Solve +3x<4x^2

$$+ 3 x < 4 x ^ { 2 }$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{3}{4},\infty\right)$$

Solution Steps

Subtract 4x^{2} from both sides.

$$3x-4x^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 3x-4x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-3x+4x^{2}>0$$

Factor out x.

$$x\left(4x-3\right)>0$$

For the product to be positive, x and x-\frac{3}{4} have to be both negative or both positive. Consider the case when x and x-\frac{3}{4} are both negative.

$$x<0 x-\frac{3}{4}<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-\frac{3}{4} are both positive.

$$x-\frac{3}{4}>0 x>0$$

The solution satisfying both inequalities is x>\frac{3}{4}.

$$x>\frac{3}{4}$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{3}{4}$$