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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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A
B
C
D
E

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

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

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New post 14 Oct 2015, 22:35
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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.


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|

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

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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.

The answer is, therefore, (E).
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Re: If k^2 = m^2, which of the following must be true? [#permalink]

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

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    New post 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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      Collection of Questions:
      PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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