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# What is the value of p of x^2+6x+p? 1) x^2+6x+p= -3 2)

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Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea
What is the value of p of x^2+6x+p? 1) x^2+6x+p= -3 2) [#permalink]

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11 Dec 2005, 06:38
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What is the value of p of x^2+6x+p?

1) x^2+6x+p=[(x+r)^2]-3
2) x^2+6x+p=(x+m)(x+n)
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Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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11 Dec 2005, 08:16
p,m,n,r are all different.
Seems like E)
Intern
Joined: 11 Sep 2005
Posts: 19
Location: Framingham, MA, USA

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11 Dec 2005, 13:15
Statement 1: Since 3 variables (x, p, r) are present, we need 3 different equations, but we only have 2 equations.
Statement 2: Since 4 variables (x, p, m, n) are present, we need 4 different equations.
Combined: Since 5 different (x, p, r, m, n) variables are present, we need to have atleast 5 different equations.

Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea

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12 Dec 2005, 10:29
Please give it one more thought.
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VP
Joined: 22 Aug 2005
Posts: 1112
Location: CA

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12 Dec 2005, 11:17
is it A?

S1:
x^2+6x+p=[(x+r)^2]-3 or

x^2+6x+ (p + 3) = [(x+r)^2]

to be perfect square (x+3)^2,
p+3 should be 9

or p = 6

S2:

x^2+6x+p=(x+m)(x+n) or
x^2+6x+p= x^2 + (m+n)x + mn

or:
m+n = 6
mn = p

as we do not know value of m,n,p
insufficient
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SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka

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12 Dec 2005, 11:34
agree with duttsit. A is the answer.
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hey ya......

Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea

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12 Dec 2005, 19:27
duttsit wrote:
is it A?

S1:
x^2+6x+p=[(x+r)^2]-3 or

x^2+6x+ (p + 3) = [(x+r)^2]

to be perfect square (x+3)^2,
p+3 should be 9

or p = 6

S2:

x^2+6x+p=(x+m)(x+n) or
x^2+6x+p= x^2 + (m+n)x + mn

or:
m+n = 6
mn = p

as we do not know value of m,n,p
insufficient

Good job, duttist.

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Auge um Auge, Zahn um Zahn !

12 Dec 2005, 19:27
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