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1. If the equations 
and 
have a common root and a,b,c belong to N, find the minimum value of a+b+c.
2. If the equation 
does not have two distinct real roots and a + c>b, the prove that f(x) >= 0, for all x belonging to R.
Comments (5)
x^2 + 3x+ 5 =0
clearly the roots are complex conjugates of each other.
so if ax^2 + bx + c = 0 and x^2 + 3x + 5 have a common root then their other roots are also equal.
so, a/1 = b/3 = c/5 = (a+b+c)/(1+3+5) = (a+b+c)/9.
a,b,c are natual numbers.
the minimum value is 5+3+1 = 9.
for the second question,
a + c - b > 0.
or, f(-1) > 0.
this might help.
i wanted to ask if a,b,c are natural numers here like the first question?
sorry, but i am too dizzy to think anything...
2. Since the equation doesn't have two distinct real roots : discriminant , D<=0
So the graph will either be above x-axis or touch it or it will be below x-axis or touch it
So, f(x) is either >=0 or <=0 for all x
f(-1) = a-b+c > 0, since for x=-1 it is positive therefore f(x)>=0
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1. equations are
2. x2 + 3x +5=0 and ax2 +bx +c=0