How to solve (x)^(2)-2x-8 >= 0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x)^(2)-2x-8 >= 0 | Equatix

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Question

$${ x }^{ 2 } -2x-8 \geq 0$$

Solve for x

$x\in (-\infty,-2]\cup [4,\infty)$

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 $ax^{2}+bx+c=0$.

$$x^{2}-2x-8=0$$

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 $1$ for $a$, $-2$ for $b$, and $-8$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$x=\frac{2±6}{2}$$

Solve the equation $x=\frac{2±6}{2}$ when $±$ is plus and when $±$ is minus.

$$x=4$$ $$x=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-4\right)\left(x+2\right)\geq 0$$

For the product to be $≥0$, $x-4$ and $x+2$ have to be both $≤0$ or both $≥0$. Consider the case when $x-4$ and $x+2$ are both $≤0$.

$$x-4\leq 0$$ $$x+2\leq 0$$

The solution satisfying both inequalities is $x\leq -2$.

$$x\leq -2$$

Consider the case when $x-4$ and $x+2$ are both $≥0$.

$$x+2\geq 0$$ $$x-4\geq 0$$

The solution satisfying both inequalities is $x\geq 4$.

$$x\geq 4$$

The final solution is the union of the obtained solutions.

$$x\leq -2\text{; }x\geq 4$$

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