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 Apr 11, 2018

Best Answer 

 #1
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Given that   \(\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2\)   where  \(a\)  and  \(b\)  are integers, find the values of  \(a\)  and  \(b\) .

 

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\(\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2\)

 

So     \(a=-3\)     and     \(b=4\)  .

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 Apr 11, 2018
edited by hectictar  Apr 11, 2018
 #1
avatar+7616 
+2
Best Answer

Given that   \(\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2\)   where  \(a\)  and  \(b\)  are integers, find the values of  \(a\)  and  \(b\) .

 

----------------

 

\(\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2\)

 

So     \(a=-3\)     and     \(b=4\)  .

hectictar Apr 11, 2018
edited by hectictar  Apr 11, 2018

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