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If $\alpha ,\beta \: \: and\: \: \gamma$ are three consecutive terms of a non-constant

G.P. such that the equations $\alpha x^{2}+2\beta x+\gamma =0$ and

$x^{2}+ x-1=0$ have a common root , then $\alpha (\beta +\gamma )$ is equal to :

Option 1)
0

Option 2)
$\alpha \beta$

Option 3)
$\alpha \gamma$

Option 4)
$\beta \gamma$

Answers (1)

$\alpha ,\beta \: \: and\: \: \gamma$ are in G.P. => $\beta ^{2}=\alpha \gamma$

For equation, $\alpha x^{2}+2\beta x+\gamma =0$

$\Delta =4\beta ^{2}-4\alpha \gamma =0$

Hence, roots are equal & equals to $-\frac{\beta }{\alpha }=-\frac{\gamma }{\beta }$

Since, given equation have common roots , hence $-\frac{\gamma }{\beta }$ must be root of $x^{2}+x-1=0$

$=> \frac{\gamma ^{2}}{\beta ^{2}}-\frac{\gamma }{\beta }-1=0$

$=> \gamma ^{2}-\gamma \beta -\beta ^{2}=0$

$=> \gamma ^{2}=\beta (\gamma +\beta )$

$=> \gamma .\frac{\beta ^{2}}{\alpha }=\beta (\gamma +\beta )$

$=> \gamma .\beta =\alpha (\gamma +\beta )$

Option 1)

0

Option 2)

$\alpha \beta$

Option 3)

$\alpha \gamma$

Option 4)

$\beta \gamma$

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Vakul

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