F.4 Maths

Let a=3k,b=4k. Using the relationships between roots and coefficients. We have a+b=7/5,ab=c/5

So 7k=7/5=>k=1/5,ab=12k^2=12/25=>c=5ab=12/5

as a and b are the roots of the quadratic equation 5x^2 - 7x + c = 0
so 5a^2-7a+c=0----(1)
5b^2-7b+c=0
as a/b=3/4 so b=4a/3----(3)
5(4a/3)^2-7(4a/3)+c=0----(2)
solving (1) and (2)
5a^2-7a+c=5(4a/3)^2-7(4a/3)+c
5a^2-7a=5(4a/3)^2-7(4a/3)
45a^2-63a=80a^2-84a
35a^2-21a=0
5a^2-3a=0
a(5a-3)=0
a=0(rejected BECAUSEa:b=3;4) or 3/5
5(3/5)^2-7(3/5)+c=0
c=2.4

Alternative method(faster)
by using sum of roots
a+b=7/5----(4)
put (3)into (4)
a+4a/3=7/5
a=3/5
so 5(3/5)^2-7(3/5)+c=0
c=2.4