If $ 3x^2 + 2\alpha xy + 2y^2 + 2ax - 4y + 1 $ can be resolved into two linear factors, then prove the following.

Question

Prove that : $ \alpha $ is a root of the equation $ x^2 + 4ax + 2a^2 + 6 = 0 $.

What does it mean by "can be resolved into two linear factors"? If it means $( ax + b ) ( cx + d )$ , is it necessary that $ a,b,c,d \in Q $ .

The solving says that: For the condition to be true , the roots of the equation must be rational.

$ \implies { \displaystyle\frac{ -2\alpha y + 2a \pm \sqrt {D} }{2} } $ must give rational roots.

For the above expression to give rational roots, $D$ must be a perfect square of a rational number.

$\implies 4 { ( \alpha^2 - 2 ) y^2 + ( 2a\alpha + 4 ) y + a^2 - 1 } $ should be a perfect square.

$\implies $ The discriminant of above expression $= 0$. By doing that, we get our required result.

**My Doubt ** :

Cant it be resolved into linear factors if $D$ is not a perfect square, i.e., $D > 0$, giving irrational roots? Is it necessary that for $ax + b$ to be called linear, $a,b \in Q$?

Answer 1

The polynomial

$$P(x,y)=ax^2+by^2+cxy+dx+ey+f=0$$

geometrically represents conic sections. When it is said that it can be resolved into linear factors, then it is meant that there is a factorization such that

$$P(x,y)=(Ax+By+C)(Ex+Fy+D)=0$$

Which geometrically represents two crossing or coinciding lines! For such a factorization to exist some relations must hold between the constants $a,b,c,d,e,f$. You can see this link for more information.

Answer 2

Let $\alpha \in \mathbb{R}$ (or $\alpha \in \mathbb{C}$?). The set of points $(x,y) \in \mathbb{R}^2$ satisfying

$$ C: 3x^2 + 2\alpha xy + 2y^2 +2ax - 4y + 1 = 0 $$

is a conic. When $C$ factors into a product of two linear factors, we say that $C$ is degenerate. Degeneracy can be tested by computing the determinant of its matrix of coefficients of its homogeneous form. So let's homogenize $C$ first:

$$ \widetilde{C} : 3X^2 + 2\alpha XY + 2Y^2 + 2a XZ - 4YZ + Z^2 = 0. $$

Just "stick a $Z$ onto the factors until all the degrees become the same". The original conic $C$ may be recovered from $\widetilde{C}$ by setting $Z=1$. The matrix of coefficients of $\widetilde{C}$ is given by

$$ M := \begin{bmatrix} 3 & \alpha & a\\ \alpha & 2 & -2 \\ a& -2 & 1 \end{bmatrix}. $$

The determinant of this matrix is

$$ \det M = \det \begin{bmatrix} 3 & \alpha & a\\ \alpha & 2 & -2 \\ a& -2 & 1 \end{bmatrix} = 6 - 2a\alpha - 2a \alpha - 2a^2 - 12 - \alpha^2 = -(\alpha^2 +4a\alpha + 2a^2+6). $$ Hence, the conic $C$ is degenerate if and only if $\det M = 0$, if and only if $\alpha^2 + 4a\alpha + 2a^2 + 6 = 0$, if and only if $\alpha$ is a root of the equation $x^2 + 4ax + 2a^2 + 6 = 0$.

Answer 3

What does it mean by "can be resolved into two linear factors"? If it means (ax+b)(cx+d) , is it necessary that a,b,c,d∈Q

Geometrically, it means that the second degree equation must represent a pair of straight line.

The general equation of a second degree is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

Given equation is: $ 3x^2 + 2\alpha xy + 2y^2 + 2ax - 4y + 1 $

On comparing you get $a=3, b=2, c=1, f=-2, g=a, h= \alpha$

For any second-degree equation to represent a straight line, determinant of:

$$\begin{pmatrix} a&h&g\\h&b&f\\g&f&c \end{pmatrix}.$$ should be zero

Thus, determinant of:

$$\begin{pmatrix} 3& \alpha &a\\\alpha&2&-2\\a&-2&1 \end{pmatrix}.$$ should be zero

Upon solving you get, $\alpha^2 + 4a\alpha + 2a^2 + 6 = 0$ Replace $\alpha$ with x and you get your solution.