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Addressing multiple solutions of eqns with absolute value

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Addressing multiple solutions of eqns with absolute value

by clawhammer » Fri Dec 03, 2010 8:35 am
What is x?

(1) |x| < 2

(2) |x| = 3x - 2

Note that, i derived both and found:
x < 2, or x > -2
x = 1 and 1/2

So basically independently or together, x could have multiple values that support any or all of the above equations. So I chose E.

Now the OA explains that, multiple solutions of an equation involving absolute values must be verified by plugging in the values again to the original equation. Which tels us in (2) that x=1/2 does not satisfy the equation. Therefore x = 1 and (B) alone is sufficient.

- Is this the official approach? I've never plugged in values back to the original equation (for such purpose) when doing problems before.
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by Rahul@gurome » Fri Dec 03, 2010 8:54 am
clawhammer wrote:What is x?

(1) |x| < 2

(2) |x| = 3x - 2

Note that, i derived both and found:
x < 2, or x > -2
x = 1 and 1/2

So basically independently or together, x could have multiple values that support any or all of the above equations. So I chose E.

Now the OA explains that, multiple solutions of an equation involving absolute values must be verified by plugging in the values again to the original equation. Which tels us in (2) that x=1/2 does not satisfy the equation. Therefore x = 1 and (B) alone is sufficient.

- Is this the official approach? I've never plugged in values back to the original equation (for such purpose) when doing problems before.
Yes, it's the methodical approach. Let's see why.

2nd Statement: |x| = (3x - 2)
We have two cases, x positive and x negative.

For x positive: x = (3x - 2) => x = 1
For x negative: x = -(3x - 2) => x = 1/2 => We have assumed x negative. Therefore, value of x cannot be 1/2. If x = 1/2, then you cannot write x = -(3x - 2) in the beginning.

As you have noted, you don't have to put the result in the original equation to check the correctness of the result. Only thing you have to remember is the assumptions you are making! This is a very common feature with absolute values. Always cross-check your result with initial assumption.
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by goyalsau » Fri Dec 03, 2010 11:22 am
Rahul@gurome wrote: For x positive: x = (3x - 2) => x = 1
For x negative: x = -(3x - 2) => x = 1/2 => We have assumed x negative. Therefore, value of x cannot be 1/2. If x = 1/2, then you cannot write x = -(3x - 2) in the beginning.

As you have noted, you don't have to put the result in the original equation to check the correctness of the result. Only thing you have to remember is the assumptions you are making! This is a very common feature with absolute values. Always cross-check your result with initial assumption.
I am trying to rephrase above statement for better understanding, Please correct if am wrong any where,

While Considering Absolute Signs, Negative value of X must be Negative, It can not be positive.

Rahul the assumption you are talking about, Does it means when we consider the +ve value , value must be +ve, and while considering the -ve value it must be -ve.
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by Rahul@gurome » Fri Dec 03, 2010 11:31 am
goyalsau wrote:Rahul the assumption you are talking about, Does it means when we consider the +ve value , value must be +ve, and while considering the -ve value it must be -ve.
Yes.

Whenever we say |x| = x (or -x), we assume x as positive (or negative). We must stick to that assumption to the end of the solution.
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by goyalsau » Fri Dec 03, 2010 11:42 am
Rahul@gurome wrote:
goyalsau wrote:Rahul the assumption you are talking about, Does it means when we consider the +ve value , value must be +ve, and while considering the -ve value it must be -ve.
Yes.

Whenever we say |x| = x (or -x), we assume x as positive (or negative). We must stick to that assumption to the end of the solution.
Thanks Rahul,
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by clawhammer » Fri Dec 03, 2010 8:38 pm
Rahul@gurome wrote:
goyalsau wrote:Rahul the assumption you are talking about, Does it means when we consider the +ve value , value must be +ve, and while considering the -ve value it must be -ve.
Yes.

