4}\cdot \frac{1}{x}-\frac{33}{2\cdot 9}
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\end{align}</math>}}
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<math>\begin{align}
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and because <math>33=3\cdot 11</math>, <math>9=3\cdot 3</math> and <math>4=2\cdot 2</math>, the whole equation can rewritten as
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& \left( \frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2} \right)-\left( \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9} \right)-\left( \frac{1}{4x^{2}}-\frac{1}{9} \right) \\
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& =\left( \frac{1}{2}-\frac{1}{4}-\frac{1}{4} \right)\frac{1}{x^{2}}+\left( \frac{1}{4}+\frac{2}{3} \right)\frac{1}{x}+\left( -\frac{3}{2}-\frac{4}{9}+\frac{1}{9} \right) \\
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& =\frac{2-1-1}{4}\frac{1}{x^{2}}+\frac{3+2\centerdot 4}{3\centerdot 4}\frac{1}{x}+\frac{-3\centerdot 9-4\centerdot 2+1\centerdot 2}{2\centerdot 9} \\
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& =\frac{11}{3\centerdot 4}\centerdot \frac{1}{x}-\frac{33}{2\centerdot 9} \\
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\end{align}</math>
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and because
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<math>33=3\centerdot 11</math>,
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<math>9=3\centerdot 3</math>
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and
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<math>4=2\centerdot 2</math>, the whole equation can rewritten as
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<math>\frac{11}{3\centerdot 2\centerdot 2}\centerdot \frac{1}{x}-\frac{3\centerdot 11}{2\centerdot 3\centerdot 3}=0</math>
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{{Displayed math||<math>\frac{11}{3\cdot 2\cdot 2}\cdot \frac{1}{x}-\frac{3\cdot 11}{2\cdot 3\cdot 3}=0\,\textrm{.}</math>}}
Taking out common factors, we get
Taking out common factors, we get
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{{Displayed math||<math>\frac{11}{3\cdot 2}\biggl(\frac{1}{2x}-1\biggr)=0</math>}}
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<math>\frac{11}{3\centerdot 2}\left( \frac{1}{2x}-1 \right)=0</math>
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and then we see that the equation has the solution <math>x=1/2</math>.
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and then we see that the equation has the solution
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<math>x={1}/{2}\;</math>.
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Finally, we substitute
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<math>x={1}/{2}\;</math>
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into the original equation to check that we have calculated correctly.
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Finally, we substitute <math>x=1/2</math> into the original equation to check that we have calculated correctly.
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \left( \frac{2}{\frac{1}{2}}-3 \right)\left( \frac{1}{4\centerdot \frac{1}{2}}+\frac{1}{2} \right)-\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{2}{3} \right)^{2}-\left( \frac{1}{2\centerdot \frac{1}{2}}+\frac{1}{3} \right)\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{1}{3} \right) \\
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& \biggl(\frac{2}{\frac{1}{2}}-3\biggr)\biggl(\frac{1}{4\cdot\frac{1}{2}} + \frac{1}{2}\biggr) - \biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{2}{3}\biggr)^{2} - \biggl(\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3}\biggr)\biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{1}{3}\biggr)\\[5pt]
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& \\
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&\qquad\quad{}= (4-3)\biggl(\frac{1}{2}+\frac{1}{2}\biggr) - \biggl(1-\frac{2}{3}\biggr)^{2} - \biggl(1+\frac{1}{3}\biggr)\biggl(1-\frac{1}{3}\biggr)\\[5pt]
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& =\left( 4-3 \right)\left( \frac{1}{2}+\frac{1}{2} \right)-\left( 1-\frac{2}{3} \right)^{2}-\left( 1+\frac{1}{3} \right)\left( 1-\frac{1}{3} \right) \\
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&\qquad\quad{}= 1 - \biggl(\frac{1}{3}\biggr)^{2} - \frac{4}{3}\cdot\frac{2}{3}\\[5pt]
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& \\
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&\qquad\quad{}= 1 - \frac{1}{9} - \frac{8}{9}\\[5pt]
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& =1-\left( \frac{1}{3} \right)^{2}-\frac{4}{3}\centerdot \frac{2}{3}=1-\frac{1}{9}-\frac{8}{9}=0 \\
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&\qquad\quad{}= 0\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Current revision

