1. ## equation inequalities

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

2. Hello quah13579
Originally Posted by quah13579
Find the solutions sets for the following inequalities.
(a) |x − 2| < 4
(b) $\displaystyle x^2 − 1 ≥ 3$
(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 $\displaystyle x \ge 2$

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) $\displaystyle x^2 − 1 ≥ 3$
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 $\displaystyle x\in (\infty,-2] \cup [2,\infty)$