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Solution 2.2:3d

From Förberedande kurs i matematik 1

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m (Lösning 2.2:3d moved to Solution 2.2:3d: Robot: moved page)
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(The exercise is taken from an actual exam in Spring Term 1945!)
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<center> [[Image:2_2_3d-1(3).gif]] </center>
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{{NAVCONTENT_STOP}}
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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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{{NAVCONTENT_START}}
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<center> [[Image:2_2_3d-2(3).gif]] </center>
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{{NAVCONTENT_STOP}}
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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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<center> [[Image:2_2_3d-3(3).gif]] </center>
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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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{{NAVCONTENT_STOP}}
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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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Collecting up terms, the left-hand side becomes
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<math>\begin{align}
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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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Taking out common factors, we get
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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
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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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<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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& \\
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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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& \\
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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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\end{align}</math>

Revision as of 15:48, 17 September 2008

(The exercise is taken from an actual exam in Spring Term 1945!)

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:


x2−314x+21=x214x−x221−314x−321=12x2+x1−34x−23=12x2+14x−2312x−322=12x2−212x32+322=14x2−23x+94


12x+3112x−31=conjugate rule=12x2−132=14x2−91 


Collecting up terms, the left-hand side becomes


12x2+14x−23−14x2−23x+94−14x2−91=21−41−411x2+41+32x1+−23−94+91=42−1−11x2+343+24x1+29−39−42+12=1134x1−3329


and because 33=311, 9=33 and 4=22, the whole equation can rewritten as


11322x1−311233=0


Taking out common factors, we get


113212x−1=0 


and then we see that the equation has the solution x=12.

Finally, we substitute x=12 into the original equation to check that we have calculated correctly.


221−31421+21−1221−322−1221+311221−31=4−321+21−1−322−1+311−31=1−312−3432=1−91−98=0

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