1. ## equation inequalities

Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) $x^2 − 1 ≥ 3$

2. Hello quah13579
Originally Posted by quah13579
Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) $x^2 − 1 ≥ 3$
(a) $|x-2| < 4$ means that the numerical value of $(x-2)$ is less than $4$. So $(x-2)$ lies between $-4$ and $4$.

$\Rightarrow -4 < x-2 < 4$

Take the LH inequality: $-4< x-2$

$\Rightarrow -2

...and the RH inequality: $x-2<4$

$\Rightarrow x < 6$

So, combining these two answers into one: $-2

(b) Although you had a LaTeX error in your original posting, I assume it's $x^2 - 1 \ge 3$

So $x^2 \ge 4$

When we take square roots, be careful with the negative values. We need to use an absolute value sign |...|, like this:

$|x| \ge 2$

$\Rightarrow x \le -2$ or $x \ge 2$

3. [quote=quah13579;348363]Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) x^2 − 1 ≥ 3

4. Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) x^2 − 1 ≥ 3

5. Originally Posted by quah13579
Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) $x^2 − 1 ≥ 3$
for the absolute value

$a)\mid x-2 \mid < 4$ in general

$\mid x-a \mid < b \Rightarrow -b
and

$\mid x-a \mid > b \Rightarrow -b>x$ or $b
so

$\mid x-2 \mid < 4 \Rightarrow -4

second one

$b)x^2-1 \geq 3$

$x^2-1-3 \geq 0$

$x^2-4 \geq 0$

$(x-2)(x+2) \geq 0$ we want the values of x which make the value of $x^2-4$ positive or zero so we need to study the sign of it
the root of it is -2 ,2 plot them in the real line like this

then take numbers from the three intervals (-infinity , -2 ) and from (-2,2) and from (2,infinity) then sub them in the formula if it was positive write positive sign if negative write negative on the real line like this

the solution is $x\in (\infty,-2] \cup [2,\infty)$