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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
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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
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Yes, you are right Bunuel.

I must explicitly mention 'When we know that this equality holds generically.'
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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.
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Re: Equating coefficients in a quadratic equation [#permalink] New post 23 Jan 2015, 03:48
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Re: Equating coefficients in a quadratic equation   [#permalink] 23 Jan 2015, 03:48
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