Quadratic cum Logarithm

for f(x) = ax² + bx + c

Let a>0 D>0....
Clearly from graph f(k₁) > 0 and f(k₂) > 0
Then for (Alpha) , (beta) to lie b/w k₁ and k₂

a(f(k₁) >0
a(f(k₂) >0
k₁ < -(b/2a) < k₂
D > 0

or

a(f(0) >0
a(f(1) >0
0 < -(b/2a) < 1
D > 0

for f(x) = ax² - bx + c

ac >0
a(a-b+c) >0
0 < (b/2a) < 1
D > 0

or

ac >0
a(a-b+c) >0
0 < b < 2a
b² - 4ac > 0

or
ac >0
a(a-b+c) >0
0 < b < 2a
b² > 4ac

also we have

a ≥ 1
b ≥ 1
c ≥ 1.....as (a,b,c) belongs to N

after this i am stuck...how to find abc = ??

Please point out my mistakes if any....

........

2) 2^p is always a multiple of 4,if p is greater than or equal to 2

now 3^q + 5^r = (4-1)^q + (4+1)^r= 4k + (-1)^q + (1)^r

if q is odd and r is anything 3^q + 5^r = 4k, so 2^p+ 3^q + 5^r is a multiple of 4,where p≥2

if q is even and r is anything then 3^q + 5^r = 4k + 2, so 2^p+ 3^q + 5^r is a multiple of 4 if and only if p=1

now u can finish it offfff