# problem of quadratic equation with two variables

• Aug 16th 2011, 07:13 PM
sumedh
problem of quadratic equation with two variables
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.

just hint is expected.
• Aug 16th 2011, 11:34 PM
FernandoRevilla
Re: problem of quadratic equation with two variables
Quote:

Originally Posted by sumedh
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.

Hint : The discriminant of the second degree equation on $x$ : $3x^2+(2\alpha y+2a)x+2y^2-4y+1=0$ must be a perfect square.
• Aug 17th 2011, 03:26 AM
HallsofIvy
Re: problem of quadratic equation with two variables
That is, if you meant "can be resolved into two linear factors" with rational coefficients.
• Aug 17th 2011, 04:28 AM
FernandoRevilla
Re: problem of quadratic equation with two variables
Quote:

Originally Posted by HallsofIvy
That is, if you meant "can be resolved into two linear factors" with rational coefficients.

Why?. The problem is equivalent to find where the given conic is degenerated. That is $\Delta=\begin{vmatrix}{3}&{\alpha}&{a}\\{\alpha}&{ 2}&{-2}\\{a}&{-2}&{1}\end{vmatrix}=0$ i.e. $\alpha^2+4a\alpha +2a^2+6=0$ . Why do we need rational coefficients ?
• Aug 17th 2011, 08:37 AM
sumedh
Re: problem of quadratic equation with two variables
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????
• Aug 17th 2011, 09:29 AM
FernandoRevilla
Re: problem of quadratic equation with two variables
The discriminant of the equation $p(x,y)=3x^2+(2\alpha y+2a)x+2y^2-4y+1=0$ is $\Delta=(2\alpha y+2a)^2-12(2y^2-4y+1)$ then,

$p(x,y)=3\left(x-\dfrac{-(2\alpha y+a)+\sqrt{\Delta}}{6}\right)\left(x-\dfrac{-(2\alpha y+a)-\sqrt{\Delta}}{6}\right)$

$\Delta$ is a perfect square iff the discriminant $\Delta_1$ of $\Delta=0$ is $0$ . Now, verify $\Delta_1=0\Leftrightarrow \ldots \Leftrightarrow \alpha^2+4\alpha a+2a^2+6=0$ . Equivalently, $\alpha$ is a root of $x^2+4ax+2a^2+6=0$ .
• Aug 19th 2011, 11:49 PM
sumedh
Re: problem of quadratic equation with two variables
thank you very much i got it