Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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