Why does equating quadratic coefficients not work here? : GMAT Quantitative Section
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# Why does equating quadratic coefficients not work here?

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Joined: 03 Nov 2012
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03 Nov 2012, 06:15
Problem: Given z and 1/z are roots of the quadratic equation 3kx^2 -18tx + (2k+3) = 0 where t and k are constants. Find the value of k.

I attempted this by using the method of equating coefficients:

If z and 1/z are roots then it follows that (x-z) and (x-1/z) must be roots.
Multiplying these roots out gives:
x^2 - (z+1/z)x + 1 = 0

If I equate the coefficients of the two quadratic equations, I get:
3k = 1, (z+1/z) = 18, and 2k+3 = 1

Problem is this gives me two different values for k:
if 3k=1, then k = 1/3 .... but if 2k+3=1, then k= -1

However, if I use the following property of quadratic roots I get yet a different answer (the correct answer).
Property: product of the roots = c/a (for an equation of the form ax^2 + bx +c = 0)
This tells me that z * (1/z) = (2k+3)/3k
=> z/z = 1 = (2k+3)/3k
=> 2k+3 = 3k
=> k = 3

Going back to equating coefficients, I do get the answer k = 1 IF I FIRST DIVIDE BOTH SIDES OF 3kx^2 -18tx + (2k+3) = 0 by 3k to get the coefficient of the x squared term the same on both sides before I start equating coefficients.
Can anyone tell me why I have to do this first? Is there a rule that says you can only equate coefficients in quadratics if the coefficient of x squared is one or something similar to this? I've googled around a bit and I can't find any clear answer to this. Any help in understanding this would be greatly appreciated.
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03 Nov 2012, 23:46
1
KUDOS
aragusha wrote:
Any help in understanding this would be greatly appreciated.

Probably multiple kudos? :D
aragusha wrote:
Problem: Given z and 1/z are roots of the quadratic equation 3kx^2 -18tx + (2k+3) = 0 where t and k are constants. Find the value of k.
I attempted this by using the method of equating coefficients:
If z and 1/z are roots then it follows that (x-z) and (x-1/z) must be roots.
Multiplying these roots out gives:
x^2 - (z+1/z)x + 1 = 0
If I equate the coefficients of the two quadratic equations, I get:
3k = 1, (z+1/z) = 18, and 2k+3 = 1
Problem is this gives me two different values for k:
if 3k=1, then k = 1/3 .... but if 2k+3=1, then k= -1
However, if I use the following property of quadratic roots I get yet a different answer (the correct answer).
Property: product of the roots = c/a (for an equation of the form ax^2 + bx +c = 0)
This tells me that z * (1/z) = (2k+3)/3k
=> z/z = 1 = (2k+3)/3k
=> 2k+3 = 3k
=> k = 3
Going back to equating coefficients, I do get the answer k = 1 IF I FIRST DIVIDE BOTH SIDES OF 3kx^2 -18tx + (2k+3) = 0 by 3k to get the coefficient of the x squared term the same on both sides before I start equating coefficients.
Can anyone tell me why I have to do this first? Is there a rule that says you can only equate coefficients in quadratics if the coefficient of x squared is one or something similar to this? I've googled around a bit and I can't find any clear answer to this. Any help in understanding this would be greatly appreciated.

equations:
5x^2 + 10x +15 =0
and
x^2 +2x +3 =0
are same and naturally are going to have same roots. right?

But does it mean? 5=1, 10=2 and 15=3?

Reason? When we equate the coefficients we should be equating equal terms. For example, in the above case if we multiply first by 1/5 or second by 5 to make atleast one term equal, Only then we can equate others.

Coming back to your question. For given question,

Equations you tried to equate for each coefficients are:
x^2 - (z+1/z)x + 1 = 0
3kx^2 -18tx + (2k+3) = 0

you have to make sure atleast one of these coefficients is same for both equations!

So try, same approach with :
x^2 - (z+1/z)x + 1 = 0
x^2 -18tx/3k + (2k+3)/3k = 0

or
x^2 - (z+1/z)x + 1 = 0
3k/(2k+3) x^2 -18tx/(2k+3) + 1 = 0

Hope it helps!
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Re: Why does equating quadratic coefficients not work here?   [#permalink] 03 Nov 2012, 23:46
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