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If b, c, and d are constants and x^2 + bx + c = (x + d)^2

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If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink]

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New post 24 Sep 2008, 18:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6

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Re: GMAT SET 20 [#permalink]

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New post 24 Sep 2008, 21:09
f b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6

Am I missing a part of this question, or is my interpretation incorrect :

Substituting the values of d & b mentioned above:

X^2 + 6x+ c = (x+3)^2
x^2 + 6x + c = x^2 + 6x + 9

Therefore c = 9

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Re: GMAT SET 20 [#permalink]

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New post 24 Sep 2008, 21:46
nikhilpoddar wrote:
If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value
of c?

(1) d = 3
(2) b = 6


x^2 + bx + c = (x + d)^2
x^2 + bx + c = x^2 + 2xd + d^2

1: if d = 3, then x^2 + bx + c = x^2 + 6x + 9
bx + c = 6x + 9
b should be 6 and c = 9. suff..

2: if b = 6, then x^2 + bx + c = x^2 + 2xd + d^2
6x + c = 2xd + d^2
x, c and d can have many values.
nsf.
A.
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Re: GMAT SET 20 [#permalink]

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New post 24 Sep 2008, 21:48
if b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6

1)x^2+bx+C= x^2+2dx+d^2

c=d^2 or 9...

2) 2dx=6x then d=3...c=d^2..9,,

i get D

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Re: GMAT SET 20 [#permalink]

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New post 24 Sep 2008, 23:24
The trick is to compare LIKE terms.
X^2 + BX + C = X^2 + 2DX + D^2

Therefore comparing:
B = 2D
C = D^2

Ans is A.

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Re: GMAT SET 20 [#permalink]

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New post 25 Sep 2008, 06:48
I am also getting D.

It is sufficient with both statements.

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Re: GMAT SET 20 [#permalink]

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New post 26 Sep 2008, 21:34
GMAT TIGER wrote:
nikhilpoddar wrote:
If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?

(1) d = 3
(2) b = 6


x^2 + bx + c = (x + d)^2
x^2 + bx + c = x^2 + 2xd + d^2

1: if d = 3, then x^2 + bx + c = x^2 + 6x + 9
bx + c = 6x + 9
b should be 6 and c = 9. suff..

2: if b = 6, then x^2 + bx + c = x^2 + 2xd + d^2
6x + c = 2xd + d^2
x, c and d can have many values.
nsf.
A.


in 2, suppose if x = 2, and d = 3
6x + c = 2xd + d^2
6*2 + c = 2*2*3 + 9
c = 9

suppose if x = 1, and d = 10
6x + c = 2xd + d^2
6 + c = 2d + d^2
6 + c = 20+100
c = 114

again, suppose if x = 5, and d = 10
6x + c = 2xd + d^2
6*5 + c = 10*10 + 100
c = 170

so C varies with x and D. how is st. (2) suff?

I guess I am missing something. anybody?
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Re: GMAT SET 20 [#permalink]

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New post 26 Sep 2008, 21:38
nikhilpoddar wrote:
If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6


from 1

x^2 + bx + c = = x^2 + 3x+9

thus c = 9

from 2

x^2 +6x + c = x^2 + dx+d^2

c = d^2 = 9.........suff

my answer is d

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Location: New York
Re: GMAT SET 20 [#permalink]

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New post 26 Sep 2008, 21:46
nikhilpoddar wrote:
If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6


x^2 + bx + c = (x + d)^2
= x^2+2dx+d^2

b=2d
c=d^2

1) d = 3 -->c=d^2=9

suffcient
2) b=6 --> d=b/2=3 --> c=9
suffcieint

Ans D
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Re: GMAT SET 20 [#permalink]

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New post 26 Sep 2008, 21:57
I think everyone agrees that ST1 is sufficient.

Using ST2:
When b=6,
=> x^2+6x+c=x^2+2dx+d^2
=> c=x(2d-6) + d^2.

Since the stem says that c is a constant, it is possible when the constant c=d^2, i.e when 2d=6 => d=3
Since, c=d^2 => c=9

Hence (D)
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Re: GMAT SET 20   [#permalink] 26 Sep 2008, 21:57
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If b, c, and d are constants and x^2 + bx + c = (x + d)^2

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