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mgmat absolute values

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by resilient » Wed May 21, 2008 11:30 am
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|


qa is a

st. 1 gives pos and neg
st.2 gives 0 and 0 so i chose b
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by senthil » Wed May 21, 2008 8:45 pm
I feel it can't be answered by both the options ! the option should be E !


Thanks
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by netigen » Wed May 21, 2008 9:42 pm
I am getting x=0 or x=2 from both the equations should be E
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by simplyjat » Wed May 21, 2008 11:31 pm
If I remember correctly this is a question from MGMAT, and they have a really good explanation for the solution. What did you not understand from the MGMAT explanation?
simplyjat
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by mandy12 » Thu May 22, 2008 12:22 am
Well a very tricky question ..but indeed the answer is A.

Here is how - I learnt from some website that the best way to solve the mod problems is to remove the mod ...i.e. square the equation

so from both 1 and 2 we get the roots as 0 and 2. But if we put these values back in the given equations we see that both 0 and 2 satisfy equation 2 but only 2 satisfies equation 1... hence A is the answer
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mgmat inequalties!

by resilient » Sat May 24, 2008 10:14 pm
Simplyjat,

As you see, I am trying to see where I went wrong my esteemed and dear friend. Your memory serves you correctly and yes this is an mgmat question. The explanation is good but I do want to learn more about absolute values. I tend to get these wrong. This is why! ALso, I am aiming at enriching the conversation on absolute values here. If you don't like the topic, simply don't participate for you are here only on your own choice and voluntary basis.
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by netigen » Sat May 24, 2008 10:53 pm
Good point mandy,

Need to be careful with abs questions. I remember now that we need to always substitute back and confirm the roots work.
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abs

by resilient » Sun May 25, 2008 7:21 am
YES,

For abs I know that you must account for pos and neg situations but I was never checking the alues into the equation to confirm. THis was why I got it wrong. I learned something very important today!
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by kiskopata » Thu May 29, 2008 9:20 am
go through this and you will discover the ancient lost art of solving absolute value equations.....
https://www.manhattangmat.com/strategy-s ... -value.cfm

:lol:
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by netigen » Thu May 29, 2008 10:04 am
There is another way to solve A

|x + 3| = 4x – 3
x = (|x + 3| + 3) / 4

which will always be +ve
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absolutes

by resilient » Thu May 29, 2008 11:06 am
nice approach i like that. makes sense too
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by zacharyz » Thu May 29, 2008 12:56 pm
Mandy12 - when do you find it appropriate (and easiest) to find the answer by squaring the equation. I had not thought about this approach before but am intrigued for it applications.
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