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M31-19

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Joined: 14 Jan 2017
Posts: 120
Location: India
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Re: M31-19  [#permalink]

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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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Re: M31-19  [#permalink]

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New post 16 Oct 2018, 23:44
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] 16 Oct 2018, 23:44

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