Question

Could I solve this inequality by completing the square? How would I do so?

Answer 1

`\large\boxed{x>-2+√(14)\ \vee\ x<-2-√(14)}\\\boxed{x\in(-\infty,\ -2-√(14))\ \cup\ (-2+√(14),\ \infty)}`

Explanation:

`x^2+4x>10\\\\x^2+2(x)(2)>10\qquad\text{add}\ 2^2=4\ \text{to both sides}\\\\x^2+2(x)(2)+2^2>10+4\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2>14\Rightarrow x+2>√(14)\ \vee\ x+2<-√(14)\qquad\text{subtract 2 from both sides}\\\\x>-2+√(14)\ \vee\ x<-2-√(14)`