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

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Manager
Joined: 06 Apr 2010
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If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink]

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23 Aug 2010, 07:51
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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
[Reveal] Spoiler: OA
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Joined: 02 Sep 2009
Posts: 43917

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23 Aug 2010, 08:14
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udaymathapati 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

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?

$$x^2 + bx + c = (x + d)^2$$ --> $$x^2+bx+c=x^2+2dx+d^2$$ --> $$bx+c=2dx+d^2$$.

Now, as above expression is true "for ALL values of $$x$$" then it must hold true for $$x=0$$ too --> $$c=d^2$$.

Next, substitute $$c=d^2$$ --> $$bx+d^2=2dx+d^2$$ --> $$bx=2dx$$ --> again it must be true for $$x=1$$ too --> $$b=2d$$.

So we have: $$c=d^2$$ and $$b=2d$$. Question: $$c=?$$

(1) $$d=3$$ --> as $$c=d^2$$, then $$c=9$$. Sufficient
(2) $$b=6$$ --> as $$b=2d$$ then $$d=3$$ --> as $$c=d^2$$, then $$c=9$$. Sufficient.

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18 Oct 2010, 11:59
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This question is based on special algebraic equations, particularly on (x + y)^2 = x^2 + 2xy + y^2.

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

Equating the co-efficients of x, 2d = b
and equating the constant terms, d^2 = c

If I have d = 3, I get b = 6 and c = 9. (Statement I alone is sufficient.)
If I have b = 6, I get d = 3 and then c = 9 (Statement II alone is sufficient.)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews BSchool Thread Master Joined: 19 Feb 2010 Posts: 390 Re: Equation DS [#permalink] ### Show Tags 18 Oct 2010, 20:31 Ha, I had this question yesterday in a practice set. I chose A but when reviewed all the answers, understood why it was D. I usually only remember the answer if I reviewed the question. I hope to get a similar one in the real exam now that I know how to solve it! Manager Joined: 30 Sep 2010 Posts: 56 Re: If b, c, and d are constants [#permalink] ### Show Tags 07 Nov 2010, 09:05 Hi, This question should be put under the DS section. Also the question should read as: If b, c, and d are constants and x2 + bx + c = (x + d)^2 for all values of x, what is the value of c? (1) d = 3 (2) b = 6 The answer will be D. As the equation is true for all the values of X. Hence corresponding constants attached to the powers of X should be equal x2 + bx + c = x2 + 2dx + d^2 so b=2d and c=d^2 That means if we know the value of b or d, we can find out c Hence answer should be D. Intern Joined: 25 Jul 2009 Posts: 11 DS: GMAT paper test: If b, c, and d are constants a [#permalink] ### Show Tags 25 Nov 2010, 14:36 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 since expanding (x + d)^2 = x^2 + 2xd + D^2 hence we have bx + c = 2xd + D^2 we need to know value of b and value of d to get the correct answer. Can you please explain if it is other wise. _________________ Failure it not and option -- Gene Kranz Current Student Status: Three Down. Joined: 09 Jun 2010 Posts: 1914 Concentration: General Management, Nonprofit Re: DS: GMAT paper test: If b, c, and d are constants a [#permalink] ### Show Tags 25 Nov 2010, 14:52 1 This post received KUDOS vrajesh wrote: If b, c, and d are constants and x^2 + bx + c = (x + d)^2 f[highlight]or all values of x[/highlight], what is the value of c? (1) d = 3 (2) b = 6 since expanding (x + d)^2 = x^2 + 2xd + D^2 hence we have bx + c = 2xd + D^2 we need to know value of b and value of d to get the correct answer. Can you please explain if it is other wise. Given $$x^2 + bx + c = (x+d)^2$$ Expanding on both sides, we get: $$x^2 + 2dx + d^2 = x^2 + bx + c.$$ We can cancel the x^2 on both sides and that leaves us with: $$2dx + d^2 = bx + c$$ And this is valid for all values of x, we are given. Let's just substitute two values of x: $$x = 1$$ Then $$2d + d^2 = b + c$$ $$x = 2$$ $$4d + d^2 = 2b + c$$ Now taking these two equations together: $$2d + d^2 = b + c 4d + d^2 = 2b + c$$ Multiplying the first equation by 2 and solving, we get: $$4d + 2d^2 = 2b + 2c 4d + d^2 = 2b + c$$ So we get: $$d^2 = c$$ Thus, from our two statements, we know that statement one is sufficient by itself. Now look at statement 2. And look at the first equation we had: $$2d + d^2 = b + c.