Solve 48x^2-32x-1leq0

$$48 x ^ { 2 } - 32 x - 1 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{\sqrt{19}}{12}+\frac{1}{3},\frac{\sqrt{19}}{12}+\frac{1}{3}\end{bmatrix}$$

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.

$$48x^{2}-32x-1=0$$

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 48 for a, -32 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 48\left(-1\right)}}{2\times 48}$$

Do the calculations.

$$x=\frac{32±8\sqrt{19}}{96}$$

Solve the equation x=\frac{32±8\sqrt{19}}{96} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{19}}{12}+\frac{1}{3} x=-\frac{\sqrt{19}}{12}+\frac{1}{3}$$

Rewrite the inequality by using the obtained solutions.

$$48\left(x-\left(\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\right)\left(x-\left(-\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(\frac{\sqrt{19}}{12}+\frac{1}{3}\right) and x-\left(-\frac{\sqrt{19}}{12}+\frac{1}{3}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\geq 0 and x-\left(-\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\leq 0.

$$x-\left(\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\geq 0 x-\left(-\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\leq 0 and x-\left(-\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\geq 0.

$$x-\left(-\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\geq 0 x-\left(\frac{\sqrt{19}}{12}+\frac{1}{3}\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{\sqrt{19}}{12}+\frac{1}{3},\frac{\sqrt{19}}{12}+\frac{1}{3}\right].

$$x\in \begin{bmatrix}-\frac{\sqrt{19}}{12}+\frac{1}{3},\frac{\sqrt{19}}{12}+\frac{1}{3}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{\sqrt{19}}{12}+\frac{1}{3},\frac{\sqrt{19}}{12}+\frac{1}{3}\end{bmatrix}$$