Question

If the difference between the corresponding roots of the equations $x^2+ax+b=0$ and $x^2+bx+b=0(a\ne b)$ is the same; then -$(a+b)=$.............

Answer 1

$\alpha -\beta =\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$
For equation $x^2+ax+b=0$-------(1)
$Sum=-a,Product=b$
For equation $x^2+bx+b=0$---------(2)
$Sum=-b,Product=b$
For equation (1) ------$Difference=\sqrt{a^2-4b}$
For equation (2) ------$Difference=\sqrt{b^2-4b}$
Now $a^2-4b=b^2-4b$
$a^2-b^2=0$
$(a-b)(a+b)=0$
$a+b=0$ can be one of the solutions