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If x2 + 3x + c = (x + a)(x + b) for all x, what is the value

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If x2 + 3x + c = (x + a)(x + b) for all x, what is the value [#permalink] New post 08 Jun 2007, 19:45
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Q10:
If x2 + 3x + c = (x + a)(x + b) for all x, what is the value of c ?
(1) a = 1
(2) b = 2
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 [#permalink] New post 08 Jun 2007, 20:04
C?

I solved it to c=3. Is this correct?
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 [#permalink] New post 09 Jun 2007, 00:14
(D) for me :)

We have:
x^2 + 3x + c
= (x + a)(x + b)
= x^2 + (a+b)*x + a*b

So,
o c = a*b
o a+b = 3

Thus, if we know one of a,b, we can conclude on both values of a & b and then on the value of c.
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 [#permalink] New post 09 Jun 2007, 00:54
Fig wrote:
(D) for me :)

We have:
x^2 + 3x + c
= (x + a)(x + b)
= x^2 + (a+b)*x + a*b

So,
o c = a*b
o a+b = 3

Thus, if we know one of a,b, we can conclude on both values of a & b and then on the value of c.


Very clever Fig - but what about a=0 and b=3 ?

x^2+3x+0 = (x+0)(x+3)

My answer is (C)

:?
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 [#permalink] New post 09 Jun 2007, 03:23
KillerSquirrel wrote:
Fig wrote:
(D) for me :)

We have:
x^2 + 3x + c
= (x + a)(x + b)
= x^2 + (a+b)*x + a*b

So,
o c = a*b
o a+b = 3

Thus, if we know one of a,b, we can conclude on both values of a & b and then on the value of c.


Very clever Fig - but what about a=0 and b=3 ?

x^2+3x+0 = (x+0)(x+3)

My answer is (C)

:?


I dont understand your question :)... The case analysed by the 2 given statments brings a = 2, b=1 & c = 2.

Even if we want to consider a=0 & b=3.... we have the value of c = 0. :)

Would u perhaps reformulate ? :)
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 [#permalink] New post 09 Jun 2007, 03:39
Fig wrote:
KillerSquirrel wrote:
Fig wrote:
(D) for me :)

We have:
x^2 + 3x + c
= (x + a)(x + b)
= x^2 + (a+b)*x + a*b

So,
o c = a*b
o a+b = 3

Thus, if we know one of a,b, we can conclude on both values of a & b and then on the value of c.


Very clever Fig - but what about a=0 and b=3 ?

x^2+3x+0 = (x+0)(x+3)

My answer is (C)

:?


I dont understand your question :)... The case analysed by the 2 given statments brings a = 2, b=1 & c = 2.

Even if we want to consider a=0 & b=3.... we have the value of c = 0. :)

Would u perhaps reformulate ? :)


Sorry for that - I now understand. Thanks Fig well done !

:-D
SVP
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Posts: 1816
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 [#permalink] New post 09 Jun 2007, 04:09
KillerSquirrel wrote:
Fig wrote:
KillerSquirrel wrote:
Fig wrote:
(D) for me :)

We have:
x^2 + 3x + c
= (x + a)(x + b)
= x^2 + (a+b)*x + a*b

So,
o c = a*b
o a+b = 3

Thus, if we know one of a,b, we can conclude on both values of a & b and then on the value of c.


Very clever Fig - but what about a=0 and b=3 ?

x^2+3x+0 = (x+0)(x+3)

My answer is (C)

:?


I dont understand your question :)... The case analysed by the 2 given statments brings a = 2, b=1 & c = 2.

Even if we want to consider a=0 & b=3.... we have the value of c = 0. :)

Would u perhaps reformulate ? :)


Sorry for that - I now understand. Thanks Fig well done !

:-D


ok ;)
  [#permalink] 09 Jun 2007, 04:09
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