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M31-19

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Bunuel
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M31-19 [#permalink] New post   09 Jun 2015, 08:32
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What is the range of all the roots of \(|x^2 - 2| = x\)?

A. 4
B. 3
C. 2
D. 1
E. 0
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D
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Bunuel
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Re M31-19 [#permalink] New post   09 Jun 2015, 08:32
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Official Solution:

What is the range of all the roots of \(|x^2 - 2| = x\)?

A. 4
B. 3
C. 2
D. 1
E. 0


First, note that since \(x\) is equal to the absolute value of some number (\(|x^2 - 2|\)), \(x\) cannot be negative.

Next, \(|x^2 - 2| = x\) implies that either \(x^2 - 2 = x\) or \(-(x^2 - 2) = x\).

The first equation, \(x^2 - 2 = x\), yields \(x = -1\) or \(x = 2\). Since \(x\) cannot be negative, we only consider \(x = 2\).

The second equation, \(-(x^2 - 2) = x\), yields \(x = -2\) or \(x = 1\). Again, since \(x\) cannot be negative, we only consider \(x = 1\).

The range is the difference between the largest and smallest values: range = 2 - 1 = 1.


Answer: D
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monika25
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Re: M31-19 [#permalink] New post   23 Jun 2016, 08:57
I think this is a high-quality question.
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aalekhoza
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Re: M31-19 [#permalink] New post   16 Oct 2018, 22:42
Hello Bunuel
For this particular question, can we not square both the sides?

|x^2 - 2| = x
(x^2 - 2)^2 = x^2 (squaring both the sides)
x^2 + 4 -4x = x^2 (expanding the equations)
4x = 4 (Subtracting x^2)
x = 1
and hence, D?

Though I completely agree and understood your above explanation and I also remember from one of you post that as we don't know whether the mod value is positive or negative we should not square and look for an alternative way to solve the question, can you elaborate on the difference between the two approaches?
Thanks.
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Bunuel
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Re: M31-19 [#permalink] New post   16 Oct 2018, 23:44
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aalekhoza wrote:
Hello Bunuel
For this particular question, can we not square both the sides?

|x^2 - 2| = x
(x^2 - 2)^2 = x^2 (squaring both the sides)
x^2 + 4 -4x = x^2 (expanding the equations)
4x = 4 (Subtracting x^2)
x = 1
and hence, D?

Though I completely agree and understood your above explanation and I also remember from one of you post that as we don't know whether the mod value is positive or negative we should not square and look for an alternative way to solve the question, can you elaborate on the difference between the two approaches?
Thanks.


1. We can square equations if both sides are non-negative (for example, if it were |x^2 - 2| = |x|, then squaring would be correct). When that's not the case squaring usually creates more roots, then there actually are.

2. Highlighted part is not correct: \((x^2 - 2)^2=x^4 - 4 x^2 + 4\), notice that x on the left hand side is in fourth power not squared.

3. If you solve \(x^4 - 4 x^2 + 4=x^2\), you'll get that x can be -2, -2, 1, or 2. But -2 and -1 do not satisfy |x^2 - 2| = x, and should be discarded. So, we are left with only x = 1 and x = 2. As you can see, squaring gave more roots, then there actually are.

Hope it helps.
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Re: M31-19 [#permalink] New post   20 Nov 2019, 16:52
VeritasKarishma,

Can we solve this question using graphs that you discussed in one of the posts? I tried using that method but the only relevant data that I could get was that we can have 2 roots. But I could not find what the roots are from the graphs. Am I missing something?
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KarishmaB
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Re: M31-19 [#permalink] New post   20 Nov 2019, 22:01
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Alok322 wrote:
VeritasKarishma,

Can we solve this question using graphs that you discussed in one of the posts? I tried using that method but the only relevant data that I could get was that we can have 2 roots. But I could not find what the roots are from the graphs. Am I missing something?



The easiest way to deal with this question is to square and find out which roots satisfy the original equation. Graphing helps you see that there are two roots. One will be between 0 and sqrt(2) and the other a bit greater than sqrt(2). Here, I could see that 1 and 2 satisfy the equation though some quadratic solving would be required if you were unable to guess the roots. Finding the roots is easier with lines because of the linear relation between x and y.
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RenanBragion
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Re: M31-19 [#permalink] New post   18 Jun 2020, 01:56
I think this is a high-quality question and I agree with explanation.
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Bunuel
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Re: M31-19 [#permalink] New post   18 Apr 2023, 15:16
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M31-19 [#permalink] New post   12 Aug 2023, 11:06
I think this is a high-quality question and I agree with explanation.
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