Lösung 2.3:2e
Aus Online Mathematik Brückenkurs 1
K (Lösning 2.3:2e moved to Solution 2.3:2e: Robot: moved page) |
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| - | {{ | + | Write the equation in normalized form by dividing both sides by |
| - | < | + | <math>5</math>, |
| - | {{ | + | |
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| + | <math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0</math> | ||
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| + | |||
| + | Completing the square on the left-hand side, | ||
| + | <math></math> | ||
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| + | <math></math> | ||
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| + | <math></math> | ||
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| + | |||
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| + | <math>\begin{align} | ||
| + | & x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\ | ||
| + | & =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\ | ||
| + | \end{align}</math> | ||
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| + | |||
| + | The equation is now rewritten as | ||
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| + | |||
| + | <math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}</math> | ||
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| + | |||
| + | and taking the root gives the solutions | ||
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| + | <math>x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5}</math> | ||
| + | because | ||
| + | <math>\left( \frac{4}{5}^{2} \right)=\frac{16}{25}</math> | ||
| + | which gives | ||
| + | <math>x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},</math> | ||
| + | |||
| + | |||
| + | <math>x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}</math> | ||
| + | which gives | ||
| + | <math>x=-\frac{1}{5}-\frac{4}{5}=-1,</math> | ||
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| + | Finally, we check the answer by substituting | ||
| + | <math>x=-\text{1 }</math> | ||
| + | and | ||
| + | <math>x=\text{3}/\text{5 }</math> | ||
| + | into the equation: | ||
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| + | <math>x=-\text{1 }</math>: LHS | ||
| + | <math>=5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0=</math> | ||
| + | RHS | ||
| + | |||
| + | <math>x=\text{3}/\text{5 }</math>: LHS | ||
| + | <math>=5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0=</math> | ||
| + | RHS | ||
Version vom 14:17, 20. Sep. 2008
Write the equation in normalized form by dividing both sides by \displaystyle 5,
\displaystyle x^{2}+\frac{2}{5}x-\frac{3}{5}=0
Completing the square on the left-hand side,
\displaystyle
\displaystyle
\displaystyle
\displaystyle \begin{align} & x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\ & =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\ \end{align}
The equation is now rewritten as
\displaystyle \left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}
and taking the root gives the solutions
\displaystyle x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5}
because
\displaystyle \left( \frac{4}{5}^{2} \right)=\frac{16}{25}
which gives
\displaystyle x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},
\displaystyle x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}
which gives
\displaystyle x=-\frac{1}{5}-\frac{4}{5}=-1,
Finally, we check the answer by substituting
\displaystyle x=-\text{1 }
and
\displaystyle x=\text{3}/\text{5 }
into the equation:
\displaystyle x=-\text{1 }: LHS
\displaystyle =5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0=
RHS
\displaystyle x=\text{3}/\text{5 }: LHS \displaystyle =5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0= RHS
