Difference between the corresponding roots of ${\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}=0 \mathrm{and} {\mathrm{x}}^2+\mathrm{bx}+\mathrm{a}=0$ is same and $\mathrm{a}\ne \mathrm{b}$, then

Let ${\mathrm{\alpha }}_1, {\mathrm{\beta }}_1$ are the roots of the equation

 ${\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}=0\Rightarrow \mathrm{x}=\frac{-\mathrm{a}\pm \sqrt{{\mathrm{a}}^2-4\mathrm{b}}}{2}$

$\Rightarrow {\mathrm{\alpha }}_1=\frac{-\mathrm{a}+\sqrt{{\mathrm{a}}^2-4\mathrm{b}}}{2}, {\mathrm{\beta }}_1=\frac{-\mathrm{a}-\sqrt{{\mathrm{a}}^2-4\mathrm{b}}}{2}$

and ${\mathrm{\alpha }}_2, {\mathrm{\beta }}_2$ are the roots of the equation

 ${\mathrm{x}}^2+\mathrm{bx}+\mathrm{a}=0$

So, ${\mathrm{\alpha }}_2=\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^2-4\mathrm{a}}}{2}, {\mathrm{\beta }}_2=\frac{-\mathrm{b}-\sqrt{{\mathrm{b}}^2-4\mathrm{a}}}{2}$

Now ${\mathrm{\alpha }}_1-{\mathrm{\beta }}_1=\sqrt{{\mathrm{a}}^2-4\mathrm{b}}; {\mathrm{\alpha }}_2-{\mathrm{\beta }}_2=\sqrt{{\mathrm{b}}^2-4\mathrm{a}}$

Given, ${\mathrm{\alpha }}_1-{\mathrm{\beta }}_1={\mathrm{\alpha }}_2-{\mathrm{\beta }}_2\Rightarrow \sqrt{{\mathrm{a}}^2-4\mathrm{b}}=\sqrt{{\mathrm{b}}^2-4\mathrm{a}}$

$\Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2=-4(\mathrm{a}-\mathrm{b})\Rightarrow \mathrm{a}+\mathrm{b}+4=0$