Solve 3a^2-2geq4[a)

$$3 a ^ { 2 } - 2 \geq 4 [ a )$$

Solve for a

$$a\in (-\infty,\frac{2-\sqrt{10}}{3}]\cup [\frac{\sqrt{10}+2}{3},\infty)$$

Steps Using the Quadratic Formula

Subtract 4a from both sides.

$$3a^{2}-2-4a\geq 0$$

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.

$$3a^{2}-2-4a=0$$

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 3 for a, -4 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{4±2\sqrt{10}}{6}$$

Solve the equation a=\frac{4±2\sqrt{10}}{6} when ± is plus and when ± is minus.

$$a=\frac{\sqrt{10}+2}{3} a=\frac{2-\sqrt{10}}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(a-\frac{\sqrt{10}+2}{3}\right)\left(a-\frac{2-\sqrt{10}}{3}\right)\geq 0$$

For the product to be ≥0, a-\frac{\sqrt{10}+2}{3} and a-\frac{2-\sqrt{10}}{3} have to be both ≤0 or both ≥0. Consider the case when a-\frac{\sqrt{10}+2}{3} and a-\frac{2-\sqrt{10}}{3} are both ≤0.

$$a-\frac{\sqrt{10}+2}{3}\leq 0 a-\frac{2-\sqrt{10}}{3}\leq 0$$

The solution satisfying both inequalities is a\leq \frac{2-\sqrt{10}}{3}.

$$a\leq \frac{2-\sqrt{10}}{3}$$

Consider the case when a-\frac{\sqrt{10}+2}{3} and a-\frac{2-\sqrt{10}}{3} are both ≥0.

$$a-\frac{2-\sqrt{10}}{3}\geq 0 a-\frac{\sqrt{10}+2}{3}\geq 0$$

The solution satisfying both inequalities is a\geq \frac{\sqrt{10}+2}{3}.

$$a\geq \frac{\sqrt{10}+2}{3}$$

The final solution is the union of the obtained solutions.

$$a\leq \frac{2-\sqrt{10}}{3}\text{; }a\geq \frac{\sqrt{10}+2}{3}$$