mwickerh mwickerh

17-11-2014
Mathematics

contestada

completing[tex]a^2-8a+15=0[/tex]
complete the square

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konrad509 konrad509

17-11-2014

[tex]a^2-8a+15=0\\ a^2-8a+16-1=0\\ (a-4)^2=1\\ a-4 =1 \vee a-4=-1\\ a=5 \vee a=3 [/tex]

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