Therefore, the solution set is $\boxed{ \left[ -4, 2 \right] }$.

[tmt1]

Solve the Ineqality

$$x^2 + 2x -8 \le 0$$I know enough to factor it like this

$$(x-4) (x+2) \le 0$$

So I get 4 and -2. I just don't know how to get to the answer from here which is:

$$x \ge -4\cup x\le 2$$

unless I'm misreading the answer incorrectly.

Thanks

 

[mathbalarka]

You have factored it incorrectly.

$$x^2 + 2x - 8 = x^2 + (4 - 2)x - 4 \cdot 2 = x^2 + 4x - 2x - 2 \cdot 4 = x(x + 4) - 2(x + 4) = (x \, {\color{red}{+}} \, 4)(x \, {\color{red}{-}} \, 2)$$

NOT $(x - 4)(x + 2)$. So your inequality is

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

Note that if product of two reals is negative (or zero) then one of the two number must be negative and another must be positive (or both zero). Can you use this?

 

[Prove It]

Solve the Ineqality

$$x^2 + 2x -8 \le 0$$I know enough to factor it like this

$$(x-4) (x+2) \le 0$$

So I get 4 and -2. I just don't know how to get to the answer from here which is:

$$x \ge -4\cup x\le 2$$

unless I'm misreading the answer incorrectly.

Thanks

The most direct way of solving quadratic inequalities is by completing the square...

$\displaystyle \begin{align*} x^2 + 2x - 8 &\leq 0 \\ x^2 + 2x &\leq 8 \\ x^2 + 2x + 1^2 &\leq 8 + 1^2 \\ \left( x + 1 \right) ^2 &\leq 9 \\ \sqrt{ \left( x + 1 \right) ^2 } &\leq \sqrt{9} \\ \left| x + 1 \right| &\leq 3 \\ -3 \leq x + 1 &\leq 3 \\ -4 \leq x &\leq 2 \end{align*}$