Solve 6x^2+5x+1leq0

$$6 x ^ { 2 } + 5 x + 1 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{1}{2},-\frac{1}{3}\end{bmatrix}$$

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.

$$6x^{2}+5x+1=0$$

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 6 for a, 5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6\times 1}}{2\times 6}$$

Do the calculations.

$$x=\frac{-5±1}{12}$$

Solve the equation x=\frac{-5±1}{12} when ± is plus and when ± is minus.

$$x=-\frac{1}{3} x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x+\frac{1}{3}\right)\left(x+\frac{1}{2}\right)\leq 0$$

For the product to be ≤0, one of the values x+\frac{1}{3} and x+\frac{1}{2} has to be ≥0 and the other has to be ≤0. Consider the case when x+\frac{1}{3}\geq 0 and x+\frac{1}{2}\leq 0.

$$x+\frac{1}{3}\geq 0 x+\frac{1}{2}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{1}{3}\leq 0 and x+\frac{1}{2}\geq 0.

$$x+\frac{1}{2}\geq 0 x+\frac{1}{3}\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{1}{2},-\frac{1}{3}\right].

$$x\in \begin{bmatrix}-\frac{1}{2},-\frac{1}{3}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{1}{2},-\frac{1}{3}\end{bmatrix}$$