36let, f(x) = ax2 + bx + c
now, since this equation dosen't have distinct real roots
so, D ≤ 0
=> b2 - 4ac ≤ 0 or, a ≥ b2/4a --- (i)
and given, a > c - b
so can this help in proving a > 0
if the equation ax2 + bx + c = 0 does not have 2 distinct real roots and a + b > c, then prove that f(x) ≥ 0, for all x E R.
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5 Answers
36
21No one will, if you post wrong questions ...
-(x^2 -x + 1)
it has no real root...a+b>c, (cuz a=-1,b=1,c= -1,so 0>-1)
But f(x)<0 for all real x, while you want us to prove f(x)≥0, so its wrong...
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if you mean a+ c>b then the question is write, and it just means f(-1)>0 , so f(x)≥0
