341If a+b+c = 0, then ax2+bx+c = 0 has real roots and hence we must have b2≥4ac.
Let f(x) = (b-x)2-4(a-x)(c-x)
Case 1: If b2=4ac, then this means both roots of ax2+bx+c = 0 are 1 and hence c=a and b = -2a.
So the second quadratic is (x+2a)2-4(x-a)2 which has roots x = 0 and x = 4a.
If a = 0, then both roots are zero. But a = 0, means b =c = 0 in this case.
Case 2: If b2>4ac, then we see that f(0) >0 but, as x→∞, f(x) → - ∞ and also as x→-∞, f(x) → - ∞.
This tells us that f(x) has two real roots one positive and the other negative and they obviously cannot be equal