Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) $\displaystyle x^2 − 1 ≥ 3$
Hello quah13579(a) $\displaystyle |x-2| < 4$ means that the numerical value of $\displaystyle (x-2)$ is less than $\displaystyle 4$. So $\displaystyle (x-2)$ lies between $\displaystyle -4$ and $\displaystyle 4$.
$\displaystyle \Rightarrow -4 < x-2 < 4$
Take the LH inequality: $\displaystyle -4< x-2$
$\displaystyle \Rightarrow -2<x$
...and the RH inequality: $\displaystyle x-2<4$
$\displaystyle \Rightarrow x < 6$
So, combining these two answers into one: $\displaystyle -2<x<6$
(b) Although you had a LaTeX error in your original posting, I assume it's $\displaystyle x^2 - 1 \ge 3$
So $\displaystyle x^2 \ge 4$
When we take square roots, be careful with the negative values. We need to use an absolute value sign |...|, like this:
$\displaystyle |x| \ge 2$
$\displaystyle \Rightarrow x \le -2$ or $\displaystyle x \ge 2$
Grandad
for the absolute value
$\displaystyle a)\mid x-2 \mid < 4$ in general
$\displaystyle \mid x-a \mid < b \Rightarrow -b<x-a <b$
and
$\displaystyle \mid x-a \mid > b \Rightarrow -b>x $ or $\displaystyle b<x$
so
$\displaystyle \mid x-2 \mid < 4 \Rightarrow -4<x-2<4 \Rightarrow -4+2 < x < 4+2 \Rightarrow -2<x<6$
second one
$\displaystyle b)x^2-1 \geq 3 $
$\displaystyle x^2-1-3 \geq 0 $
$\displaystyle x^2-4 \geq 0 $
$\displaystyle (x-2)(x+2) \geq 0 $ we want the values of x which make the value of $\displaystyle x^2-4$ 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 $\displaystyle x\in (\infty,-2] \cup [2,\infty)$