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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
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
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