Solve 4a^2-8a-8geq0

$$4 a ^ { 2 } - 8 a - 8 \geq 0$$

Solve for a

$$a\in (-\infty,1-\sqrt{3}]\cup [\sqrt{3}+1,\infty)$$

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.

$$4a^{2}-8a-8=0$$

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 4 for a, -8 for b, and -8 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{8±8\sqrt{3}}{8}$$

Solve the equation a=\frac{8±8\sqrt{3}}{8} when ± is plus and when ± is minus.

$$a=\sqrt{3}+1 a=1-\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(a-\left(\sqrt{3}+1\right)\right)\left(a-\left(1-\sqrt{3}\right)\right)\geq 0$$

For the product to be ≥0, a-\left(\sqrt{3}+1\right) and a-\left(1-\sqrt{3}\right) have to be both ≤0 or both ≥0. Consider the case when a-\left(\sqrt{3}+1\right) and a-\left(1-\sqrt{3}\right) are both ≤0.

$$a-\left(\sqrt{3}+1\right)\leq 0 a-\left(1-\sqrt{3}\right)\leq 0$$

The solution satisfying both inequalities is a\leq 1-\sqrt{3}.

$$a\leq 1-\sqrt{3}$$

Consider the case when a-\left(\sqrt{3}+1\right) and a-\left(1-\sqrt{3}\right) are both ≥0.

$$a-\left(1-\sqrt{3}\right)\geq 0 a-\left(\sqrt{3}+1\right)\geq 0$$

The solution satisfying both inequalities is a\geq \sqrt{3}+1.

$$a\geq \sqrt{3}+1$$

The final solution is the union of the obtained solutions.

$$a\leq 1-\sqrt{3}\text{; }a\geq \sqrt{3}+1$$