Question

If the equation $\frac{a{x}^2-24x+b}{x^2-1}=x$ has exactly two real solutions and their sum is 12 then find the value of $(a-b)$.

Answer 1

Answer: 5854

$2\alpha +\beta =a$ .....(1)
${\alpha }^2+2\alpha \beta =23$ .....(2)
$\text{ and }{\alpha }^2\beta =\beta$ .....(3)
$\text{ Also given }\alpha +\beta =12$ .....(4)
$\text{ from (2) and (4) }$
${\alpha }^2+2\alpha (12-\alpha )=23$
${\alpha }^2+24\alpha -2{\alpha }^2=23$
${\alpha }^2-24\alpha +23=0$
$\alpha =1\text{ (rejected) }$
$\text{ since }x\ne \pm 1$
$\therefore \alpha =23;$
$\therefore \beta =-11$
$\therefore a=35\text{ from }(4)$
$\text{ and }b={\alpha }^2\beta =529\times -11$
$\Rightarrow b=-5819$
$\Rightarrow a-b=35-(-5819)=5854$