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, $a,b \in Q$?
Best Answer
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 $x^2 + 4ax + 2a^2 + 6 = 0$.