Question

If the equation $6x^2-\alpha xy-3y^2-24x+3y+\beta =0$ represents a pair of straight lines that intersect on the $x-$axis, then the value of $20\alpha -\beta$ is

Answer 1

As the lines intersect at the $x-$axis, let point of intersection of lines be $P(k,0)$
Therefore point P must satisfy the given equation
$\Rightarrow 6k^2-24k+\beta =0$

There is only one such $k$ possible for the intersection of the two lines and so the above equation must have repreated roots.
$\Rightarrow b^2-4ac=0$
$\Rightarrow \beta =24$ and $k=2$

Substitute $(2,y)$ back into the equation
$6(2{)}^2-\alpha (2)y-3y^2-24(2)+3y+24=0\Rightarrow y^2+(3-2\alpha )y=0$
But we know that $y=0$ is the only solution to the above equation, so even this equation has repeated roots.
$\Rightarrow b^2-4ac=0\Rightarrow \alpha =\frac{3}{2}$

Hence $20\alpha -\beta =20\times \frac{3}{2}-24=6$