If $2{\sin }^2\frac{\mathrm{\pi }}{8}$ is a root of the equation x² + ax +b= 0, where a and b are rational numbers, then a - b is equal to

$\because 2{\sin }^2\frac{\mathrm{\pi }}{8}=1-\cos \frac{\mathrm{\pi }}{4}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}} (\mathrm{irrationalroot})$
So, other root is $\frac{\sqrt{2}+1}{\sqrt{2}}$
sum of roots $=-\mathrm{a}=1-\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}=2\Rightarrow \mathrm{a}=-2$
Productof roots  $=1-\frac{1}{2}=\frac{1}{2}=\mathrm{b}$ 
$\mathrm{a}-\mathrm{b}=-2-\frac{1}{2}=-\frac{5}{2}$