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03 Nov 2012, 07:15
Kudos
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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.
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.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
04 Nov 2012, 00:46
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aragusha
Probably multiple kudos?
aragusha
Before I answer your question, let me ask one question:
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?
Well the answer is No.
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!
PS : Kudos <<
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
