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.
Why?. The problem is equivalent to find where the given conic is degenerated. That is $\displaystyle \Delta=\begin{vmatrix}{3}&{\alpha}&{a}\\{\alpha}&{ 2}&{-2}\\{a}&{-2}&{1}\end{vmatrix}=0$ i.e. $\displaystyle \alpha^2+4a\alpha +2a^2+6=0$ . Why do we need rational coefficients ?
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????
The discriminant of the equation $\displaystyle p(x,y)=3x^2+(2\alpha y+2a)x+2y^2-4y+1=0$ is $\displaystyle \Delta=(2\alpha y+2a)^2-12(2y^2-4y+1)$ then,
$\displaystyle p(x,y)=3\left(x-\dfrac{-(2\alpha y+a)+\sqrt{\Delta}}{6}\right)\left(x-\dfrac{-(2\alpha y+a)-\sqrt{\Delta}}{6}\right)$
$\displaystyle \Delta$ is a perfect square iff the discriminant $\displaystyle \Delta_1$ of $\displaystyle \Delta=0$ is $\displaystyle 0$ . Now, verify $\displaystyle \Delta_1=0\Leftrightarrow \ldots \Leftrightarrow \alpha^2+4\alpha a+2a^2+6=0$ . Equivalently, $\displaystyle \alpha$ is a root of $\displaystyle x^2+4ax+2a^2+6=0$ .