Question

The set of values of 'c' for which the equation $x^2-4x-c-\sqrt{8x^2-32x-8c}=0$ has exactly two distinct real solution is $(a,b)$ then find the value of $(b-a)$.

Answer 1

$x^2-4x-c-\sqrt{8}\sqrt{x^2-4x-c}=0$

$\sqrt{x^2-4x-c}(\sqrt{x^2-4x-c}-\sqrt{8})=0$

$x^2-4x-c=0$ & $x^2-4x-c=8$

$D=b^2-4acD=b^2-4ac$

$D=16+4cD=16+4c+32$

$16+4c>048+4c>0$

$c>-4c>-12$

If $c>-4$ then $4$ solution

$-12<c<-4$ then two distinct solutions

$(a,b)=(-12,-4)$

$\therefore b-a=-4+12=8$