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

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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, 01:32
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A
B
C
D
E

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  5% (low)

Question Stats:

89% (00:21) correct 11% (00:11) wrong based on 65 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, 08:09
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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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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, 18:32
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   [#permalink] 04 Nov 2017, 18:32
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