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Re: If b, c, and d are constants and x^2 + bx + c = (x + [#permalink]
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fortunetellerz
Galiya
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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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]
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]
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]
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]
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Sidharth003
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]
galiya
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]
dykinlee
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]
Expert Reply
mlindak
dykinlee
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]
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