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If x^2 + 2x – 8 = 0, then x is either –4 or

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Joined: 02 Sep 2009
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If x^2 + 2x – 8 = 0, then x is either –4 or  [#permalink]

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New post 29 Oct 2017, 00:32
1
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

98% (00:32) correct 2% (00:32) wrong based on 83 sessions

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Re: If x^2 + 2x – 8 = 0, then x is either –4 or  [#permalink]

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New post 29 Oct 2017, 07:09
1
Bunuel wrote:
If x^2 + 2x – 8 = 0, then x is either –4 or

(A) –2
(B) –1
(C) 0
(D) 2
(E) 8


\(x^2 + 2x – 8 = 0\)

Or, \(x^2 + 4x - 2x - 8 = 0\)

Or, \(x(x + 4 ) -2 (x + 4) = 0\)

Or, x = 2 and x = -4

Thus, answer will be (D) 2
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Re: If x^2 + 2x – 8 = 0, then x is either –4 or  [#permalink]

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New post 01 Nov 2017, 07:44
1
x^2+2x–8 =0
: x^2+4x−2x−8=0
: x(x+4)−2(x+4)=0
: (x-2)(x+4)=0
x=2, x= -4
ans. D.
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Re: If x^2 + 2x – 8 = 0, then x is either –4 or  [#permalink]

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New post 04 Nov 2017, 17:32
So, we're starting off with

\(x^2\)+ 2x – 8 = 0

There is no need to dive into a series of long steps. We just need to focus on the 2 numbers highlighted below:

\(x^2\) + 2x – 8 = 0

Two numbers ADDED should give you 2, and MULTIPLIED should give you -8
4 and -2 work.

To get the final solution you alway 'flip' the signs. So the two solutions are -4 and 2.

Answer is D
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Re: If x^2 + 2x – 8 = 0, then x is either –4 or &nbs [#permalink] 04 Nov 2017, 17:32
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