Solve 4x^2-4x-3geq0

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

Solve for x

$$x\in (-\infty,-\frac{1}{2}]\cup [\frac{3}{2},\infty)$$

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.

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

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

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

Do the calculations.

$$x=\frac{4±8}{8}$$

Solve the equation x=\frac{4±8}{8} when ± is plus and when ± is minus.

$$x=\frac{3}{2} x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)\geq 0$$

For the product to be ≥0, x-\frac{3}{2} and x+\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when x-\frac{3}{2} and x+\frac{1}{2} are both ≤0.

$$x-\frac{3}{2}\leq 0 x+\frac{1}{2}\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{1}{2}.

$$x\leq -\frac{1}{2}$$

Consider the case when x-\frac{3}{2} and x+\frac{1}{2} are both ≥0.

$$x+\frac{1}{2}\geq 0 x-\frac{3}{2}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{3}{2}.

$$x\geq \frac{3}{2}$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{1}{2}\text{; }x\geq \frac{3}{2}$$