Solve 9x^2+6x-9<0

$$9 { x }^{ 2 } +6x-9 < 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{10}-1}{3},\frac{\sqrt{10}-1}{3}\right)$$

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.

$$9x^{2}+6x-9=0$$

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 9 for a, 6 for b, and -9 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-9\right)}}{2\times 9}$$

Do the calculations.

$$x=\frac{-6±6\sqrt{10}}{18}$$

Solve the equation x=\frac{-6±6\sqrt{10}}{18} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{10}-1}{3} x=\frac{-\sqrt{10}-1}{3}$$

Rewrite the inequality by using the obtained solutions.

$$9\left(x-\frac{\sqrt{10}-1}{3}\right)\left(x-\frac{-\sqrt{10}-1}{3}\right)<0$$

For the product to be negative, x-\frac{\sqrt{10}-1}{3} and x-\frac{-\sqrt{10}-1}{3} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{10}-1}{3} is positive and x-\frac{-\sqrt{10}-1}{3} is negative.

$$x-\frac{\sqrt{10}-1}{3}>0 x-\frac{-\sqrt{10}-1}{3}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{10}-1}{3} is positive and x-\frac{\sqrt{10}-1}{3} is negative.

$$x-\frac{-\sqrt{10}-1}{3}>0 x-\frac{\sqrt{10}-1}{3}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{10}-1}{3},\frac{\sqrt{10}-1}{3}\right).

$$x\in \left(\frac{-\sqrt{10}-1}{3},\frac{\sqrt{10}-1}{3}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{10}-1}{3},\frac{\sqrt{10}-1}{3}\right)$$