Algebra> If alpha is a real root of the equation a...

If alpha is a real root of the equation ax2+bx+c and beta is a real root of equation -ax2+bx+c.Show that there exists a root gama of the equation (a/2)x2+bx+c which lies between alpha and beta.feiends and experts plz help me.thanks

Ankit Jaiswal , 8 Years ago

Grade 12

7 Answers

Ajay

Last Activity: 8 Years ago

From the question it seems alpha and beta are roots of same equation ax2+bx+c. =0.

Is this true or mistake in question?

Ankit Jaiswal

Last Activity: 8 Years ago

no alpha is the root

Ajay

Last Activity: 8 Years ago

$\large let\quad f(x)=\frac { a }{ 2 } x^{ 2 }+bx+c=0\\ if\quad a\quad root\quad ofequation\quad f(x)=0\quad \quad lies\quad between\quad \alpha \quad and\quad \beta \quad then\quad f(\alpha ).f(\beta )<0.\\ This\quad is\quad easy\quad to\quad visualise\quad if\quad you\quad draw\quad graph\quad for\quad y=f(x).\\ Now\quad f(\alpha ).f(\beta )=\quad \left( \frac { a }{ 2 } \alpha ^{ 2 }+b\alpha +c \right) \left( \frac { a }{ 2 } \beta ^{ 2 }+b\beta +c \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 1 }{ 4 } \left( a\alpha ^{ 2 }+2(b\alpha +c) \right) \left( a\beta ^{ 2 }+2(b\beta +c) \right) ------------(1)\quad \quad \quad \\ Given\quad \alpha \quad is\quad root\quad of\quad equation\quad ax^{ 2 }+bx+c\quad =\quad 0\quad hence\quad a\alpha ^{ 2 }+b\alpha +c=0\\ or\quad b\alpha +c=-a\alpha ^{ 2 }-------------(2)$ $\large Given\quad \beta \quad is\quad root\quad of\quad equation\quad -ax^{ 2 }+bx+c\quad =\quad 0\quad hence\quad -a\beta ^{ 2 }+b\beta +c=0\\ or\quad b\beta +c=a\beta ^{ 2 }-------------------(3)\\ substuting\quad from\quad (2)\quad and\quad (3)\quad into\quad (1)\\ f(\alpha ).f(\beta )\quad =\quad \frac { 1 }{ 4 } \left( a\alpha ^{ 2 }-2a\alpha ^{ 2 } \right) \left( a\beta ^{ 2 }+2a\beta ^{ 2 } \right) \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 3 }{ 2 } { a }^{ 2 }\alpha ^{ 2 }\beta \\ or\quad f(\alpha ).f(\beta )<0\quad which\quad is\quad required\quad condition\\ Hence\quad Proved$

Ankit Jaiswal

Last Activity: 8 Years ago

no actually alpha is the roo

Ajay

Last Activity: 8 Years ago

Small Mistake in last para posting again..............................................................................................................

$\large f(\alpha ).f(\beta )\quad =\quad \frac { 1 }{ 4 } \left( a\alpha ^{ 2 }-2a\alpha ^{ 2 } \right) \left( a\beta ^{ 2 }+2a\beta ^{ 2 } \right) \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 3 }{ 2 } { a }^{ 2 }\alpha ^{ 2 }\beta ^{ 2 }\\ or\quad f(\alpha ).f(\beta )<0\quad which\quad is\quad required\quad condition\\ Hence\quad Proved$