Find b and then solve the equation: d (b−5)x2−(b−2)x+b=0, if one of its roots is 1/2

ANSWER[tex]b = \frac{1}{3} [/tex][tex]x = \frac{1}{2} \: or \: x = - \frac{1}{7} [/tex]EXPLANATIONThe 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]