Solve 8x^2+6x+1<0

$$8 x ^ { 2 } + 6 x + 1 < 0$$

Solve for x

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

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.

$$8x^{2}+6x+1=0$$

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

$$x=\frac{-6±\sqrt{6^{2}-4\times 8\times 1}}{2\times 8}$$

Do the calculations.

$$x=\frac{-6±2}{16}$$

Solve the equation x=\frac{-6±2}{16} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$8\left(x+\frac{1}{4}\right)\left(x+\frac{1}{2}\right)<0$$

For the product to be negative, x+\frac{1}{4} and x+\frac{1}{2} have to be of the opposite signs. Consider the case when x+\frac{1}{4} is positive and x+\frac{1}{2} is negative.

$$x+\frac{1}{4}>0 x+\frac{1}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{1}{2} is positive and x+\frac{1}{4} is negative.

$$x+\frac{1}{2}>0 x+\frac{1}{4}<0$$

The solution satisfying both inequalities is x\in \left(-\frac{1}{2},-\frac{1}{4}\right).

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

The final solution is the union of the obtained solutions.

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