Solve x^2-1/4x+8/9geq0

$$x ^ { 2 } - \frac { 1 } { 4 } x + \frac { 8 } { 9 } \geq 0$$

Solve for x

$$x\in \mathrm{R}$$

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.

$$x^{2}-\frac{1}{4}x+\frac{8}{9}=0$$

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, -\frac{1}{4} for b, and \frac{8}{9} for c in the quadratic formula.

$$x=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\left(-\frac{1}{4}\right)^{2}-4\times 1\times \frac{8}{9}}}{2}$$

Do the calculations.

$$x=\frac{\frac{1}{4}±\sqrt{-\frac{503}{144}}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression x^{2}-\frac{1}{4}x+\frac{8}{9} has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$0^{2}-\frac{1}{4}\times 0+\frac{8}{9}=\frac{8}{9}$$

The value of the expression x^{2}-\frac{1}{4}x+\frac{8}{9} is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$