On average you will see at least one question on the Revised GRE dealing with absolute values. You may even see a few. Yet, absolute value gets lost in the prep fray amongst the more popular concepts. So if you don\u2019t want this relatively innocuous concept to surprise you test day read on.<\/p>\n
<\/p>\n
What do -4 and 4 have in common? They are both four units from the 0 on a number line. Think of absolute value has how far from the zero a given number is on a number line.<\/p>\n
For positive numbers finding the absolute value is easy \u2013 it is always the number between the absolute value signs, which look like this \\({|}{ }{|}\\).<\/p>\n
\\({|}{5}{|}=5\\)<\/p>\n
\\({|}{1\/2}{|}=1\/2\\)<\/p>\n
\\({|}{100}{|}=100\\)<\/p>\n
<\/p>\n
When we take the absolute value of a negative number, we drop the negative, and the absolute value sign.<\/p>\n
\\({|}{-5}{|}=5\\)<\/p>\n
\\({|}{-1\/2}{|}=1\/2\\)<\/p>\n
\\({|}{-100}{|}={100}\\)<\/p>\n
<\/p>\n
What is the value of x in the equation \\({|}{x}{|}=5\\)? Well, the absolute value of both \\(5\\) and \\(-5\\) is \\(5\\), so \\(x = 5\\) or \\(-5\\).<\/p>\n
Now take a look at the following: \\({|}{x}{|}={-5}\\)<\/p>\n
Anything seem off? Well, the absolute value of any number can never be a negative therefore there is no value for x.<\/p>\n
To solve for a variable inside an absolute value sign, we want to remove the absolute value sign and solve the equation. However, there is a slight twist: you will want to create two separate equations. For one remove the absolute value signs and solves for x. For the other, make the side of the equation not inside the absolute value equal to a negative. Let\u2019s try a practice problem:<\/p>\n
\\({|}{x-2}{|}=4\\)?<\/p>\n
Our two equations are:<\/p>\n
\\(x-2=4\\)<\/p>\n
\\(x-2=-4\\)<\/p>\n
If this seems strange, think of it this way: when you find a value for x that makes the equation equal -4, you have also solved for positive four. Remember, the absolute value of -4 is 4. Solving for x we get:<\/p>\n
\\(x-2=4\\), \\(x=6\\)<\/p>\n
\\(x-2=-4\\), \\(x=-2\\).<\/p>\n
Therefore x can equal 6 or -2.<\/p>\n
<\/p>\n
Now let\u2019s complicate things a little and throw an inequality in to the picture. Have a look:<\/p>\n
\\({|}{x-4}{|}<3\\)<\/p>\n
<\/p>\n
We turn the inequality sign into an equal sign and solve for x the way we did above.<\/p>\n
\\(x-4=3\\)<\/p>\n
\\(x=7\\)<\/p>\n
so \\(x<7\\).<\/p>\n
<\/p>\n
However, when we turn 3 into -3, we have to reverse the sign. This is the one critical step.<\/p>\n
\\(x-4=-3\\)<\/p>\n
\\(x=1\\)<\/p>\n
so \\(x>1\\).<\/p>\n
<\/p>\n
Therefore, \\(1<x<y\\). You can plug in different values for x to see how this is the case.<\/p>\n
<\/p>\n
This is a basic overview to absolute value and should help you with most of the sub-150 problems. For the harder problems, however, you will want to make sure to practice with more advanced problems.<\/p>\n","protected":false},"excerpt":{"rendered":"
On average you will see at least one question on the Revised GRE dealing with absolute values. You may even see a few. Yet, absolute value gets lost in the prep fray amongst the more popular concepts. So if you don\u2019t want this relatively innocuous concept to surprise you test day read on. What […]<\/p>\n","protected":false},"author":10,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[41,25],"tags":[],"ppma_author":[12264],"class_list":["post-8233","post","type-post","status-publish","format-standard","hentry","category-arithmetic","category-math"],"acf":[],"yoast_head":"\n