Solve 2m^2-3m+1geq0

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

Solve for m

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

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.

$$2m^{2}-3m+1=0$$

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

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

Do the calculations.

$$m=\frac{3±1}{4}$$

Solve the equation m=\frac{3±1}{4} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

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

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

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

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

The solution satisfying both inequalities is m\geq 1.

$$m\geq 1$$

The final solution is the union of the obtained solutions.

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