Problem of quadratic equation with two variables

  • Thread starter Sumedh
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Homework Statement



If 3x2+2αxy+2y2+2ax-4y+1 can be resolved into two linear factors, prove that α is the root of the equation x2+ 4ax+2a2+6=0.

please don't solve the problem
just hint is expected.
 

Answers and Replies

  • #2
NascentOxygen
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please don't solve the problem
just hint is expected.

Try re-arranging as (......)y2 + (.....)y + C
and proceed to solve that binomial
 
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  • #3
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then what about 3x2 and 2ax
 
  • #4
NascentOxygen
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then what about 3x2 and 2ax

They will be within the term C

You re-arrange it as:

(......)y2 + (.....)y + (.....)
 
  • #5
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after arranging the original eq. with respect to x
i got this (...)x2 +(...)x+c
by finding its discriminant and equating it to zero
on solving i got
α^2 y^2+a^2+2aαy =-6y^2-3+12y [α means alpha]
α^2 y^2+a^2+2aαy -9 (y-3)^2
after equating what should i do???

could you please tell me the concept for solving this????
 
  • #6
NascentOxygen
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after arranging the original eq. with respect to x
i got this (...)x2 +(...)x+c
by finding its discriminant and equating it to zero
on solving i got

....etc....

after equating what should i do???

You are headed in the right direction. :redface: So, for the original equation to have two real roots, the square root of the discriminant must exist. That means, the discriminant is factorable as the product of two identical terms. To find what these terms are, you (as always) need to solve the binomial of that discriminant. And continue ....

Now, I recommended that you re-arrange the expression as(......)y2 + (.....)y + (.....)

But you instead re-arranged it as (......)x2 + (.....)x + (.....)
which gives the discriminant as a function of y

If you do it your way, you may not end up with the function of x that the question asks you to show. You will end up with a function of y. Equally valid, but not what the question is seeking.
 
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  • #7
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thank you very much i got it:smile:
 

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