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Question

# If the equation ax^2+bx+c=0 does not have 2 distinct real roots and a+c>b, then prove that f(x)>=0,for all x belongs to real number.

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Solution

## f(x) = ax² + bx + c? I'll assume so. I'll also assume a ≠ 0 Since f(x) has 0 or 1 distinct real roots, then f(x) ≥ 0 for all x iff a > 0 Since f(x) doesn't have 2 distinct real root, discriminant must be ≤ 0. b² − 4ac ≤ 0 0 ≤ b² ≤ 4ac If c = 0, then b = 0, and since a+c > b ---> a+0 > 0 ----> a > 0 So f(x) ≥ 0 for all real x If c ≠ 0, since 4ac ≥ 0, then either a and c are both positive, or a and c are both negative. If a and c are both positive, then f(x) ≥ 0 for all real x So the only case that remains is if a and c are both negative. If a = c, then 0 ≤ b² ≤ 4ac ---> b² ≤ 4a² b < a+c < 0 ---> b < 2a < 0 ---> b² > 4a² But b² cannot both be ≤ 4a² and > 4a² If a ≠ c, then a−c ≠ 0 (a−c)² > 0 a² − 2ac + c² > 0 a² − 2ac + c² + 4ac > 4ac a² + 2ac + c² > 4ac (a+c)² > 4ac b < a+c < 0 ---> b² > (a+c)² > 4ac But this contradicts: b² ≤ 4ac Therefore, a cannot be < 0 Since a > 0 and f(x) has at most 1 real distinct root, then f(x) ≥ 0 for all real x

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