Find b and then solve the equation:(b−5)x2−(b−2)x+b=0, if one of its roots is0.5

b = ⅓

x = ½, -1/7

Step-by-step explanation:

(b−5)x² − (b−2)x + b = 0

(b - 5)(0.5)² - (b - 2)(0.5) + b = 0

0.25b - 1.25 - 0.5b + 1 + b = 0

0.75b = 0.25

b = ⅓

(⅓−5)x² − (⅓−2)x + ⅓ = 0

(-14/3)x² + (5/3)x + 1/3 = 0

14x² - 5x - 1 = 0

14x² - 7x + 2x - 1 = 0

7x(2x - 1) + (2x - 1) = 0

(7x + 1)(2x - 1) = 0

x = 0.5, -1/7

b = 1/3    x = -1/7 and -1/2

Step-by-step explanation:

(b−5)x^2−(b−2)x+b=0

Since we know one of the solutions, we can solve for b

(b−5)(.5)^2−(b−2)(.5)+b=0

(b−5)(.25)−(b−2)(.5)+b=0

Distribute

.25b -1.25 -.5b+1 +b = 0

Combine like terms

.75b -.25 =0

Add .25 to each side

.75b = .25

Divide by .75

b = .25/.75

b =1/3

Substituting this back into the equation

(1/3−5)(x)^2−(1/3−2)(x)+1/3=0

Getting a common denominator

(1/3−15/3)(x)^2−(1/3−6/3)(x)+1/3=0

-14/3 x^2  +5/3x +1/3 =0

Multiplying by 3 to get rid of the fractions

-14x^2 +5x +1 =0

using the quadratic formula

-b ±sqrt( b^2-4ac)

  2a

-5 ±sqrt( 5^2-4(-14)1)

  2(-14)

-5 ±sqrt( 81)

  2(-14)

-5 ±9

  -28

-5+9                  -5-9

  and  

-28                   -28

-4/28     and  14/28

-1/7           and  1/2

many reason,

step-by-step explanation:

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