1. ## 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.

2. ## Re: problem of quadratic equation with two variables

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.

3. ## Re: problem of quadratic equation with two variables

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

4. ## Re: problem of quadratic equation with two variables

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 ?

5. ## 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????

6. ## 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 $x^2+4ax+2a^2+6=0$ .

7. ## Re: problem of quadratic equation with two variables

thank you very much i got it