+0  
 
0
31
1
avatar+270 

https://vle.mathswatch.co.uk/images/questions/question1918.png

qualitystreet  Apr 11, 2018

Best Answer 

 #1
avatar+6943 
+2

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
Sort: 

1+0 Answers

 #1
avatar+6943 
+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

11 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details