Solve 9x-4<2x^2=

$$9 x - 4 < 2 x ^ { 2 } =$$

Solve for x

$$x\in \left(-\infty,\frac{1}{2}\right)\cup \left(4,\infty\right)$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$9x-4-2x^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 9x-4-2x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-9x+4+2x^{2}>0$$

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.

$$-9x+4+2x^{2}=0$$

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 2 for a, -9 for b, and 4 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{9±7}{4}$$

Solve the equation x=\frac{9±7}{4} when ± is plus and when ± is minus.

$$x=4 x=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-4\right)\left(x-\frac{1}{2}\right)>0$$

For the product to be positive, x-4 and x-\frac{1}{2} have to be both negative or both positive. Consider the case when x-4 and x-\frac{1}{2} are both negative.

$$x-4<0 x-\frac{1}{2}<0$$

The solution satisfying both inequalities is x<\frac{1}{2}.

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

Consider the case when x-4 and x-\frac{1}{2} are both positive.

$$x-\frac{1}{2}>0 x-4>0$$

The solution satisfying both inequalities is x>4.

$$x>4$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{2}\text{; }x>4$$