-1
$\begingroup$

Solve : $|x-4|>a$.
Case 1: $a>0$; Case 2: $a<0$

Progress

I am getting answers which look similar in both cases:

  • Let $a>0$ so $x>4+a$ or $x<4-a$ ,
  • Let $a<0$ so $x>4+a$ or $x<4-a$ .

Though I know that both answers' meaning is different I am unable to find out how the points included in both cases are different

I wish to know why it is so and how different both answers are when plotted on a number line.

$\endgroup$
  • $\begingroup$ can u show your work? $\endgroup$ – Bhargav Jun 4 '12 at 10:12
  • $\begingroup$ let a>0 so x>4+a or x<4-a , let a<0 so x>4+a or x<4-a .Though i know that both answer's meaning is different i am unable to find out how the points included in both cases are different $\endgroup$ – mgh Jun 4 '12 at 10:14
  • $\begingroup$ Related: math.stackexchange.com/questions/152869/… $\endgroup$ – TMM Jun 4 '12 at 11:07
1
$\begingroup$

If $a \lt 0$, all $x$ will satisfy it as all absolute values are $ \ge 0$. If $a \gt 0$ you need the points more than $a$ from $4$.

$\endgroup$
  • $\begingroup$ :i was having problem to solve:|x-2|+|x-5|=3 . I was trying to post this question but could not(i was being said that it does not meet our quality standards).So posted it here. $\endgroup$ – mgh Jun 4 '12 at 15:43
  • $\begingroup$ @meg_1997: when you have two absolute value signs, it is easiest to consider each region of $x$ and resolve the signs. So for $x\le 2$ both expressions are negative and need to be inverted. You are left with $7-2x=3$ AND $x \le 2$. You solve the equality and see if it meets the inequality. Then there are two more sections of the real line to consider the same way. $\endgroup$ – Ross Millikan Jun 4 '12 at 21:15
0
$\begingroup$

Consider multiple cases

Case 1: $a = 0$

Then, any number other than $4$ will satisfy your inequality.

Case 2: $a < 0$

Then, any $x$ will satisfy your inequality since absolute values are $\ge 0$

Case 3: $a > 0$

Then $|x - 4| > a$ if and only if $x$ is farther than $a$ units from $4$. Hence, $x - 4 > a$ or $x - 4 < -a$. So the set of all real numbers that satisfy your inequality is $$(-\infty, 4 - a) \cup (4+a, \infty)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.