Solve 81x^2+36x-2geq0

$$81 x ^ { 2 } + 36 x - 2 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{-\sqrt{6}-2}{9}]\cup [\frac{\sqrt{6}-2}{9},\infty)$$

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.

$$81x^{2}+36x-2=0$$

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 81 for a, 36 for b, and -2 for c in the quadratic formula.

$$x=\frac{-36±\sqrt{36^{2}-4\times 81\left(-2\right)}}{2\times 81}$$

Do the calculations.

$$x=\frac{-36±18\sqrt{6}}{162}$$

Solve the equation x=\frac{-36±18\sqrt{6}}{162} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{6}-2}{9} x=\frac{-\sqrt{6}-2}{9}$$

Rewrite the inequality by using the obtained solutions.

$$81\left(x-\frac{\sqrt{6}-2}{9}\right)\left(x-\frac{-\sqrt{6}-2}{9}\right)\geq 0$$

For the product to be ≥0, x-\frac{\sqrt{6}-2}{9} and x-\frac{-\sqrt{6}-2}{9} have to be both ≤0 or both ≥0. Consider the case when x-\frac{\sqrt{6}-2}{9} and x-\frac{-\sqrt{6}-2}{9} are both ≤0.

$$x-\frac{\sqrt{6}-2}{9}\leq 0 x-\frac{-\sqrt{6}-2}{9}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{-\sqrt{6}-2}{9}.

$$x\leq \frac{-\sqrt{6}-2}{9}$$

Consider the case when x-\frac{\sqrt{6}-2}{9} and x-\frac{-\sqrt{6}-2}{9} are both ≥0.

$$x-\frac{-\sqrt{6}-2}{9}\geq 0 x-\frac{\sqrt{6}-2}{9}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{6}-2}{9}.

$$x\geq \frac{\sqrt{6}-2}{9}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{-\sqrt{6}-2}{9}\text{; }x\geq \frac{\sqrt{6}-2}{9}$$