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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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