# Solve for x: ABS (x^2 + x) 12

## Question:

Solve for x:

{eq}ABS (x^2 + x) \leq 12 {/eq}

## Absolute Value:

{eq}\text{For }\alpha \le \beta {/eq} the set {eq}\{ x\in \mathbb{R} : (x-\alpha)(x- \beta) \le 0 \} {/eq} is the set {eq}\{ x\in \mathbb{R} : \alpha \le x \le \beta \} =[\alpha, \beta] {/eq}

To get this, we observe that for {eq}(x-\alpha)(x- \beta) \le 0 {/eq}, the terms {eq}(x-\alpha) {/eq} and {eq}(x- \beta) \ {/eq} need to be of opposite sign, which is the case only when x lies in the specified interval.

{eq}| x^2 + x | \leq 12\\ \Rightarrow -12 \le x^2+x \le 12 \\ \Rightarrow x^2+x\le -12\quad \text{ and }\quad x^2+x\le 12 {/eq}

Now {eq}x^2+x\ge...

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