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# If x = 9a2 and a < 0, then squareroot(x) = 3a 3a 9a 9a 81

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Manager
Joined: 14 Oct 2008
Posts: 160
If x = 9a2 and a < 0, then squareroot(x) = 3a 3a 9a 9a 81 [#permalink]

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05 Nov 2008, 03:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x = 9a2 and a < 0, then squareroot(x) =

–3a
3a
9a
–9a
81
It looks simple , but i am not convinced by the answer kaplan gave for this.
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Manager
Joined: 30 Sep 2008
Posts: 111
Re: Looks simple, is the qs correct ? [#permalink]

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05 Nov 2008, 03:17
If x = 9a^2 => sqrt(x) = 3a or sqrt(x) = -3a

sqrt(x) must be greater than 0, we have a<0 => sqrt(x) = -3a

Manager
Joined: 14 Oct 2008
Posts: 160
Re: Looks simple, is the qs correct ? [#permalink]

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05 Nov 2008, 03:35
Yup, the sqrt must be taken as positive .
SVP
Joined: 17 Jun 2008
Posts: 1507
Re: Looks simple, is the qs correct ? [#permalink]

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05 Nov 2008, 04:05
sqrt(a^2) will be |a| and since a is negative, |a| = -a.

Hence, sqrt(9a^2) = -3a.
Re: Looks simple, is the qs correct ?   [#permalink] 05 Nov 2008, 04:05
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# If x = 9a2 and a < 0, then squareroot(x) = 3a 3a 9a 9a 81

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