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

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Manager
Joined: 11 Jan 2007
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If x2 + 3x + c = (x + a)(x + b) for all x, what is the value [#permalink]  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
_________________

cool

Intern
Joined: 03 Aug 2006
Posts: 22
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C?

I solved it to c=3. Is this correct?
SVP
Joined: 01 May 2006
Posts: 1798
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Kudos [?]: 102 [0], given: 0

(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.
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 136 [0], given: 0

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)

SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

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)

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 ?
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 136 [0], given: 0

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)

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 !

SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

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)

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 !

ok
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