Solve 4a^2-21a+4geq0

$$4 a ^ { 2 } - 21 a + 4 \geq 0$$

Solve for a

$$a\in (-\infty,\frac{21-\sqrt{377}}{8}]\cup [\frac{\sqrt{377}+21}{8},\infty)$$

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.

$$4a^{2}-21a+4=0$$

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 4 for a, -21 for b, and 4 for c in the quadratic formula.

$$a=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Do the calculations.

$$a=\frac{21±\sqrt{377}}{8}$$

Solve the equation a=\frac{21±\sqrt{377}}{8} when ± is plus and when ± is minus.

$$a=\frac{\sqrt{377}+21}{8} a=\frac{21-\sqrt{377}}{8}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(a-\frac{\sqrt{377}+21}{8}\right)\left(a-\frac{21-\sqrt{377}}{8}\right)\geq 0$$

For the product to be ≥0, a-\frac{\sqrt{377}+21}{8} and a-\frac{21-\sqrt{377}}{8} have to be both ≤0 or both ≥0. Consider the case when a-\frac{\sqrt{377}+21}{8} and a-\frac{21-\sqrt{377}}{8} are both ≤0.

$$a-\frac{\sqrt{377}+21}{8}\leq 0 a-\frac{21-\sqrt{377}}{8}\leq 0$$

The solution satisfying both inequalities is a\leq \frac{21-\sqrt{377}}{8}.

$$a\leq \frac{21-\sqrt{377}}{8}$$

Consider the case when a-\frac{\sqrt{377}+21}{8} and a-\frac{21-\sqrt{377}}{8} are both ≥0.

$$a-\frac{21-\sqrt{377}}{8}\geq 0 a-\frac{\sqrt{377}+21}{8}\geq 0$$

The solution satisfying both inequalities is a\geq \frac{\sqrt{377}+21}{8}.

$$a\geq \frac{\sqrt{377}+21}{8}$$

The final solution is the union of the obtained solutions.

$$a\leq \frac{21-\sqrt{377}}{8}\text{; }a\geq \frac{\sqrt{377}+21}{8}$$