Theory Of Equations Ques 1
- If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant GP 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
(2019 Main, 12 April II)
(a) $0$
(b) $\alpha \beta$
(c) $\alpha \gamma$
(d) $\beta \gamma$
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Answer:
Correct Answer: 1.(d)
Solution: (d) Given $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant GP.
Let $\quad \alpha=\alpha, \beta=\alpha r, \gamma=\alpha r^2,\{r \neq 0,1\}$
and given quadratic equation is
$ \alpha x^2+2 \beta x+\gamma=0 $ $\quad$ ……..(i)
On putting the values of $\alpha, \beta, \gamma$ in Eq. (i), we get
$ \begin{aligned} & \alpha x^2+2 \alpha e r x+\alpha r^2=0 \\ & \Rightarrow \quad x^2+2 r x+r^2=0 \\ & \Rightarrow \quad(x+r)^2=0 \\ & \Rightarrow \quad x=-r \\ \end{aligned} $
$\because \quad$ The quadratic equations $\alpha x^2+2 \beta x+\gamma=0$ and $x^2+x-1=0$ have a common root, so
$x=-r$ must be root of equation $x^2+x-1=0$, so
$ r^2-r-1=0 $
Now,
$ \begin{aligned} \alpha(\beta+\gamma) & =\alpha\left(\alpha r+\alpha r^2\right) \\ & =\alpha^2\left(r+r^2\right) \end{aligned} $
From the options,
$ \begin{aligned} & \quad \beta \gamma=\alpha r \cdot \alpha r^2=\alpha^2 r^3=\alpha^2\left(r+r^2\right) \\ & \therefore \quad \alpha\left(\because r^2-r-1=0 \Rightarrow r^3=r+r^2\right] \\ & \therefore \quad \alpha(\beta)=\beta \gamma \end{aligned} $
BETA
