Solve m^2-m-3/4geq0

$$m ^ { 2 } - m - \frac { 3 } { 4 } \geq 0$$

Solve for m

$$m\in (-\infty,-\frac{1}{2}]\cup [\frac{3}{2},\infty)$$

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.

$$m^{2}-m-\frac{3}{4}=0$$

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 1 for a, -1 for b, and -\frac{3}{4} for c in the quadratic formula.

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

Do the calculations.

$$m=\frac{1±2}{2}$$

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

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

Rewrite the inequality by using the obtained solutions.

$$\left(m-\frac{3}{2}\right)\left(m+\frac{1}{2}\right)\geq 0$$

For the product to be ≥0, m-\frac{3}{2} and m+\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when m-\frac{3}{2} and m+\frac{1}{2} are both ≤0.

$$m-\frac{3}{2}\leq 0 m+\frac{1}{2}\leq 0$$

The solution satisfying both inequalities is m\leq -\frac{1}{2}.

$$m\leq -\frac{1}{2}$$

Consider the case when m-\frac{3}{2} and m+\frac{1}{2} are both ≥0.

$$m+\frac{1}{2}\geq 0 m-\frac{3}{2}\geq 0$$

The solution satisfying both inequalities is m\geq \frac{3}{2}.

$$m\geq \frac{3}{2}$$

The final solution is the union of the obtained solutions.

$$m\leq -\frac{1}{2}\text{; }m\geq \frac{3}{2}$$