Solve 0<8x^2-28x+8

$$0 < 8 x ^ { 2 } - 28 x + 8$$

Solve for x

$$x\in \left(-\infty,\frac{7-\sqrt{33}}{4}\right)\cup \left(\frac{\sqrt{33}+7}{4},\infty\right)$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$8x^{2}-28x+8>0$$

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.

$$8x^{2}-28x+8=0$$

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 8 for a, -28 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 8\times 8}}{2\times 8}$$

Do the calculations.

$$x=\frac{28±4\sqrt{33}}{16}$$

Solve the equation x=\frac{28±4\sqrt{33}}{16} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{33}+7}{4} x=\frac{7-\sqrt{33}}{4}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(x-\frac{\sqrt{33}+7}{4}\right)\left(x-\frac{7-\sqrt{33}}{4}\right)>0$$

For the product to be positive, x-\frac{\sqrt{33}+7}{4} and x-\frac{7-\sqrt{33}}{4} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{33}+7}{4} and x-\frac{7-\sqrt{33}}{4} are both negative.

$$x-\frac{\sqrt{33}+7}{4}<0 x-\frac{7-\sqrt{33}}{4}<0$$

The solution satisfying both inequalities is x<\frac{7-\sqrt{33}}{4}.

$$x<\frac{7-\sqrt{33}}{4}$$

Consider the case when x-\frac{\sqrt{33}+7}{4} and x-\frac{7-\sqrt{33}}{4} are both positive.

$$x-\frac{7-\sqrt{33}}{4}>0 x-\frac{\sqrt{33}+7}{4}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{33}+7}{4}.

$$x>\frac{\sqrt{33}+7}{4}$$

The final solution is the union of the obtained solutions.

$$x<\frac{7-\sqrt{33}}{4}\text{; }x>\frac{\sqrt{33}+7}{4}$$