Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 134

If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink]
Show Tags
23 Aug 2010, 08:51
Question Stats:
46% (01:02) correct 54% (01:57) wrong based on 305 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 46035

Re: Equation DS [#permalink]
Show Tags
23 Aug 2010, 09:14
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. Answer: D.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8097
Location: Pune, India

Re: Equation DS [#permalink]
Show Tags
18 Oct 2010, 12:59
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 coefficients 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.)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



BSchool Thread Master
Joined: 19 Feb 2010
Posts: 367

Re: Equation DS [#permalink]
Show Tags
18 Oct 2010, 21:31
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: 55

Re: If b, c, and d are constants [#permalink]
Show Tags
07 Nov 2010, 10:05
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, 15:36
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



SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1869
Concentration: General Management, Nonprofit

Re: DS: GMAT paper test: If b, c, and d are constants a [#permalink]
Show Tags
25 Nov 2010, 15:52
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: 14
Location: India
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, 23:32
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: 24 Jun 2013
Posts: 11
Location: India
Concentration: General Management, Economics
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, 21:46
The discriminant of \(x^2+bx+c\) should be zero > \(b^24c=0\) 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: 103

Re: If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink]
Show Tags
17 Sep 2017, 01:59
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^24ac=0 (6)^2  4(1)(c) = 0 c=9
sufficient.
Statement 2
b=6
b^24ac=0 (6)^2  4(1)(c) = 0 c=9
sufficient.
Answer D
Experts pls comment.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8097
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, 22:00
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 coefficients as shown in my post above: https://gmatclub.com/forum/ifbcandd ... ml#p802314Depending 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
Veritas Prep Reviews



Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 275

Re: If b, c, and d are constants and x^2 + bx + c = (x + d)^2 [#permalink]
Show Tags
26 Sep 2017, 08:00
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 coefficients as shown in my post above: https://gmatclub.com/forum/ifbcandd ... ml#p802314Depending 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, 08:00






