if c is positive and 2ax2 +3bx + 5c = 0, then prove that 2a -3b + 5c>0

REMARK: PLEASE TYPE THE FULL QUESTION(THE ROOTS ARE UNREAL AND c>0)

f(x)=2ax²+3bx+5c=0

SINCE, ROOTS ARE UNREAL.

CASE 1:a>0;D<0 =>9b²-40ac<0  =>9b²<40ac (i)

clearly the rhs of inequality is always positive and so is the lhs

CASE2:a<0;9b²<40ac  (ii)

lhs>0 & rhs<0 so not possible

proceeding with case1

we have

f(x),a upper parabola whose vertex is>0,which is also here the minimum value of f(x)

so parabola is always positive

therefore, f(-1)=2a-3b+5c>0

proved

Approved

 i found the correct answer buddy that is -

we have two cases f(x)>0 or f(x)<0

now, for F(0)=5c>0

therefroe for all real values of F(x)>0

for F(1)>0

2a -3b +5c >0, hence proved