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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, 08:51
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75% (hard)

Question Stats:

46% (01:02) correct 54% (01:55) wrong based on 320 sessions

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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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Joined: 02 Sep 2009
Posts: 49320

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23 Aug 2010, 09:14
7
7
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, 12:59
7
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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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!
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Re: If b, c, and d are constants  [#permalink]

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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

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

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Joined: 25 Jul 2009
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DS: GMAT paper test: If b, c, and d are constants a  [#permalink]

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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.
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Re: DS: GMAT paper test: If b, c, and d are constants a  [#permalink]

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25 Nov 2010, 15:52
1
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.
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If b, c, and d are constants and x^2 + bx + c = (x + d)^2  [#permalink]

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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)

Is this approach correct?
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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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05 Jul 2016, 21:46
1
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.
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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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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^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.

Experts pls comment.
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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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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 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.
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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, 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 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 &nbs [#permalink] 26 Sep 2017, 08:00
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# If b, c, and d are constants and x^2 + bx + c = (x + d)^2

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