Can someone help me to do this mathematic homework

Given that α and β are the roots of (2m-1)x^2 + (1+m)x+1=0, find 'm' if:
a. α = β
b.α= 1/β
c. α = 2
d. α + β = 2αβ

x = [-m - 1 ± √((m + 1)² - 4(2m - 1))] / [2(2m - 1)]
x = [-m - 1 ± √(m² + 2m + 1 - 8m + 4)] / (4m - 2)
x = [-m - 1 ± √(m² - 6m + 5)] / (4m - 2)
x = [-m - 1 ± √((m - 1)(m - 5))] / (4m - 2)

a. If α = β, (m - 1)(m - 5) = 0; therefore, m = 1 or m = 5.

d. -(m + 1)/(2m - 1) = ((m + 1)² - (m - 1)(m - 5))/(2(2m - 1)²)
-2(m + 1)(2m - 1) = (m² + 2m + 1 - m² + 6m - 5)
-2(2m² + m - 1) = (8m - 4)
-2m² - m + 1 = 4m - 2
0 = 2m² + 5m - 3
0 = (2m - 1)(m + 3)
m = 1/2, -3

b. (2m - 1)/(2m - 1)² = 1
1/(2m - 1) = 1
2m - 1 = 1
2m = 2
m = 1

c. [-m - 1 ± √((m - 1)(m - 5))] / (4m - 2) = 2
-m - 1 ± √((m - 1)(m - 5)) = 8m - 4
±√((m - 1)(m - 5)) = 9m - 3
m² - 6m + 5 = 3(9m² - 6m + 1)
m² - 6m + 5 = 27m² - 18m + 3
0 = 26m² - 12m - 2
0 = 13m² - 6m - 1
m = [6 ± √(36 + 52)] / 26
m = [3 ± √22] / 13

Given that α and ß are roots ot ax²+bx+c=0, (x-α)(x-ß)=0
α+ß=-b/a and αß=c/a
from
▲= b ² -4ac x=(-b ± √ ▲)/2a
▲= (1+m)²-4(2m-1)(1)=
▲=(1+m)²-8m+4
if α=ß, the equation has one solution, that is ▲=0
▲= (1+m)²-4(2m-1)(1)=0
▲=m²-6m+5=0
solve for m
(m-5)(m-1)=0
m=5 or m=1

b)plug α=1/ß in
αß=c/a=1
or c=a
or 2m-1=1
m=0
can you solve the rest? using similar method.