
answers:
ahmedmohamed say: ####

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

Annie say: Unfortunately, there has never been a time in my life where I could have figured this out.

Carl say: x= 5/3, b=5

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

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

Mike G say: Let the roots be 1/2 and p
(1/2+p) = (b2)/(b5)
p/2 = b/(b5)
p = 2b/(b5)
1/2+2b/(b5) = (b2)/(b5)
(b5)/2+2b = b2
b5+4b = 2b4
3b = 1
b = 1/3
(14/3)x^2(5/3)x+1/3 = 0
14x^2+5x+1 = 0
14x^25x1) = 0
(7x+1)(2x1) = 0
x = 1/7 or 1/2
See Graph
https://www.desmos.com/calculator/y0etas...

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)

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.
