Solve 13-3x^2>0

$$13 - 3 x ^ { 2 } > 0$$

Solve for x

$$x\in \left(-\frac{\sqrt{39}}{3},\frac{\sqrt{39}}{3}\right)$$

Solution Steps

Multiply the inequality by -1 to make the coefficient of the highest power in 13-3x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-13+3x^{2}<0$$

Add \frac{13}{3} to both sides.

$$x^{2}<\frac{13}{3}$$

Calculate the square root of \frac{13}{3} and get \frac{\sqrt{39}}{3}. Rewrite \frac{13}{3} as \left(\frac{\sqrt{39}}{3}\right)^{2}.

$$x^{2}<\left(\frac{\sqrt{39}}{3}\right)^{2}$$

Inequality holds for |x|<\frac{\sqrt{39}}{3}.

$$|x|<\frac{\sqrt{39}}{3}$$

Rewrite |x|<\frac{\sqrt{39}}{3} as x\in \left(-\frac{\sqrt{39}}{3},\frac{\sqrt{39}}{3}\right).

$$x\in \left(-\frac{\sqrt{39}}{3},\frac{\sqrt{39}}{3}\right)$$