Solve 9x^2+6x+1>0

$$9 { x }^{ 2 } +6x+1 > 0$$

Solve for x

$$x\neq -\frac{1}{3}$$

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.

$$9x^{2}+6x+1=0$$

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 9 for a, 6 for b, and 1 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\times 1}}{2\times 9}$$

Do the calculations.

$$x=\frac{-6±0}{18}$$

Solutions are the same.

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

Rewrite the inequality by using the obtained solutions.

$$9\left(x+\frac{1}{3}\right)^{2}>0$$

Inequality holds for x\neq -\frac{1}{3}.

$$x\neq -\frac{1}{3}$$