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Simple Algebra

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Manager
Manager
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Joined: 14 May 2007
Posts: 184

Kudos [?]: 87 [0], given: 12

Location: India
Simple Algebra [#permalink]

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New post 31 Jan 2009, 22:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi Guys,

I got a simple question and I was kinda stuck. The ques is -

Q-1) 5(x+2)^2 = 180 . What can be a value for 2x + 5

(a) 6
(b) 12
(c) -9
(d) -11
(e) 17

How should I go about solving this question ?

Thanks,

GR

Kudos [?]: 87 [0], given: 12

SVP
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Joined: 29 Aug 2007
Posts: 2473

Kudos [?]: 832 [0], given: 19

Re: Simple Algebra [#permalink]

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New post 31 Jan 2009, 23:01
greatchap wrote:
5 (x+2)^2 = 180. What can be a value for 2x + 5

(a) 6
(b) 12
(c) -9
(d) -11
(e) 17



Always be careful for can/could be:


5 (x+2)^2 = 180
(x+2)^2 = 36
x^2 + 4x + 4 = 36
x^2 + 4x - 32 = 0
x^2 + 8x - 4x - 32 = 0
x (x + 8) (x - 4) = 0
x = -8 or 4.

2x + 5 = -11 or 13. 13 is not in the answer choices, so it is -11.
D.
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GT

Kudos [?]: 832 [0], given: 19

Manager
Manager
User avatar
Joined: 14 May 2007
Posts: 184

Kudos [?]: 87 [0], given: 12

Location: India
Re: Simple Algebra [#permalink]

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New post 01 Feb 2009, 01:08
Thanks a lot. I got it. This method slipped my mind. Cheers :)

Kudos [?]: 87 [0], given: 12

Re: Simple Algebra   [#permalink] 01 Feb 2009, 01:08
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