principal square root

Posted on October 1, 2012 by skxsephiroth

Does \displaystyle\sqrt{a^2}=\pm{a} ?

\displaystyle\sqrt{a^2}=|a|.

\displaystyle a^2=4 \Rightarrow |a|=2 \Rightarrow a=\pm{2}.

Consider solving \displaystyle 3+\sqrt{x-5}=0

\displaystyle \Rightarrow \sqrt{x-5}=-3 \Rightarrow x-5=9 \Rightarrow x=14?

\displaystyle 3+\sqrt{x-5}=0 \Rightarrow \sqrt{x-5}=-3 \Rightarrow \sqrt{x-5}<0 \Rightarrow contradiction.

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