Solve -x^2+5/2x-1>0

$$- x ^ { 2 } + \frac { 5 } { 2 } x - 1 > 0$$

Solve for x

$$x\in \left(\frac{1}{2},2\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+\frac{5}{2}x-1 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-\frac{5}{2}x+1<0$$

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.

$$x^{2}-\frac{5}{2}x+1=0$$

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 1 for a, -\frac{5}{2} for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{\frac{5}{2}±\frac{3}{2}}{2}$$

Solve the equation x=\frac{\frac{5}{2}±\frac{3}{2}}{2} when ± is plus and when ± is minus.

$$x=2 x=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-2\right)\left(x-\frac{1}{2}\right)<0$$

For the product to be negative, x-2 and x-\frac{1}{2} have to be of the opposite signs. Consider the case when x-2 is positive and x-\frac{1}{2} is negative.

$$x-2>0 x-\frac{1}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{1}{2} is positive and x-2 is negative.

$$x-\frac{1}{2}>0 x-2<0$$

The solution satisfying both inequalities is x\in \left(\frac{1}{2},2\right).

$$x\in \left(\frac{1}{2},2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{1}{2},2\right)$$