Solve 1/4x^2-1/2x-2geq0

$$\frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x - 2 \geq 0$$

Solve for x

$$x\in (-\infty,-2]\cup [4,\infty)$$

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.

$$\frac{1}{4}x^{2}-\frac{1}{2}x-2=0$$

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 \frac{1}{4} for a, -\frac{1}{2} for b, and -2 for c in the quadratic formula.

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

Do the calculations.

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

Solve the equation x=\frac{\frac{1}{2}±\frac{3}{2}}{\frac{1}{2}} when ± is plus and when ± is minus.

$$x=4 x=-2$$

Rewrite the inequality by using the obtained solutions.

$$\frac{1}{4}\left(x-4\right)\left(x+2\right)\geq 0$$

For the product to be ≥0, x-4 and x+2 have to be both ≤0 or both ≥0. Consider the case when x-4 and x+2 are both ≤0.

$$x-4\leq 0 x+2\leq 0$$

The solution satisfying both inequalities is x\leq -2.

$$x\leq -2$$

Consider the case when x-4 and x+2 are both ≥0.

$$x+2\geq 0 x-4\geq 0$$

The solution satisfying both inequalities is x\geq 4.

$$x\geq 4$$

The final solution is the union of the obtained solutions.

$$x\leq -2\text{; }x\geq 4$$