A = x² + 8x + 1 

= x² + 2.x.4 + 16 - 15

= \(\left(x+4\right)^2-15\ge-15\)

=> Minimum of A = -15 at \(\left(x+4\right)^2=0\) and x = -4

B = x² - x + 1 

= \(x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=> Minimun of B = \(\dfrac{3}{4}\) at \(\left(x-\dfrac{1}{2}\right)^2=0\) and x = \(\dfrac{1}{2}\)