Solve 48x^2+24x-1geq0

$$48 x ^ { 2 } + 24 x - 1 \geq 0$$

Solve for x

$$x\in (-\infty,-\frac{\sqrt{3}}{6}-\frac{1}{4}]\cup [\frac{\sqrt{3}}{6}-\frac{1}{4},\infty)$$

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.

$$48x^{2}+24x-1=0$$

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

$$x=\frac{-24±\sqrt{24^{2}-4\times 48\left(-1\right)}}{2\times 48}$$

Do the calculations.

$$x=\frac{-24±16\sqrt{3}}{96}$$

Solve the equation x=\frac{-24±16\sqrt{3}}{96} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{3}}{6}-\frac{1}{4} x=-\frac{\sqrt{3}}{6}-\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$48\left(x-\left(\frac{\sqrt{3}}{6}-\frac{1}{4}\right)\right)\left(x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{4}\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\frac{\sqrt{3}}{6}-\frac{1}{4}\right) and x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{4}\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\frac{\sqrt{3}}{6}-\frac{1}{4}\right) and x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{4}\right) are both ≤0.

$$x-\left(\frac{\sqrt{3}}{6}-\frac{1}{4}\right)\leq 0 x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{4}\right)\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{\sqrt{3}}{6}-\frac{1}{4}.

$$x\leq -\frac{\sqrt{3}}{6}-\frac{1}{4}$$

Consider the case when x-\left(\frac{\sqrt{3}}{6}-\frac{1}{4}\right) and x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{4}\right) are both ≥0.

$$x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{4}\right)\geq 0 x-\left(\frac{\sqrt{3}}{6}-\frac{1}{4}\right)\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{3}}{6}-\frac{1}{4}.

$$x\geq \frac{\sqrt{3}}{6}-\frac{1}{4}$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{\sqrt{3}}{6}-\frac{1}{4}\text{; }x\geq \frac{\sqrt{3}}{6}-\frac{1}{4}$$