#### Answer

(a) yes
(b) no

#### Work Step by Step

(a) Using $x=\displaystyle \frac{b}{2}$, we get:
Left side:
$(\displaystyle \frac{b}{2})^{2}-b(\frac{b}{2})+\frac{1}{4}b^{2}=\frac{b^{2}}{4}-\frac{b^{2}}{2}+\frac{b^{2}}{4}=0$
Right side:
$0$
The sides match, so $x=b/2$ is a solution.
(b) Using $x=\displaystyle \frac{1}{b}$, we get:
Left side:
$(\displaystyle \frac{1}{b})^{2}-b(\frac{1}{b})+\frac{1}{4}b^{2}=\frac{1}{b^{2}}-1+\frac{b^{2}}{4}$
Right side:
$0$
The sides do not match, so $x=\displaystyle \frac{1}{b}$ is not a solution.