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If \alpha is a real root of ax^{2}+bx+c= 0 and \beta  is a real root of -ax^{2}+bx+c= 0 , Then a root of equation ax^{2}+2bx+2c= 0 lies between \left ( a\alpha \beta \, \neq \, 0\, ,a,b,c\, \epsilon \, R \right )

  • Option 1)

    0\: and\: \alpha

  • Option 2)

    0\: and\: \beta

  • Option 3)

    \alpha \: and\: \beta

  • Option 4)

    2\alpha \: and\: 2\beta

 

Answers (1)

best_answer

Let f\left ( x \right )=ax^{2}+2bx+2c

\Rightarrow \: f\left ( x \right )=ax^{2}+2bx+2c

\because \: \alpha is  root of ax^{2}+bx+c=0 , So b\alpha +c=-a\alpha ^{2}

\therefore \: f\left ( \alpha \right )=a\alpha ^{2}+2\left ( -a\alpha ^{2} \right )=-a\alpha ^{2}\cdots \cdots \left ( 1 \right )

f\left ( \beta \right )=a\beta ^{2}+2b\beta+2c

\because \: \beta is root of -ax^{2}+bx+c=0, So b\beta +c=a\beta ^{2}

\therefore \: f\left ( \beta \right )=a\beta ^{2}+2\left ( a\beta ^{2} \right )=3a\beta ^{2}\cdots \cdots \left ( 2 \right )

f\left ( \alpha \right )\cdot f\left ( \beta \right )=-3a^{2}\alpha ^{2}\beta ^{2}< 0

\Rightarrow \: f\left ( \alpha \right ) & f\left ( \beta \right ) are of opposite sign 

So f\left ( x \right )=0 has a real root between \alpha & \beta

\therefore Option (C)

 

f (α ) & f (β) are of opposite signs -

Here, f\left ( x \right )= ax^{2}+bx+c

f\left ( x \right )= 0 for one x\in \left ( \alpha ,\beta \right )

- wherein

 

  


Option 1)

0\: and\: \alpha

This is incorrect

Option 2)

0\: and\: \beta

This is incorrect

Option 3)

\alpha \: and\: \beta

This is correct

Option 4)

2\alpha \: and\: 2\beta

This is incorrect

Posted by

Himanshu

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