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answers:
ahmedmohamed say: ####
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Ian H say: Substituting x = 1/2 we get
(b - 5)/4 - (b - 2)/2 + b = 0
3b/4 = 1/4, b = 1/3
sum of roots = (b – 2)/(b – 5) = 5/14 = 1/2 + t
The other root is (5 – 7)/14 = -1/7
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Annie say: Unfortunately, there has never been a time in my life where I could have figured this out.
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Carl say: x= 5/3, b=5
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sepia say: (b − 5)x^2 − (b − 2)x + b = 0
b x^2 - b x + b - 5 x^2 + 2 x = 0
b (x^2 - x + 1) = 5 x^2 - 2 x
b = x(5x - 2)/(x^2 - x + 1)
one of its roots is 1/2
b = 1/2(5/2 - 2)/(1/4 - 1/2 + 1) = 1/3
(b − 5)x^2 − (b − 2)x + b = 0
(1/3 − 5)x^2 − (1/3 − 2)x + 1/3 = 0
-1/3 (2 x - 1) (7 x + 1) = 0
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llaffer say: If we have:

(b - 5)x² - (b - 2)x + b = 0

And we know that x = 1/2 is one root, then we can substitute (1/2) in for x and solve for b:

(b - 5)(1/2)² - (b - 2)(1/2) + b = 0
(b - 5)(1/4) - (b - 2)(1/2) + b = 0

Let's multiply both sides by 4 to get rid of the fractions:

b - 5 - 2(b - 2) + 4b = 0

Continue to simplify:

b - 5 - 2b + 4 + 4b = 0
3b - 1 = 0
3b = 1
b = 1/3
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Mike G say: Let the roots be 1/2 and p
(1/2+p) = (b-2)/(b-5)
p/2 = b/(b-5)
p = 2b/(b-5)
1/2+2b/(b-5) = (b-2)/(b-5)
(b-5)/2+2b = b-2
b-5+4b = 2b-4
3b = 1
b = 1/3
(-14/3)x^2-(-5/3)x+1/3 = 0
-14x^2+5x+1 = 0
14x^2-5x-1) = 0
(7x+1)(2x-1) = 0
x = -1/7 or 1/2
See Graph
https://www.desmos.com/calculator/y0etas...
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ted s say: thus : b ( 1/4 - 1 + 1/2) = 5 / 4 - 2 OR b( - 1 / 4) = - 3 / 4 ===> b = 3......- 2 x²- x + 3 = 0 = ( - 2 x + 1)(x + 3)
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az_lender say: (b - 5)*(1/4) - (b - 2)(1/2) + b = 0 =>
b(1/4 - 1/2 + 1) - 5/4 + 1 = 0 =>
(3/4) b = 1/4 => b = 1/3.
So the original equation becomes
-(14/3)x^2 + (5/3)x + (1/3) = 0 =>
14x^2 - 5x - 1 = 0 =>
(2x - 1)(7x + 1) = 0 =>
x = 1/2 or -1/7.
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