ANSWER
[tex]b = \frac{1}{3} [/tex]
[tex]x = \frac{1}{2} \: or \: x = - \frac{1}{7} [/tex]
EXPLANATION
The given expression is
[tex](b - 5) {x}^{2} - (b - 2)x + b = 0[/tex]
If
[tex]x = \frac{1}{2} [/tex]
is a root, then it must satisfy the given equation.
[tex](b - 5) {( \frac{1}{2} )}^{2} - (b - 2)( \frac{1}{2} )+ b = 0[/tex]
[tex](b - 5) {( \frac{1}{4} )} - (b - 2)( \frac{1}{2} )+ b = 0[/tex]
Multiply through by 4,
[tex](b - 5)- 2(b - 2)+4 b = 0[/tex]
Expand:
[tex]b - 5- 2b + 4+4 b = 0[/tex]
Group similar terms;
[tex]b - 2b + 4b = 5 - 4[/tex]
[tex]3b = 1[/tex]
[tex]b = \frac{1}{3} [/tex]
Our equation then becomes:
[tex]( \frac{1}{3} - 5) {x}^{2} - ( \frac{1}{3} - 2)x + \frac{1}{3} = 0[/tex]
[tex]( - \frac{14}{3} ) {x}^{2} - ( - \frac{5}{3} )x + \frac{1}{3} = 0[/tex]
[tex] - 14{x}^{2} + 5x + 1= 0[/tex]
Factor:
[tex](2x - 1)(7x + 1) = 0[/tex]
[tex]x = \frac{1}{2} \: or \: x = - \frac{1}{7} [/tex]
