Solve 4x^2-3geq-11x

$$4 x ^ { 2 } - 3 \geq - 11 x$$

Solve for x

$$x\in (-\infty,-3]\cup [\frac{1}{4},\infty)$$

Steps Using the Quadratic Formula

Add 11x to both sides.

$$4x^{2}-3+11x\geq 0$$

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.

$$4x^{2}-3+11x=0$$

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

$$x=\frac{-11±\sqrt{11^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-11±13}{8}$$

Solve the equation x=\frac{-11±13}{8} when ± is plus and when ± is minus.

$$x=\frac{1}{4} x=-3$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be ≥0, x-\frac{1}{4} and x+3 have to be both ≤0 or both ≥0. Consider the case when x-\frac{1}{4} and x+3 are both ≤0.

$$x-\frac{1}{4}\leq 0 x+3\leq 0$$

The solution satisfying both inequalities is x\leq -3.

$$x\leq -3$$

Consider the case when x-\frac{1}{4} and x+3 are both ≥0.

$$x+3\geq 0 x-\frac{1}{4}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{1}{4}.

$$x\geq \frac{1}{4}$$

The final solution is the union of the obtained solutions.

$$x\leq -3\text{; }x\geq \frac{1}{4}$$