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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| Answer E
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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|

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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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.

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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Help me make my explanation better by providing a logical feedback.

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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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|.

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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