Solve x^2-9x+9<0

$$x ^ { 2 } - 9 x + 9 < 0$$

Solve for x

$$x\in \left(\frac{9-3\sqrt{5}}{2},\frac{3\sqrt{5}+9}{2}\right)$$

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.

$$x^{2}-9x+9=0$$

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, -9 for b, and 9 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{9±3\sqrt{5}}{2}$$

Solve the equation x=\frac{9±3\sqrt{5}}{2} when ± is plus and when ± is minus.

$$x=\frac{3\sqrt{5}+9}{2} x=\frac{9-3\sqrt{5}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{3\sqrt{5}+9}{2}\right)\left(x-\frac{9-3\sqrt{5}}{2}\right)<0$$

For the product to be negative, x-\frac{3\sqrt{5}+9}{2} and x-\frac{9-3\sqrt{5}}{2} have to be of the opposite signs. Consider the case when x-\frac{3\sqrt{5}+9}{2} is positive and x-\frac{9-3\sqrt{5}}{2} is negative.

$$x-\frac{3\sqrt{5}+9}{2}>0 x-\frac{9-3\sqrt{5}}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{9-3\sqrt{5}}{2} is positive and x-\frac{3\sqrt{5}+9}{2} is negative.

$$x-\frac{9-3\sqrt{5}}{2}>0 x-\frac{3\sqrt{5}+9}{2}<0$$

The solution satisfying both inequalities is x\in \left(\frac{9-3\sqrt{5}}{2},\frac{3\sqrt{5}+9}{2}\right).

$$x\in \left(\frac{9-3\sqrt{5}}{2},\frac{3\sqrt{5}+9}{2}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{9-3\sqrt{5}}{2},\frac{3\sqrt{5}+9}{2}\right)$$