Whenever we say |x| = x (or -x), we assume x as positive (or negative). We must stick to that assumption to the end of the solution.
So I could not check for all values, but if I find a POSITIVE value when I assumed X is NEGATIVE, only then I check with that value (or vice versa)? Or should I always check both values, regardless of what I find?
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by goyalsau » Fri Dec 03, 2010 8:57 pm
clawhammer wrote:
So I could not check for all values, but if I find a POSITIVE value when I assumed X is NEGATIVE, only then I check with that value (or vice versa)? Or should I always check both values, regardless of what I find?
Thanks Buddy, for posting such a good question, I never knew that we have such kind of a rule as well, as +ve value when +ve x or -ve value when -ve x

What i a able to understand is that, On almost all the occasions we get a +ve value when considering x as positive and -ve value when considering x as Negative,

But if something different is happening then its always preferred to recheck it once again, Because i think in this particular Test Maker was checking this knowledge only,
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by Rahul@gurome » Fri Dec 03, 2010 10:37 pm
clawhammer wrote:So I could not check for all values, but if I find a POSITIVE value when I assumed X is NEGATIVE, only then I check with that value (or vice versa)? Or should I always check both values, regardless of what I find?
It's better to check always. If you remember your assumptions then you don't have to plug the results into the original equation. For example in this case, for positive x we got x = 1 (which is positive, thus correct) and for negative x, we got x = 1/2 (which is not negative, thus incorrect). How much time did it take it to determine whether they satisfies initial assumption? Maximum of 5 seconds may be! So why don't check it always.
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by Night reader » Sat Dec 04, 2010 12:48 am
clawhammer wrote:What is x?

(1) |x| < 2

(2) |x| = 3x - 2

Note that, i derived both and found:
x < 2, or x > -2
x = 1 and 1/2

So basically independently or together, x could have multiple values that support any or all of the above equations. So I chose E.

Now the OA explains that, multiple solutions of an equation involving absolute values must be verified by plugging in the values again to the original equation. Which tels us in (2) that x=1/2 does not satisfy the equation. Therefore x = 1 and (B) alone is sufficient.

- Is this the official approach? I've never plugged in values back to the original equation (for such purpose) when doing problems before.
great post and thanks to Rahul for explaining +ve and -ve, pre-answer plug in checks
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by rishab1988 » Sat Dec 04, 2010 5:32 am
Hi Rahul

This is a question for you.I noticed that you eliminated value of x=1/2 for it produced a positive and we assumed x to be negative.This causes some doubt,for I have seen in GMATFocus question that the test makers did not do so [atleast it is not visible to my eyes].Perhaps I am missing something.Please shed some light on the same.I am attaching the image.
Image

I believe when x^2+4x+4= |x+2| and x+2 is assumed to be positive then x+2 must yield a +ve value right?

But GMAT solved x=-2 and x=-1.With x=-1 I agree x+2 would yield a positive value 1,but when x=-2 is plugged in x+2,it yields x=0.

Now here is the problem 0 is neither positive not -ve.
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by Rahul@gurome » Sat Dec 04, 2010 6:12 am
rishab1988 wrote:I believe when x^2+4x+4= |x+2| and x+2 is assumed to be positive then x+2 must yield a +ve value right?

But GMAT solved x=-2 and x=-1.With x=-1 I agree x+2 would yield a positive value 1,but when x=-2 is plugged in x+2,it yields x=0.

Now here is the problem 0 is neither positive not -ve.
And the solution is we can treat 0 accordingly. :)

For x = 0, we can take |x| = x or -x anything as we wish. Most of the time it is taken as x and |x| is defined accordingly, i.e. |x| = x for x ≥ 0 and |x| = -x for x < 0. In fact this the actual definition of |x|.

Now the question you mentioned the solution is correct if not perfect! Let's see why. The equation has one crucial point which is x = -2. We will analyze the problem in two region x < -2 (same as (x + 2) < 0) and x ≥ -2 (same as (x + 2) ≥ 0)

x < -2
  • Solution can be found as given i.e. x = -3 and x = -2
    Now in this point we should discard x = -2, as we assumed x < -2. Only solution for this region x = -3
x ≥ -2
  • Solution can be found as given i.e. x = -2 and x = -1
    In this case both are correct as we have assumed x ≥ -2.
This should have been the perfect method for solving the problem. But that doesn't mean what your provided solution has done is not perfect. They have treated x = -2 in both positive and negative cases. That means they have considered |x| = x and also -x for x = 0. It's only how you define |x|.
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