Solve 25x^2geq4

$$25 { x }^{ 2 } \geq 4$$

Solve for x

$$x\in (-\infty,-\frac{2}{5}]\cup [\frac{2}{5},\infty)$$

Solution Steps

Divide both sides by 25. Since 25 is positive, the inequality direction remains the same.

$$x^{2}\geq \frac{4}{25}$$

Calculate the square root of \frac{4}{25} and get \frac{2}{5}. Rewrite \frac{4}{25} as \left(\frac{2}{5}\right)^{2}.

$$x^{2}\geq \left(\frac{2}{5}\right)^{2}$$

Inequality holds for |x|\geq \frac{2}{5}.

$$|x|\geq \frac{2}{5}$$

Rewrite |x|\geq \frac{2}{5} as x\leq -\frac{2}{5}\text{; }x\geq \frac{2}{5}.

$$x\leq -\frac{2}{5}\text{; }x\geq \frac{2}{5}$$