$$ But then $$c = d^2$$, which means that $$2d = b$$. So, if b is given, you can find d and hence c. Thus statement 2 is also sufficient. Thus the trick here is to read the question carefully. If it's valid for all values of x, it's valid for any two values of x. You can arbitrarily pick x. In fact, picking x = 0 might be even better and help you solve the problem much faster, in hindsight. Intern Joined: 07 Jun 2014 Posts: 17 Location: India GMAT 1: 720 Q49 V38 GPA: 2.91 WE: Consulting (Energy and Utilities) If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink] ### Show Tags 14 Aug 2014, 22:32 What if I solve it the way described below? will it be wrong? Given x^2 + bx + c = (x + d)^2 When these two equations will equate to zero - From (x+d)^2=0 we can determine that x will only have one unique value i.e. x=-d - if x will have only one unique value, then from x^2 + bx + c = 0 we can say b^2 - 4ac = 0, where a=1 which means b^2 = 4c -> c= (b^2)/4 ....(a) also when b^2 - 4ac = 0 then x= -b/2a (using x = (-b ±√(b^2 - 4ac))/2a ) where x=d hence d=-b/2 b=-2d ...(b) using (b) on (a) c=d ...(c) Now, evaluating statement (1), value of d is sufficient to determine value of c using (c) evaluating statement (2), value of b is sufficient to determine value of c using (a) Hence the answer is D Is this approach correct? _________________ I would rather "crash and burn" than "sulk and cry"!! Intern Joined: 23 Jun 2013 Posts: 11 Location: India Concentration: General Management, Economics Schools: Tuck '19 GPA: 3.2 WE: Information Technology (Consulting) Re: If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink] ### Show Tags 05 Jul 2016, 20:46 The discriminant of $$x^2+bx+c$$ should be zero -> $$b^2-4c=0$$ Equating the roots of expression $$-b/2=-d$$ Hence if either b or d is known c can be calculated. Manager Joined: 27 Mar 2014 Posts: 111 Schools: ISB '19, IIMA , IIMB GMAT 1: 660 Q49 V30 Re: If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink] ### Show Tags 17 Sep 2017, 00:59 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 ...................... (1) using equation 1 , we can say : x^2 + bx + c can be converted into whole square only when b^2 - 4ac = 0 OR -b/2a = -d Statement 1 : d = 3 ; also we know a=1 -b/2a = -d -b/2(1) = -(3) b=6 b^2-4ac=0 (6)^2 - 4(1)(c) = 0 c=9 sufficient. Statement 2 b=6 b^2-4ac=0 (6)^2 - 4(1)(c) = 0 c=9 sufficient. Answer D Experts pls comment. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7960 Location: Pune, India Re: If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink] ### Show Tags 25 Sep 2017, 21:00 Responding to a pm: Quote: As we know, in the equation in the form "ax^2 + bx + c", b means summation of the roots when a = 1, and c means product of the roots when a = 1. Here in this DS, the roots are the same, so, as, d = 3, c= 9, and as, b = 6, root is 6/2 = 3, and c = 9, that's okay ... In problems, however, where roots are different, is there any elegant way to work with ....? Equate the co-efficients as shown in my post above: https://gmatclub.com/forum/if-b-c-and-d ... ml#p802314 Depending on what is asked and what is given, you can answer the question. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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26 Sep 2017, 07:00
VeritasPrepKarishma wrote:
Responding to a pm:
Quote:
As we know, in the equation in the form "ax^2 + bx + c", b means summation of the roots when a = 1, and c means product of the roots when a = 1. Here in this DS, the roots are the same, so, as, d = 3, c= 9, and as, b = 6, root is 6/2 = 3, and c = 9, that's okay ...

In problems, however, where roots are different, is there any elegant way to work with ....?

Equate the co-efficients as shown in my post above:
https://gmatclub.com/forum/if-b-c-and-d ... ml#p802314

Depending on what is asked and what is given, you can answer the question.

thanks mam
Re: If b, c, and d are constants and x^2 + bx + c = (x + d)^2   [#permalink] 26 Sep 2017, 07:00
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