Solve 19x^2geq36

$$19 x ^ { 2 } \geq 36$$

Solve for x

$$x\in (-\infty,-\frac{6\sqrt{19}}{19}]\cup [\frac{6\sqrt{19}}{19},\infty)$$

Solution Steps

Divide both sides by 19. Since 19 is positive, the inequality direction remains the same.

$$x^{2}\geq \frac{36}{19}$$

Calculate the square root of \frac{36}{19} and get \frac{6\sqrt{19}}{19}. Rewrite \frac{36}{19} as \left(\frac{6\sqrt{19}}{19}\right)^{2}.

$$x^{2}\geq \left(\frac{6\sqrt{19}}{19}\right)^{2}$$

Inequality holds for |x|\geq \frac{6\sqrt{19}}{19}.

$$|x|\geq \frac{6\sqrt{19}}{19}$$

Rewrite |x|\geq \frac{6\sqrt{19}}{19} as x\leq -\frac{6\sqrt{19}}{19}\text{; }x\geq \frac{6\sqrt{19}}{19}.

$$x\leq -\frac{6\sqrt{19}}{19}\text{; }x\geq \frac{6\sqrt{19}}{19}$$