Solve x^2+3x-4geq0

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

Solve for x

$$x\in (-\infty,-4]\cup [1,\infty)$$

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.

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

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±5}{2}$$

Solve the equation x=\frac{-3±5}{2} when ± is plus and when ± is minus.

$$x=1 x=-4$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-1\right)\left(x+4\right)\geq 0$$

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

$$x-1\leq 0 x+4\leq 0$$

The solution satisfying both inequalities is x\leq -4.

$$x\leq -4$$

Consider the case when x-1 and x+4 are both ≥0.

$$x+4\geq 0 x-1\geq 0$$

The solution satisfying both inequalities is x\geq 1.

$$x\geq 1$$

The final solution is the union of the obtained solutions.

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