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# Equating coefficients in a quadratic equation

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Senior Manager
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In a quadratic equation of the form: x^2 +bx+c = x^2 + [#permalink]

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29 Oct 2010, 06:30
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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?)
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

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29 Oct 2010, 08:07
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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).

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

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

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.

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.

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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29 Oct 2010, 09:23
Yes, you are right Bunuel.

I must explicitly mention 'When we know that this equality holds generically.'
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29 Oct 2010, 11:18
Yes - Bunuel, its the same question you posted above.
Bunuel & Karishma - thanks for the explanations.

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23 Jan 2015, 04:48
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21 Nov 2016, 07:40
Hello from the GMAT Club BumpBot!

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Re: Equating coefficients in a quadratic equation   [#permalink] 21 Nov 2016, 07:40
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