Solution 3.1:7b

From Förberedande kurs i matematik 1

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We multiply the top and bottom of the fraction by the conjugate of the denominator,
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<center> [[Image:3_1_7b.gif]] </center>
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<math>\sqrt{7}+\sqrt{5}</math>, and see what it leads to,
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{{Displayed math||<math>\begin{align}
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\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}
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&= \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\[10pt]
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&= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt]
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&= \frac{5\sqrt{7}\cdot\sqrt{7}+5\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{5}}{7-5}\\[10pt]
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&= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt]
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&= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt]
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&= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt]
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&= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt]
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&= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt]
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&= -\sqrt{35}\,\textrm{.}
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\end{align}</math>}}

Current revision

We multiply the top and bottom of the fraction by the conjugate of the denominator, \displaystyle \sqrt{7}+\sqrt{5}, and see what it leads to,

\displaystyle \begin{align}

\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} &= \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\[10pt] &= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt] &= \frac{5\sqrt{7}\cdot\sqrt{7}+5\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{5}}{7-5}\\[10pt] &= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt] &= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt] &= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt] &= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt] &= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt] &= -\sqrt{35}\,\textrm{.} \end{align}

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