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Comment on Absolute Value Equations
Why the absolute value of x
I think it should be written as |-x|= x
If x < 0, then x is negative,
If x < 0, then x is negative, and -x is positive. For example, if x = -5, then -x = -(-5) = 5.
Let's test your suggestion (|-x|= x), by plugging in a negative value for x.
Let's say x = -3. When we replace x with -3, we get |-(-3)| = -3.
When we simplify it, we get |3| = -3, which isn't true.
hello dears
Hi!
Hi!
I got my answer.
Thanks for your prompt reply.
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Thank you again.
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Thanks!
Thanks!
Asma. I had the same question
I don't understand the 3rd
You're referring to this
You're referring to this question: http://gre.myprepclub.com/forum/x-y-2399.html
Notice that if x = 1 and y = -1, get get:
Quantity A:|x + y| = |(-1) + 1| = |0| = 0
Quantity B:|x| + |y| = |(-1)| |+ 1| = 1 + 1 = 2
So, as you can see, the quantities are not necessarily equal.
Great explanation. Thanks.
Hello Brent,
To solve the absolute equation, it is given in Khan's academy to isolate absolute values at one side and check whether b>=0. In this particular question −2∣x−14∣+5=−6∣x−14∣−1, we have same terms inside absolute value. So Im able to isolate them to one side and find that no solutions exist.
How to do this check when we have different absolute terms.
https://www.khanacademy.org/math/algebra-home/alg-absolute-value/alg-absolute-value-equations/e/absolute_value_equations
Thanks :)
Can you give me an example?
Can you give me an example?
Do we need to check for
We don't need to check for
We don't need to check for extraneous roots in quadratic equations.
However, if if the equation involves variables inside square roots or absolute values, we do need to check for extraneous roots.
https://gre.myprepclub.com/forum
Question link: https:/
Question link: https://gre.myprepclub.com/forum/x-6-y-x-3667.html
You're conflating two different components:
1) The part INSIDE the absolute value symbols
2) The number that the absolute value expression EVALUATES to be.
For example, the expression |-5| has a -5 INSIDE the absolute value symbols
And the expression |-5| EVALUATES to be 5 (i.e., |-5| = 5)
Some additional background: The absolute value of a number may be thought of as its DISTANCE from zero.
Since distances cannot be negative, we can see that the absolute value of any number will always evaluate to be greater than or equal to zero.
For example,
|4| = 4
|-3.1| = 3.1
|0| = 0
|-76| = 76
|25.78| = 25.78
|-4| = 4
Notice that ANY value (positive or negative) can be placed INSIDE the absolute value symbols.
However, when we EVALUATE that expression, the outcome is always greater than or equal to zero.
Now notice the first and last examples above. We see that |4| = 4 AND |-4| = 4
So, if I tell you that |x| = 4, what can we conclude about the value of x?
Well, since |4| = 4 and |-4| = 4, x can be EITHER 4 OR -4
Does that help?
Cheers,
Brent
https://gre.myprepclub.com/forum
How does the second equation tells us y cannot be positive
Question link: https:/
Question link: https://gre.myprepclub.com/forum/y-y-and-y-y-2874.html
The second equation reads: y = -|y|
We know that |y| is always GREATER THAN OR EQUAL TO ZERO.
So, -|y| must be LESS THAN OR EQUAL TO ZERO.
Since y = -|y|, we know that y must be LESS THAN OR EQUAL TO ZERO.
Another way to look at it....
Let's test some positive and negative values of y.
Test y = -3
Plug value into equation to get: -3 = -|-3| WORKS!!
Test y = -8
Plug value into equation to get: -8 = -|-8| WORKS!!
Test y = 0
Plug value into equation to get: 0 = -|0| WORKS!!
Test y = 2
Plug value into equation to get: 2 = -|2| DOESN'T WORK
Test y = 11
Plug value into equation to get: 11 = -|11| DOESN'T WORK
Does that help?
Cheers,
Brent
I thought the absolute value
Number can never be negative
Yes, that's correct. However
Yes, that's correct. However if you place a negative symbol in front of that positive value, the result is negative.
For example, |5| = 5, and -|5| = -5.
Likewise, |-7| = 7, and -|-7| = -7.
Does that help?
Cheers,
Brent
hi brent! will we ever be
Yes, I think that sort of
Yes, I think that sort of question would definitely be within the scope of the GRE.
Here's a hypothetical question.
Given: |x² + x| = x
QUANTITY A: x
QUANTITY B: 0
-----------------------
Applying our strategy, we see that:
EITHER x² + x = x OR x² + x = -x
If x² + x = x, then x = 0, so the quantities are equal
If x² + x = -x, then either x = 0 or x = -2, which means.....WAIT
When we test x = -2, we see that it's an extraneous root.
So, x = 0 is the ONLY solution, which means the correct answer is C
Does that help?
Cheers,
Brent
https://gre.myprepclub.com/forum
Could you solve and explain this question for me?
Thanks
|1 - 5| = |5 - m|, Hey brent
We can define |x| as the
We can define |x| as the POSITIVE distance from x to zero on the number line.
Some examples:
|5| = 5
|-7| = 7
|3.8| = 3.8
|-9.427| = 9.427
|0| = 0
|7| = 7
Also, if |x| = k (where k ≥ 0), then x = k or x = -k
Some examples:
If |x| = 4, then x = 4 or x = -4
If |x| = 3.1, then x = 3.1 or x = -3.1
If |x| = 8.888, then x = 8.888 or x = -8.888
If |x| = 0, then x = 0
Does that help?
Ok so what i gather is if |x|
Since |x| is defined as the
Since |x| is defined as the POSITIVE distance from x to zero on the number line, |x| can't have a negative value.
That is, if |x| = k, then it must be the case that k ≥ 0.
Can you please rephrase your question ("...whenever there is a variable between mod symbol is than when we +- x?")
ok let me try to rephrase....
That's correct; if |x| = 5,
That's correct; if |x| = 5, this would mean x = +5 and -5 are the possible solutions.
Hey Brent can u help please
in this question i squared both side got 2 as answer but
my question is can u square mod equation for example |x^2 -5x| = 1?
Question link: https://www
Question link: https://www.reddit.com/r/GRE/comments/fjf2jo/gre_math_question_of_the_da...
Q: How many different solutions does |x² - 5x| = 1 have?
I wouldn't do any squaring.
I'd use the property that says: If |something| = k (where k ≥ 0), then either something = k or something = -k
So, here we get either x² - 5x = 1 or x² - 5x = -1
If x² - 5x = 1, then x² - 5x - 1 = 0.
If x² - 5x = -1, then x² - 5x + 1 = 0.
No need to actually solve here, since the question just asked for the NUMBER of solutions.
Instead, check out the value of the discriminant b² - 4ac from the quadratic equation.
If b² - 4ac < 0, then the equation has no solutions.
If b² - 4ac = 0, then the equation has 1 solution.
If b² - 4ac > 0, then the equation has 2 solutions.
For both possibly equations, b² - 4ac > 0, which means each equation has two different solutions, giving us a total of four solutions.
For more on this, see https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi... (starting at 7:40)