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# If k^2 = m^2, which of the following must be true?

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If k^2 = m^2, which of the following must be true? [#permalink]

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14 Oct 2015, 21:24
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If k^2 = m^2, which of the following must be true?

(A) k = m
(B) k = −m
(C) k = |m|
(D) k = −|m|
(E) |k| = |m|

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[Reveal] Spoiler: OA

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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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14 Oct 2015, 21:28
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|k| = |m| suffice the condition of k^2 = m^2.

so ans: E

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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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14 Oct 2015, 22:21
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k^2=m^2
Taking the non negative square root of both sides of the equation ,
|k| = |m|
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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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14 Oct 2015, 22:35
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Expert's post
Bunuel wrote:
If k^2 = m^2, which of the following must be true?

(A) k = m
(B) k = −m
(C) k = |m|
(D) k = −|m|
(E) |k| = |m|

Kudos for a correct solution.

With Even powers of variables the signs can't be predicted about them

Hence, We can't say anything about the sign of k and m being positive or negative
therefore, Option A, B, C and D are ruled out as all these options are hinting towards specific and known sign of k and/or m

For any sign of k and m, their absolute values must be same because k^2 = m^2, therefore, |k| = |m|

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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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15 Oct 2015, 00:11
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If k^2 = m^2, which of the following must be true?

(A) k = m
(B) k = −m
(C) k = |m|
(D) k = −|m|
(E) |k| = |m|

Since k^2=m^2 we have 0=k^2 – m^2 =(k-m)*(k+m). So k=m or k=-m.
So only (A) and only (B) cannot be an answer.
The choice (C) tells us that k should be greater than or equal to 0.
Similarly the choice (D) tells us that k should be less than or equal to 0.
So neither (C) nor (D) cannot be the answer.

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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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24 Oct 2015, 02:18
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Bunuel wrote:
If k^2 = m^2, which of the following must be true?

(A) k = m
(B) k = −m
(C) k = |m|
(D) k = −|m|
(E) |k| = |m|

Kudos for a correct solution.

As squaring hides the sign of a number, k^2 will equal m^2 for every possible positive/negative combination of the two values.

The only statement we can make for sure is therefore (E) |k| = |m|.

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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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06 Sep 2017, 06:29
Hello Bunuel

I inferred that square or even power of a number is always positive. Therefore k^2 and m^2 are positive. So |k|=|m|, no matter what the signs are.
I don't see why in the official guide its mentioned that we need to take root of k^2 and m^2 when we know that ^2 always gives positive number.
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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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06 Sep 2017, 22:39
Shiv2016 wrote:
Hello Bunuel

I inferred that square or even power of a number is always positive. Therefore k^2 and m^2 are positive. So |k|=|m|, no matter what the signs are.
I don't see why in the official guide its mentioned that we need to take root of k^2 and m^2 when we know that ^2 always gives positive number.

Do not follow what you mean bu the point is that $$\sqrt{x^2}=|x|$$, so if we take the square root from k^2 = m^2, we'll get |k| = |m|.
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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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06 Sep 2017, 22:59
Bunuel wrote:
Shiv2016 wrote:
Hello Bunuel

I inferred that square or even power of a number is always positive. Therefore k^2 and m^2 are positive. So |k|=|m|, no matter what the signs are.
I don't see why in the official guide its mentioned that we need to take root of k^2 and m^2 when we know that ^2 always gives positive number.

Do not follow what you mean bu the point is that $$\sqrt{x^2}=|x|$$, so if we take the square root from k^2 = m^2, we'll get |k| = |m|.

Hi. Sorry for being not so clear

I meant to say that even powers ^2,^4, etc. always gives positive outcome no matter what the sign of the base is. For example:
(-2)^2= 4
and
(2)^2= 4

So what I inferred is that k^2 and m^2 are positive (no matter what the sign of k and m are). Therefore we can say that |k|=|m|.

Is this good?

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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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06 Sep 2017, 23:05
Shiv2016 wrote:
Bunuel wrote:
Shiv2016 wrote:
Hello Bunuel

I inferred that square or even power of a number is always positive. Therefore k^2 and m^2 are positive. So |k|=|m|, no matter what the signs are.
I don't see why in the official guide its mentioned that we need to take root of k^2 and m^2 when we know that ^2 always gives positive number.

Do not follow what you mean bu the point is that $$\sqrt{x^2}=|x|$$, so if we take the square root from k^2 = m^2, we'll get |k| = |m|.

Hi. Sorry for being not so clear

I meant to say that even powers ^2,^4, etc. always gives positive outcome no matter what the sign of the base is. For example:
(-2)^2= 4
and
(2)^2= 4

So what I inferred is that k^2 and m^2 are positive (no matter what the sign of k and m are). Therefore we can say that |k|=|m|.

Is this good?

Apart from knowing that the even roots and absolute values give non-negative result, we should also deduce that from k^2 = m^2 we can get |k| = |m|. Else, what would you say if one of the options were k^4 = |m|? Here both sides are also non-negative, but can we say from k^2 = m^2 that k^4 = |m| is true? No.
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Re: If k^2 = m^2, which of the following must be true?   [#permalink] 06 Sep 2017, 23:05
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