Solve 2x^2-x+1/8leq0

$$2 x ^ { 2 } - x + \frac { 1 } { 8 } \leq 0$$

Solve for x

$$x=\frac{1}{4}=0.25$$

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.

$$2x^{2}-x+\frac{1}{8}=0$$

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, -1 for b, and \frac{1}{8} for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\times \frac{1}{8}}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±0}{4}$$

Solutions are the same.

$$x=\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{1}{4}\right)^{2}\leq 0$$

Inequality holds for x=\frac{1}{4}.

$$x=\frac{1}{4}$$