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Equating coefficients in a quadratic equation

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In a quadratic equation of the form: x^2 +bx+c = x^2 + [#permalink] New post 29 Oct 2010, 05:30
In a quadratic equation of the form:
x^2 +bx+c = x^2 + d^2+2dx
can we directly equate coeff's on both sides and say
b=2d and c=d^2 (I think we can - any views?)
Why ask this question?
I came across a d.s problem where I had to find c and the reduced equation was of the form:
bx + c= 2dx +d^2
1) d=3
2) b=6
now the conventional way of solving for c would be to re arrange:
c = (2d-b)x + d^2
For each value of x (b,d given) we will have a unique value of c. Hence Answer (E)
On the flip side , if we equate coefficients ( b=2d and c=d^2), we get:
1: d=3, c=d^2=9
2: b=6, b=2d => d=3, c=d^2 = 9
Answer (D)
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Re: Equating coefficients in a quadratic equation [#permalink] New post 29 Oct 2010, 07:07
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VeritasPrepKarishma wrote:
vicksikand wrote:
In a quadratic equation of the form:
x^2 +bx+c = x^2 + d^2+2dx
can we directly equate coeff's on both sides and say
b=2d and c=d^2 (I think we can - any views?)

Yes, you can if you know that b, c and d are constants. Usually, an accepted assumption (but there may be exceptions).

Why ask this question?
I came across a d.s problem where I had to find c and the reduced equation was of the form:
bx + c= 2dx +d^2
1) d=3
2) b=6
now the conventional way of solving for c would be to re arrange:
c = (2d-b)x + d^2
For each value of x (b,d given) we will have a unique value of c. Hence Answer (E)

If you know that c is not an algebraic expression but a constant, the answer will not be (E), it will be (D).
The x term has to be 0, since c is a constant. So 2d=b. If we know d = 3, c =9.
Again, for statement (2), if b = 6, it will mean d = 3 and hence c = 9.
Point is, if b, c and d are constants, then b has to be equal to 2d and c has to be equal to d^2.


On the flip side , if we equate coefficients ( b=2d and c=d^2), we get:
1: d=3, c=d^2=9
2: b=6, b=2d => d=3, c=d^2 = 9
Answer (D)


I don't think the above is correct.

Question 1:
vicksikand wrote:
In a quadratic equation of the form:
x^2 +bx+c = x^2 + d^2+2dx
can we directly equate coeff's on both sides and say
b=2d and c=d^2 (I think we can - any views?)


First of all x^2 will cancel out and we'll have: bx+c=d^2+2dx. Now if we are told that this equation is true for all x-es then yes: c=d^2 and b=2d. But if we are not told that then no, we can not say this. For example bx+c=d^2+2dx can be 4x+2=2x+1, which means that x=-\frac{1}{2}, b=4, c=2, d=1. Generally the solution for x would be x=\frac{d^2-c}{b-2d} and for different values of the constants you'll get different values of x.


Question 2:
vicksikand wrote:
I came across a d.s problem where I had to find c and the reduced equation was of the form:
bx + c= 2dx +d^2
1) d=3
2) b=6
now the conventional way of solving for c would be to re arrange:
c = (2d-b)x + d^2
For each value of x (b,d given) we will have a unique value of c. Hence Answer (E)
On the flip side , if we equate coefficients ( b=2d and c=d^2), we get:
1: d=3, c=d^2=9
2: b=6, b=2d => d=3, c=d^2 = 9
Answer (D)


Are you talking about this problem:
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

Note that we are told that x^2+bx+c=(x +d)^2 for all values of x.

So, as x^2+bx+c=(x+d)^2 is true "for ALL values of x" then it must hold true for x=0 too --> 0^2+b*0+c=(0+d)^2 --> c=d^2.

Next, substitute c=d^2 --> x^2+bx+d^2=x^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.

BUT if we were not told that x^2+bx+c=(x+d)^2 is true "for ALL values of x" then the answer would be C, not D or E.

First of all x^2+bx+c=(x+d)^2=x^2+2dx+d^2 --> bx+c=2dx+d^2.

(1) d=3 --> bx+c=2dx+d^2=6x+9 --> c=x(6-b)+9. Not sufficient as if x=0 than c=9 but if x=1 and b=1 then c=14.

(2) b=6 --> bx+c=6x+c=2dx+d^2. The same here. Not sufficient.

For (1)+(2) we would have 6x+c=6x+9 --> c=9. Sufficient.

Answer: C.

So generally if b, c, d,and e are constants and we have bx+c=dx+e we can not say that b=d and c=e.

Hope it helps.
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Re: Equating coefficients in a quadratic equation [#permalink] New post 29 Oct 2010, 08:23
Expert's post
Yes, you are right Bunuel.

I must explicitly mention 'When we know that this equality holds generically.'
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Senior Manager
Senior Manager
User avatar
Joined: 29 Sep 2009
Posts: 396
Location: Ann Arbor, MI
Schools: ROSS PT 2012
WE 1: Mech Engineer - General Electric - 2yrs
WE 2: Lead Mech Engineer - Ingersoll Rand - 4 yrs
Followers: 2

Kudos [?]: 25 [0], given: 5

GMAT Tests User
Re: Equating coefficients in a quadratic equation [#permalink] New post 29 Oct 2010, 10:18
Yes - Bunuel, its the same question you posted above.
Bunuel & Karishma - thanks for the explanations.
Re: Equating coefficients in a quadratic equation   [#permalink] 29 Oct 2010, 10:18
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