If the pair of lines $6x^2-\alpha xy-3y^2-24 x+3y+\beta=0$ intersect on $x-$axis,then find the value of $20\alpha-\beta$.
I let the pair of lines cut at $(\lambda,0)$,putting in the equation gives.
$6\lambda^2-24\lambda+\beta=0$,but there seems no further way to find $\alpha$ and $\beta.$Please help me.
In general, substituting $y=0$ in the equation of a pair of straight lines
should give the points of intersection of the two individual straight lines
with the $\bf{x}$ axis. But in this case, these points of intersection are
one and the same, i.e. $(λ,0),$
So the quadratic you obtained must have repeated roots, i.e. it's discriminant should be zero.
So, you get β from that. Putting the value of β in the quadratic of λ,
solve for λ, and then substitute (λ,0) and the value of β
in the equation of the pair of straight lines, to get α, and hence 20α−β.
Hope that helps.
