Solve 9/4m^2-5m+1>0

$$\frac { 9 } { 4 } m ^ { 2 } - 5 m + 1 > 0$$

Solve for m

$$m\in \left(-\infty,\frac{2}{9}\right)\cup \left(2,\infty\right)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$\frac{9}{4}m^{2}-5m+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute \frac{9}{4} for a, -5 for b, and 1 for c in the quadratic formula.

$$m=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times \frac{9}{4}\times 1}}{2\times \frac{9}{4}}$$

Do the calculations.

$$m=\frac{5±4}{\frac{9}{2}}$$

Solve the equation m=\frac{5±4}{\frac{9}{2}} when ± is plus and when ± is minus.

$$m=2 m=\frac{2}{9}$$

Rewrite the inequality by using the obtained solutions.

$$\frac{9}{4}\left(m-2\right)\left(m-\frac{2}{9}\right)>0$$

For the product to be positive, m-2 and m-\frac{2}{9} have to be both negative or both positive. Consider the case when m-2 and m-\frac{2}{9} are both negative.

$$m-2<0 m-\frac{2}{9}<0$$

The solution satisfying both inequalities is m<\frac{2}{9}.

$$m<\frac{2}{9}$$

Consider the case when m-2 and m-\frac{2}{9} are both positive.

$$m-\frac{2}{9}>0 m-2>0$$

The solution satisfying both inequalities is m>2.

$$m>2$$

The final solution is the union of the obtained solutions.

$$m<\frac{2}{9}\text{; }m>2$$