Solve for k
k\in \left(-\infty,-2\sqrt{3}\right)\cup \left(2\sqrt{3},\infty\right)
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k^{2}>12
Add 12 to both sides.
k^{2}>\left(2\sqrt{3}\right)^{2}
Calculate the square root of 12 and get 2\sqrt{3}. Rewrite 12 as \left(2\sqrt{3}\right)^{2}.
|k|>2\sqrt{3}
Inequality holds for |k|>2\sqrt{3}.
k<-2\sqrt{3}\text{; }k>2\sqrt{3}
Rewrite |k|>2\sqrt{3} as k<-2\sqrt{3}\text{; }k>2\sqrt{3}.
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