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


What does \(k^2 = m^2\) imply?

Only that |k| = |m|

(A) k = m
Not necessary e.g. k = 5, m = -5

(B) k = −m
Not necessary e.g. k = 5, m = 5

(C) k = |m|
Not necessary e.g. k = -5, m = 5

(D) k = −|m|
Not necessary e.g. k = 5, m = 5

(E) |k| = |m|
Always necessary. Note that absolute values of the two most be equal for the square to be equal. We cannot say anything about their signs though.

Answer (E)
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Re: If k^2 = m^2, which of the following must be true? [#permalink]
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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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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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Attached is a visual that should help.
Attachments

Screen Shot 2016-04-11 at 4.31.20 PM.png
Screen Shot 2016-04-11 at 4.31.20 PM.png [ 80.7 KiB | Viewed 61484 times ]

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


The question asks us what MUST be true. So, if we can find a case where a statement is not true, we can eliminate that answer choice.

So, for example, one solution to the equation (k² = m²) is k = 1 and m = 1
Now let's check the answer choices.
A. k = m. Test: 1 = 1. Works. Keep A.
B. k = -m. Test: 1 = -1. DOESN'T WORK. ELIMINATE B.
C. k = |m|. Test: 1 = |1|. Works. Keep C.
D. k = - |m|. Test: 1 = -|1|. DOESN'T WORK. ELIMINATE D.
E. |k| = |m|. Test: |1| = |1|. Works. Keep E.

Okay, so the correct answer is A, C or E

Let's try another case. Another solution to the equation (k² = m²) is k = -1 and m = 1
Now let's check the remaining answer choices.
A. k = m. Test: -1 = 1. DOESN'T WORK. ELIMINATE A.
C. k = |m|. Test: -1 = |1|. DOESN'T WORK. ELIMINATE C.
E. |k| = |m|. Test: |-1| = |1|. Works. Keep E.

By the process of elimination, the correct answer is E

Cheers,
Brent
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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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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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Using the rule that \(\sqrt{x^2}=|x|\), the equation can be changed to:

\(\sqrt{k^2} = \sqrt{m^2} \rightarrow |k| = |m|\)

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

      If K squared equals an M squared, that means that a K is equal to plus or minus M.

      That basically solves the question, because K is not always M;

      it's not always minus M; it's not equal to the absolute value of M because K might

      be a negative, so negative cannot be something that's positive.

      D says that K, which can either be positive or negative, is always equal to

      minus of a positive number; well that's not always correct,

      and E is the correct one because it says that plus a equals a plus.
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      Re: If k^2 = m^2, which of the following must be true? [#permalink]
      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.


      Given: k^2 = m^2

      Asked: Which of the following must be true?

      If k^2 = m^2
      Taking root on both sides
      |k| = |m|

      IMO E
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      Re: If k^2 = m^2, which of the following must be true? [#permalink]
      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 \(k^2 = m^2\) which means that one positive equals the other positive i.e.
      + = +

      So we have two possibilities to show that '+ = +'
      1. Using even powers : \((+/-)^2 = (+/-)^2 \)
      2. Using Mods : |k| = |m| (since \(\sqrt{k^2}\) = |k|

      Among the options only E satisfies. Other options involve one or the other variable which can take either positive or negative values. Here we need only positive terms.

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


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      Answer: Option E

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