While solving GMAT quant questions, always remember that your one goal is to identify the correct answer as efficiently as possible, and not to please your former math teachers.
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Comment on Absolute Value Equations
https://www.beatthegmat.com
I can’t understand the question text ie (Statement 2) |x| = 4x − 15
Pls help
Beat The GMAT recently made a
Beat The GMAT recently made a few changes to their website, and now several mathematical symbols are showing up wrong.
I've changed some of the links so they now go to the same question on GMAT Club.
The link to the question you're asking about is here: https://gmatclub.com/forum/what-is-the-value-of-x-161134.html
How to solve |x+2|=|Y+2| what
If |x+2| = |y+2|, what is the value of x + y ?
(1) xy <0
(2) x > 2 and y < 2
Here's my full solution:
Here's my full solution: https://gmatclub.com/forum/x-2-y-2-what-is-the-value-of-x-y-172994.html#...
Hi Brent, I am referring to
What is the value of m?
(1) |m| = −36/m
(2) 2m+2|m| = 0
I am clear on statement 1, however when you evaluate statement 2 after simplifying it to |m| = -m , you have concluded that statement two could take any value, I can't quite get it. Please can you help explain it further? Thanks.
Statement 2: 2m + 2|m| = 0
Subtract 2m from both sides to get: 2|m| = -2m
Divide both sides by 2 to get: |m| = -m
Upon inspection we might see that there are several possible solutions, including m = 0, m = -1, m = -2 and so on.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Let's start from here: |m| =
Let's start from here: |m| = -m
Aside: We can also write: |m| = (-1)m
|m| must be greater than or equal to 0 (a property of absolute values), which means (-1)m must be greater than or equal to 0
In other words, If |m| ≥ 0, then it must also be true that -m ≥ 0
Add m to both sides of the inequality to get: 0 ≥ m
Let's test some values of m that satisfy the condition that 0 ≥ m
If m = -3, then |m| = -m becomes |-3| = -(-3). Works!
If m = -8, then |m| = -m becomes |-8| = -(-8). Works!
If m = 0, then |m| = -m becomes |0| = -0. Works!
Does that help??
Thanks Brent. This is Very
Hi Brent, I am referring to
What is the sum of all the real values of x for which |x-4|^2 + |x-4| = 30?
A) 16
B) 11
C) 9
D) 8
E) 7
Once you reach the conclusion that Since u = |x - 4|, we can know that that either |x - 4| = -6 or |x - 4| = 5.
Then you have discarded the first equation on following ground.|x - 4| = -6
Since the absolute value is always greater than or equal to 0, it's IMPOSSIBLE for |something| = -6
So, this equation has no solution
Then on to the next one, you have again assumed two possibilities +5 and -5.
|x - 4| = 5
This means that x - 4 = 5 or x - 4 = -5
If x - 4 = 5, then x = 9
If x - 4 = -5, then x = -1
My question is, earlier you have discarded the equation which contained negative value |x - 4| = -6. Now how can we assume that |x - 4| = -5? I am assuming that this can only be +5. Please can you explain?
You're referring to my
You're referring to my solution here: https://gmatclub.com/forum/what-is-the-sum-of-all-the-real-values-of-x-f...
Q: How can we assume that |x - 4| = -5?
A: We're not assuming that |x - 4| = -5. We're saying that, if |x - 4| = 5, then either x - 4 = 5 or x - 4 = -5
This is much different from saying |x - 4| = -5
Does that help?
Hi Brent,
This concept is quite difficult to fully understand.
Anyways, I do need some clarifications.
Question 1 (I will use this question to illustrate: https://gmatclub.com/forum/what-is-the-sum-of-all-possible-solutions-of-the-equation-x-85988.html).
1. When solving, I get the roots (x-8)(x+6) when assuming that |x-4| is positive and (x+16)(x+2) when negative. Therefore x=8 or x=2 and x=-16 or x=-2.
a. To check whether these 4 values for x work, do we have to plug them back in |x + 4|^2 - 10|x + 4| = 24 as the only way to verify, or we also can plug them in the expanded expression (x²+16+8x...)?
b. Do we have to check the values of x we got for |x + 4| is negative and therefore plug them in -x-4 (in this case we only check -16 and -2) and same for values resulting when |x + 4| is positive (x+4)?
Or we can check them all by plugging the 4 values only when |x + 4| is positive (or negative)?
Question 2 (illustration: 2. https://gmatclub.com/forum/x-2-y-2-what-is-the-value-of-x-y-172994.html)
2. When solving, many people square the two sides of the equation. I saw that you didn't. Is there a particular reason? Any particular type of question that suggests squaring over simple solving?
Question 1 link: https:/
Question 1 link: https://gmatclub.com/forum/what-is-the-sum-of-all-possible-solutions-of-...
Once you find your 4 solutions (x = -6, x = 8, x = -2 and x = -16), you must check to see whether each one is a valid solution by plugging them into the original equation |x + 4|^2 - 10|x + 4| = 24.
By the way, here's a MUCH faster solution to that question: https://gmatclub.com/forum/what-is-the-sum-of-all-possible-solutions-of-...
Question 2 link: https://gmatclub.com/forum/x-2-y-2-what-is-the-value-of-x-y-172994.html
There are several different ways to solve this question.
