Solve 99x<41x^2

$$99x < 41 { x }^{ 2 }$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{99}{41},\infty\right)$$

Solution Steps

Subtract 41x^{2} from both sides.

$$99x-41x^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 99x-41x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-99x+41x^{2}>0$$

Factor out x.

$$x\left(41x-99\right)>0$$

For the product to be positive, x and x-\frac{99}{41} have to be both negative or both positive. Consider the case when x and x-\frac{99}{41} are both negative.

$$x<0 x-\frac{99}{41}<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-\frac{99}{41} are both positive.

$$x-\frac{99}{41}>0 x>0$$

The solution satisfying both inequalities is x>\frac{99}{41}.

$$x>\frac{99}{41}$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{99}{41}$$