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if equation x 3 +ax+b=0 have only real roots, then prove that 4a 3 +27b 2 ≤ 0.

4 years ago

Answers : (1)

Ajay
209 Points
							
{ x }^{ 3 }+ax+b\quad =\quad 0\quad =>\quad \quad { x }^{ 3 }+ax\quad =\quad -b\\ { x }^{ 3 }+ax+b\quad =\quad 0\quad \quad will\quad have\quad all\quad real\quad roots\quad if\quad \\ curve\quad y=\quad f(x)\quad =\quad { x }^{ 3 }+ax\quad intersects\quad line\quad y=g(x)\quad =\quad -b\quad at\quad three\quad points.\\ find\quad the\quad lacal\quad maxima\quad and\quad minima\quad of\quad f(x)\quad or\quad { f` }^{ }(x)\quad =\quad 0\\ \quad this\quad gives\quad 3{ x }^{ 2 }+a\quad =\quad 0\quad or\quad x=\pm \sqrt { \frac { -a }{ 3 } } \quadLet\quad m\quad =\quad \sqrt { \frac { -a }{ 3 } } \quad so\quad that\quad x=\pm m\quad are\quad points\quad of\quad local\quad maxima\quad and\quad minima\quad for\quad f(x)\\ if\quad g(x)\quad =\quad -b\quad intersects\quad f(x)\quad at\quad three\quad points\quad then\quad -b\quad lies\quad between\quad f(-m)\quad and\quad f(m)\\ f(m)\quad =\quad { m }^{ 3 }+am\quad =\quad { \left( \sqrt { \frac { -a }{ 3 } } \right) }^{ 2 }+a\left( \sqrt { \frac { -a }{ 3 } } \right) \quad =\quad 2{ \left( \frac { -a }{ 3 } \right) }^{ 3/2 }similarly\quad f(-m)\quad =\quad -\quad 2{ \left( \frac { -a }{ 3 } \right) }^{ 3/2 }the\quad required\quad condition\quad is\quad \\ -2{ \left( \frac { -a }{ 3 } \right) }^{ 3/2 }\quad <\quad -b\quad <\quad 2{ \left( \frac { -a }{ 3 } \right) }^{ 3/2 }\quad or\\ \left| b \right| \quad <\quad 2{ \left( \frac { -a }{ 3 } \right) }^{ 3/2 }\quad or\\ squaring\quad both\quad sides\quad withh\quad give\quad us\quad the\quad required\quad condition\\ 27{ b }^{ 2 }+4{ a }^{ 3 }<0
4 years ago
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