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# Can anyone solve this? Urgent ??

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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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Oct 15, 2018
#2
+100439
+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

Oct 15, 2018