Solve x^2geq12^21

$$x ^ { 2 } \geq 12 ^ { 21 }$$

Solve for x

$$x\in (-\infty,-123834728448\sqrt{3}]\cup [123834728448\sqrt{3},\infty)$$

Solution Steps

Calculate 12 to the power of 21 and get 46005119909369701466112.

$$x^{2}\geq 46005119909369701466112$$

Calculate the square root of 46005119909369701466112 and get 123834728448\sqrt{3}. Rewrite 46005119909369701466112 as \left(123834728448\sqrt{3}\right)^{2}.

$$x^{2}\geq \left(123834728448\sqrt{3}\right)^{2}$$

Inequality holds for |x|\geq 123834728448\sqrt{3}.

$$|x|\geq 123834728448\sqrt{3}$$

Rewrite |x|\geq 123834728448\sqrt{3} as x\leq -123834728448\sqrt{3}\text{; }x\geq 123834728448\sqrt{3}.

$$x\leq -123834728448\sqrt{3}\text{; }x\geq 123834728448\sqrt{3}$$