For Data Sufficiency questions, I typically like to list/examine the possible cases before dealing with the statements.
I find that this makes it easier/faster to analyze each statement,
Need help with this question
For Case 1, I solved RHS = = +ve LHS and got solutions x = 8 or -6. After testing in original equation, only x = 8 is valid.
However, for Case 2 when I choose RHS = -ve LHS I got solutions x = 0 and 2. After testing in original equation, both x = 2 and 0 are valid. However, this is not the correct answer. My working for Case 2 is as below:
(x+4)^2 -10(x+4) = -24
x^2+8x+16 -10x-40 = -24
x^2-2x-24 = -24
x^2-2x=0
x(x-2) = 0
x = 0 or x = 2
Testing x = 0,
|0+4|^2 - 10 |0+4| = -24
16 - 40 = -24
-24=-24
Hence, x = 0 is a root.
Testing x = 2,
|2 + 4|^2 - 10|2+4| = -24
36 - 60 = -24
-24 = -24
Hence, x = 2 is a root.
Therefore, sum of roots is 8 + 0 + 2 = 10.
There are a few mistakes with
There are a few mistakes with your solution.
Keep in mind that the rules says, "If |x| = k, then either x = k or x = -k"
Another way to write this as as follows: "If |x| = k, then either x = k or -x = k"
Notice that both of these rules involve a single absolute value.
So, for our original equation, we must consider |x + 4| = x + 4 and |x + 4| = -(x + 4)
Our two cases are as follows
Case i: (x + 4)² - 10(x + 4) = 24
Case ii: [-(x + 4)]² - 10[-(x + 4)] = 24
When we solve for x, we get:
Case i: x = 8 and x = -6
Case ii: x = -2 and x = -16
At this point, we must plug all four possible values into the ORIGINAL equation |x + 4|² - 10 |x + 4| = 24
When we do this, we see that x = -2 and x = -6 are extraneous roots
Does that help?
Thanks for the quick response
A few follow up questions as the concept is not 100% clear to me.
1. For the rule |x| = k, then either x = k or x = -k I am confused how to apply when there is more than one absolute value. For simplicity sake, I put all absolute value on RHS and simply applied | RHS | = +ve LHS OR -ve LHS.
Can you help me understand the logic behind why my approach in the question above is wrong?
Also, Is there an easier way to apply the rule |x| = k, then either x = k or x = -k, when there is more than one absolute value in the question?
2. When testing for extraneous roots, do we also use case 2 roots ( i.e. x = -k ) for the ORIGINAL equation? Or must it be tested in the new equation where x = -k?
Sorry for so many questions- I need to understand the above to clarify the concepts.
1. If you have an equation
1. If you have an equation with more than one absolute value expression, AND each absolute value expression is the SAME (e.g., |2x+1|), then you must analyze two cases:
case a: |2x+1| = 2x+1
case b: |2x+1| = -(2x+1)
Having said that, if you have more one identical absolute value expressions, it's quite likely that u-substitution will yield the fastest solution. Here's what I mean: https://gmatclub.com/forum/what-is-the-sum-of-all-possible-solutions-of-...
2. To check for extraneous roots, all solutions must be plugged into the ORIGINAL equation (before any manipulations or replacements)
Thank you, it is crystal
I'm confused, so the absolute
That way, shouldn't the absolute value of -3 is -3?
If x < 0, then |x| = -x
If x < 0, then |x| = -x
Let's test a few values to see what's going on when x < 0.
If x = -3, we can replace x with -3 to get: |-3| = -(-3)
Simplify the right side: |-3| = 3. Works
If x = -7, we can replace x with -7 to get: |-7| = -(-7)
Simplify the right side: |-7| = 7. Works
Does that help?
Hi Brent, at video 0:33 and 0
In this OG question, |-R| = -r. Can we say when it comes to absolute value with negative number, it will always be negative value. So it's like without absolute funtion here.
So Absolute value only matter when it comes to positive values that we will need to calculate both positive and negative values on removing the absolute function. Could you help clarify? Thanks Brent
https://gmatclub.com/forum/the-number-line-shown-contains-three-points-r-s-and-t-whose-coordin-113036.html#p915379
In the video I say the
In the video I say the following:
Property #1: If x ≥ 0, then |x| = x
Property #2: If x < 0, then |x| = -x
So, if x = -3, then we apply property #2 (|x| = -x) to get: |-3| = -(-3), which evaluates to be: |-3| = 3
I'm not sure how you reached the conclusion that -3 > 0.
Can you elaborate?
You might find it easier to think of |k| as being the positive distance from k to zero on the number line.
For example, to find the value of |-5|, we will ask "What is the distance from -5 to 0 on the number line?"
The distance is 5, we can say that |-5| = 5
Question link: https://gmatclub.com/forum/the-number-line-shown-contains-three-points-r...
For this question, the points R, S and T on the number line tell us that R < 0, S > 0 and T > 0.
So, if |R| = r, then that means r is positive (and R is negative)
In order to make r the same value as R, we must negate r to get -r.
At this point we can say that R = -r
You can also verify this with an example.
If R = -6, then |R| = r becomes |-6| = 6, which means when R = -6, r = 6
Notice that, at the moment R and r are not equal.
To make them equal, we will negate r to get -r, in other words, -6 (the value of -r) now equal to the value of R (-6)
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