There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side

\displaystyle \begin{align}

\biggl(\frac{2}{x}-3\biggr)\biggl(\frac{1}{4x}+\frac{1}{2}\biggr) &= \frac{2}{x}\cdot\frac{1}{4x} - \frac{2}{x}\cdot\frac{1}{2} - 3\cdot\frac{1}{4x} - 3\cdot\frac{1}{2}\\[5pt] &= \frac{1}{2x^{2}} + \frac{1}{x} - \frac{3}{4x} - \frac{3}{2}\\[5pt] &= \frac{1}{2x^{2}} + \frac{1}{4x} - \frac{3}{2}\,,\\[15pt] \biggl(\frac{1}{2x}-\frac{2}{3}\biggr)^{2} &= \frac{1}{(2x)^{2}} - 2\cdot\frac{1}{2x}\cdot\frac{2}{3} + \biggl(\frac{2}{3}\biggr)^{2}\\[5pt] &= \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\,,\\[15pt] \biggl(\frac{1}{2x}+\frac{1}{3}\biggr)\biggl(\frac{1}{2x}-\frac{1}{3}\biggr) &= \bigl\{\,\text{difference of two squares}\,\}\\[5pt] &= \frac{1}{(2x)^{2}}-\frac{1}{3^{2}}\\[5pt] &= \frac{1}{4x^{2}}-\frac{1}{9}\,\textrm{.} \end{align}

Collecting up terms, the left-hand side becomes

\displaystyle \begin{align}

&\biggl(\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{1}{9}\biggr)\\[7pt] &\qquad\quad{}= \biggl(\frac{1}{2}-\frac{1}{4}-\frac{1}{4}\biggr)\frac{1}{x^{2}} + \biggl(\frac{1}{4}+\frac{2}{3}\biggr)\frac{1}{x} + \biggl(-\frac{3}{2}-\frac{4}{9}+\frac{1}{9}\biggr)\\[7pt] &\qquad\quad{}= \frac{2-1-1}{4}\,\frac{1}{x^{2}} + \frac{3+2\cdot 4}{3\cdot 4}\,\frac{1}{x} + \frac{-3\cdot 9 - 4\cdot 2 + 1\cdot 2}{2\cdot 9}\\[7pt] &\qquad\quad{}=\frac{11}{3\cdot 4}\cdot \frac{1}{x}-\frac{33}{2\cdot 9} \end{align}

and because \displaystyle 33=3\cdot 11, \displaystyle 9=3\cdot 3 and \displaystyle 4=2\cdot 2, the whole equation can rewritten as

\displaystyle \frac{11}{3\cdot 2\cdot 2}\cdot \frac{1}{x}-\frac{3\cdot 11}{2\cdot 3\cdot 3}=0\,\textrm{.}

Taking out common factors, we get

\displaystyle \frac{11}{3\cdot 2}\biggl(\frac{1}{2x}-1\biggr)=0

and then we see that the equation has the solution \displaystyle x=1/2.

Finally, we substitute \displaystyle x=1/2 into the original equation to check that we have calculated correctly.

\displaystyle \begin{align}

& \biggl(\frac{2}{\frac{1}{2}}-3\biggr)\biggl(\frac{1}{4\cdot\frac{1}{2}} + \frac{1}{2}\biggr) - \biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{2}{3}\biggr)^{2} - \biggl(\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3}\biggr)\biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{1}{3}\biggr)\\[5pt] &\qquad\quad{}= (4-3)\biggl(\frac{1}{2}+\frac{1}{2}\biggr) - \biggl(1-\frac{2}{3}\biggr)^{2} - \biggl(1+\frac{1}{3}\biggr)\biggl(1-\frac{1}{3}\biggr)\\[5pt] &\qquad\quad{}= 1 - \biggl(\frac{1}{3}\biggr)^{2} - \frac{4}{3}\cdot\frac{2}{3}\\[5pt] &\qquad\quad{}= 1 - \frac{1}{9} - \frac{8}{9}\\[5pt] &\qquad\quad{}= 0\,\textrm{.} \end{align}

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