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

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galiya
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If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   05 Aug 2011, 06:16
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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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D
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fluke
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   05 Aug 2011, 07:32
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Galiya 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


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

\(x^2 + bx + c = x^2 + d^2 + 2xd\) {:Note: (a+b)^2=a^2+b^2+2ab}

\(bx + c = d^2 + 2xd\) {:Note: (a+b)^2=a^2+b^2+2ab}


(1) d=3

\(bx + c = 9 + 6x\)

LHS=RHS for all x;

So, for x=0

\(b*0 + c = 9 + 6*0\)

\(c = 9\)

Sufficient.

(2) b=6

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

LHS=RHS for all x;

So, for x=0

\(c=d^2\)----------1

for x=1

\(b + c = d^2 + 2d\)---------2

Using 1 and 2:
\(b = 2d\)

\(6=2d\)

\(d=3\)

Using 1:
\(c=d^2=3^2=9\)
Sufficient.

Ans: "D"
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   05 Aug 2011, 07:21
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Galiya 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

Show SpoilerComment
I got A instead of OA D
could you explain me please why it so?


ok here is what i think is the reasoning ...

expand the eqn .. \(x^2 + bx + c = x^2 + 2dx + d^2\)

cancelling the first term and comparing like terms we get

\(bx = 2dx << i.e. >> \\
\\
b=2d\)

and \(c = d^2\)

now check the options we'd get both A and B to suffice ....
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   12 Sep 2017, 05:27
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fortunetellerz wrote:
Galiya 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

Show SpoilerComment
I got A instead of OA D
could you explain me please why it so?


ok here is what i think is the reasoning ...

expand the eqn .. \(x^2 + bx + c = x^2 + 2dx + d^2\)

cancelling the first term and comparing like terms we get

\(bx = 2dx << i.e. >> \\
\\
b=2d\)

and \(c = d^2\)

now check the options we'd get both A and B to suffice ....


Good to know - thanks. I forgot this strategy while solving the question.
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   26 Jul 2019, 07:15
why is X=0 in the solution? I cannot figure out. please help!
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   29 Jul 2019, 11:50
If the original question did not require that b, c, and d are constants, would the answer be C?

With d=3 the formula simplifies to 6x+9=bx+c. NS due to B and C being unknown.
Add that b=6 and then the equation reduces to c=9.

Does anyone know of another question that tests the difference between constants and variables within an equation?
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   30 Sep 2021, 07:41
VeritasKarishma Bunuel needed some help on this one. Statement 2 says b=6, I arrived at 6x+c=2xd+d^2 and got stuck up. Not following how to proceed and prove its sufficient. Please guide as to what am I missing. Would be a big help. Thanks.
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   30 Sep 2021, 21:28
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Sidharth003 wrote:
VeritasKarishma Bunuel needed some help on this one. Statement 2 says b=6, I arrived at 6x+c=2xd+d^2 and got stuck up. Not following how to proceed and prove its sufficient. Please guide as to what am I missing. Would be a big help. Thanks.

Sidharth003

We know that \((x + y)^2 = x^2 + 2xy + y^2\)

When y = d,

\((x + d)^2 = x^2 + 2xd + d^2 \)

Now we are given \((x+d)^2 = x^2 + bx + c\)
We know that c is a constant so it will have no x term.

So 2d = b
and d^2 = c

From stmnt 2, when b = 6, then d = 3. Then c = 9.
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   06 Sep 2022, 09:15
galiya 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


I don't think this is a 700 level problem, it is way below 500. Anyways, answering the question,

expanding the equation, x^2+bx+c = x^2+2dx+d^2
that implies, b=2d & c= d^2.

Now coming to statement 1, d=3, then c= d^2 = 9. A is sufficient.
Statement 2, b=6, then d=3, then c=d^2 = 9. B is also sufficent.

Hence D is the answer.
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   20 Jun 2023, 13:14
dykinlee wrote:
why is X=0 in the solution? I cannot figure out. please help!


Hey Dykinlee, did you figure it out? I also don't understand. feel that its a critical part of the question
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink] New post   20 Jun 2023, 23:56
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mlindak wrote:
dykinlee 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

why is X=0 in the solution? I cannot figure out. please help!


Hey Dykinlee, did you figure it out? I also don't understand. feel that its a critical part of the question


We are given that \(x^2 + bx + c = (x + d)^2\) for ALL values of x, so it must also be true for x = 0. Setting x to 0 allows us to simplify the equation and get c = d^2.
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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink]
20 Jun 2023, 23:56
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