+0  
 
0
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2
avatar+701 

http://prntscr.com/l6c0nf

 

the Norwegian part is saying "show that when a>0" 

 

-b/2a±sqrt(b^2/4a^2-c/a)=(-b±sqrt(b^2-4ac))/2a

when a >0

 Oct 15, 2018
edited by critical  Oct 15, 2018
edited by critical  Oct 15, 2018
edited by critical  Oct 15, 2018
edited by critical  Oct 15, 2018
 #1
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0

Why did you upload your question vertically? It is very difficult to read!!!!.

 Oct 15, 2018
 #2
avatar+98061 
+2

On the left side....Let's simplify under the radical, first

 

b^2             c

_____  -    ___       multiplyy the second fraction by 4a  on top and bottom

4a^2          a 

 

b^2            4ac

____    -  ____          and this gives us

4a^2        4a^2

 

 

[ b^2  - 4ac ]  

__________       taking the square root    of top and bottom  we have

    4a^2

 

√ [ b^2 - 4ac ]

____________

 √ [ 4a^2 ]

 

 

√ [ b^2  - 4ac ] 

____________

       2a

 

So...combine tis with the first fraction on the left and we end up with

 

              

-b               √ [ b^2  - 4ac ] 

__    ±       ____________        which gives us

2a                   2a

 

 

-b   ±  √ [ b^2  - 4ac ] 

________________       =    [the right side ]

        2a

 

 

cool cool cool

 Oct 15, 2018

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