+0  
 
0
187
1
avatar+412 

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

 Apr 11, 2018

Best Answer 

 #1
avatar+7348 
+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     \(b=4\)  .

.
 Apr 11, 2018
edited by hectictar  Apr 11, 2018
 #1
avatar+7348 
+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

24 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.