Solve 8x^2+8x+(-1)leq0

$$8 x ^ { 2 } + 8 x + ( - 1 ) \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\end{bmatrix}$$

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.

$$8x^{2}+8x-1=0$$

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 8 for a, 8 for b, and -1 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Do the calculations.

$$x=\frac{-8±4\sqrt{6}}{16}$$

Solve the equation x=\frac{-8±4\sqrt{6}}{16} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{6}}{4}-\frac{1}{2} x=-\frac{\sqrt{6}}{4}-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right) and x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 and x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0.

$$x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0 and x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0.

$$x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\right].

$$x\in \begin{bmatrix}-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\end{bmatrix}$$