Solve 4x^2-3x-1leq0

$$4 { x }^{ 2 } -3x-1 \leq 0$$

Solve for x

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

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.

$$4x^{2}-3x-1=0$$

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 4 for a, -3 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{3±5}{8}$$

Solve the equation x=\frac{3±5}{8} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$4\left(x-1\right)\left(x+\frac{1}{4}\right)\leq 0$$

For the product to be ≤0, one of the values x-1 and x+\frac{1}{4} has to be ≥0 and the other has to be ≤0. Consider the case when x-1\geq 0 and x+\frac{1}{4}\leq 0.

$$x-1\geq 0 x+\frac{1}{4}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-1\leq 0 and x+\frac{1}{4}\geq 0.

$$x+\frac{1}{4}\geq 0 x-1\leq 0$$

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

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

The final solution is the union of the obtained solutions.

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