The question isn't clearly phrased. I think what's being asked is:
Take the two-equation system #x^2-bx+q=0# and #x^2-x+b=0#. If we know that the two equations have a common root, and that the second equation has two identical roots, then deduce #b# and #q#.
If the second equation has two equal roots, then it's of the form #(x-alpha)^2=0#, where the root is #alpha#. This multiplies out as
#x^2-2alphax+alpha^2=0#. Comparing this to the above second equation, we see that #2alpha=1# from the first coefficient (so #alpha=1/2#), and that #alpha^2=b# from the second coefficient (so #b=1/4#). Thus the second equation is #x^2-x+1/4=0#, or, factorised, #(x-1/2)^2=0#.
If the two equations have a common root, then one of the above roots must be a root of the first equation. But the two above roots are both equal, so we know the shared root is also #x=1/2#. If the other root is #beta#, then the equation has the form #(x-1/2)(x-beta)=0#, which multiplies out as #x^2-(beta+1/2)x+1/2beta=0#.
The #x# coefficient lets us deduce #beta# via #beta+1/2=b=1/4#. So #beta=-1/4#. Then we can deduce #q# from the constant coefficient: #q=1/2beta=1/2*(-1/4)=-1/8#. Thus the first equation is #x^2-1/4x-1/8=0#, or, factorised, #(x-1/2)(x+1/4)=0#.
