1. ## Solving Absolute value quadratic inequalities

Solve | x2 - 4| < 8
I have not done any absolute value quadratic inequalities, here is my attempt. Please check my work, thank you!

|x2 - 4| < 8
|x2| +4 < 8
|x2| - 4 < 0
(|x2| -2 ) ( |x2| + 2) < 0

Do we stop here? What do we do next?
So, a. -x2 - 2 < 0 and b. x2 -2 < 0 & c. - x2 +2 < 0 and d. x2 + 2 < 0

2. ## Re: Solving Absolute value quadratic inequalities

Originally Posted by sakonpure6
Solve | x2 - 4| < 8
I have not done any absolute value quadratic inequalities, here is my attempt. Please check my work, thank you!
None of that is correct.
$\displaystyle \\|x^2-4|<8\\-8<x^2-4<8\\-4<x^2<12\\-\sqrt{12}<x<\sqrt{12}$

Have a look at this

3. ## Re: Solving Absolute value quadratic inequalities

how would you solve |x^2 - 4| > 8 ?

4. ## Re: Solving Absolute value quadratic inequalities

Originally Posted by sakonpure6
how would you solve |x^2 - 4| > 8 ?
Can you solve $\displaystyle x^2>12\text{ or }x^2<-4~?$

5. ## Re: Solving Absolute value quadratic inequalities

Originally Posted by sakonpure6
how would you solve |x^2 - 4| > 8 ?
Copy what Plato did, but change the $\displaystyle <$ sign to a $\displaystyle >$ sign.

$\displaystyle |x^2-4|>8$

$\displaystyle (x^2-4)<-8$ or $\displaystyle 8<(x^2-4)$

$\displaystyle x^2<-4$ or $\displaystyle 12 < x^2$

There are no real values of $\displaystyle x$ with $\displaystyle x^2<-4$, so
$\displaystyle x< -\sqrt{12}$ or $\displaystyle \sqrt{12}< x$.

6. ## Re: Solving Absolute value quadratic inequalities

Oh okay thank you guys!

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# quadratic inequalities with absolute value

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