problem of quadratic equation with two variables

**sumedh** · August 16th 2011, 07:13 PM

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.

**FernandoRevilla** · August 16th 2011, 11:34 PM

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.

**HallsofIvy** · August 17th 2011, 03:26 AM

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

**FernandoRevilla** · August 17th 2011, 04:28 AM

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 ?

**sumedh** · August 17th 2011, 08:37 AM

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????

**FernandoRevilla** · August 17th 2011, 09:29 AM

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

**sumedh** · August 19th 2011, 11:49 PM

thank you very much i got it

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