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If k^2 = m^2, which of the following must be true? [#permalink]
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14 Oct 2015, 20:24
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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, 20:28
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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, 21:21
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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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14 Oct 2015, 21:35
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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14 Oct 2015, 23:11
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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 =(km)*(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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24 Oct 2015, 01:18
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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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If k^2 = m^2, which of the following must be true? [#permalink]
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11 Apr 2016, 15:32
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Attached is a visual that should help.
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Screen Shot 20160411 at 4.31.20 PM.png [ 80.7 KiB  Viewed 7646 times ]
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Re: If k^2 = m^2, which of the following must be true? [#permalink]
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03 May 2016, 07:49
mkarthik1 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 In the above question, Both C and E seem to be correct to me .
The official answer is E. Why should it not be C? can someone please explain.
what is the difference between C and E Solution: We are given that k^2 = m^2, and we can start by simplifying the equation by taking the square root of both sides. √k^2 = √m^2 When we take the square root of a variable squared, the result is the absolute value of that variable. Thus: √k^2 = √m^2 is k = m Note that answer choices A through D could all be true, but each of them would be true only under specific circumstances. Answer choice E is the only one that is universally true. Answer: E
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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, 05:29
Hello BunuelI 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, 21:39



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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, 21:59
Bunuel wrote: Shiv2016 wrote: Hello BunuelI 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, 22:05
Shiv2016 wrote: Bunuel wrote: Shiv2016 wrote: Hello BunuelI 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 nonnegative 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 nonnegative, but can we say from k^2 = m^2 that k^4 = m is true? No.
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