Question

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 find the value of $\alpha \left( \beta +\gamma \right)$ is equal to
(a) $\beta \gamma$
(b) 0
(c) $\alpha \gamma$
(d) $\alpha \beta$

Answer 1

In this question, we are given with two equations $\alpha {{x}^{2}}+2\beta x+\gamma =0$ and ${{x}^{2}}+x-1=0$ . And $\alpha ,\beta$ 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. Now since $\alpha ,\beta$ and $\gamma$ are three consecutive terms of a non-constant G.P, therefore the common ratio between the consecutive terms $\alpha$, $\beta$ and $\gamma$ is equal. We will then for expression for $\beta$ and $\gamma$ in form of $\alpha$ and substitute the same values in the equation $\alpha {{x}^{2}}+2\beta x+\gamma =0$. We will then solve to get the desired answer.

Complete step by step answer:
We are given with two equations $\alpha {{x}^{2}}+2\beta x+\gamma =0$ and ${{x}^{2}}+x-1=0$.
We are also given that $\alpha ,\beta$ 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.
That is, we have a common ratio between the consecutive terms $\alpha$, $\beta$ and $\gamma$ say $r$.
Then since the terms of a geometric series is given by $a,ar,a{{r}^{2}},....$, hence we must have
$\beta =\alpha r$ and $\gamma =\alpha {{r}^{2}}$.
Now on substituting the values $\beta =\alpha r$ and $\gamma =\alpha {{r}^{2}}$ in the given equation $\alpha {{x}^{2}}+2\beta x+\gamma =0$, we will have
$\alpha {{x}^{2}}+2\alpha rx+\alpha {{r}^{2}}=0$
Now on taking $\alpha$ common in the above equation, we have
$\alpha \left( {{x}^{2}}+2rx+{{r}^{2}} \right)=0$
Now since $\alpha$ is a term of G.P, therefore the value of $\alpha$ cannot be zero.
Thus by dividing the equation $\alpha \left( {{x}^{2}}+2rx+{{r}^{2}} \right)=0$ by $\alpha$, we get
${{x}^{2}}+2rx+{{r}^{2}}=0$
This implies
${{\left( x+r \right)}^{2}}=0$
Since we know that ${{x}^{k}}=0$ implies $x=0$ for any natural number $k$.
Therefore we have ${{\left( x+r \right)}^{2}}=0$ implies $x+r=0$.
That implies
$x=-r$
Now since we are given that the root of the equations $\alpha {{x}^{2}}+2\beta x+\gamma =0$ and ${{x}^{2}}+x-1=0$ are equal .
Therefore we have that $x=-r$ satisfies the equation ${{x}^{2}}+x-1=0$.
Thus on substituting $x=-r$ in the equation ${{x}^{2}}+x-1=0$, we get

& {{\left( -r \right)}^{2}}+\left( -r \right)-1=0 \\\ & \Rightarrow {{r}^{2}}-r-1=0...........(1) \\\ \end{aligned}$$ Now on substituting the values $$\beta =\alpha r$$ and $$\gamma =\alpha {{r}^{2}}$$ in the expression $$\alpha \left( \beta +\gamma \right)$$, we get $$\begin{aligned} & \alpha \left( \beta +\gamma \right)=\alpha \left( \alpha r+\alpha {{r}^{2}} \right) \\\ & ={{\alpha }^{2}}\left( r+{{r}^{2}} \right)...(2) \end{aligned}$$ Now from equation (1), we get $${{r}^{2}}=r+1$$ On multiplying above with $$r$$, we get $$\begin{aligned} & {{r}^{3}}=r\left( r+1 \right) \\\ & ={{r}^{2}}+r...(3) \end{aligned}$$ Now on substituting the value of equation (3) in equation (2), we get $$\begin{aligned} & \alpha \left( \beta +\gamma \right)={{\alpha }^{2}}\left( r+{{r}^{2}} \right) \\\ & ={{\alpha }^{2}}{{r}^{3}} \\\ & =\left( ar \right)\left( a{{r}^{2}} \right) \\\ & =\beta \gamma \end{aligned}$$ Therefore we have that the value of $$\alpha \left( \beta +\gamma \right)$$ is equal to $$\beta \gamma $$. **So, the correct answer is “Option A”.** **Note:** In this problem, in order to determine the value of $$\alpha \left( \beta +\gamma \right)$$ where $$\alpha ,\beta $$ and $$\gamma $$ are three consecutive terms of a non-constant G.P, therefore the common ratio between the consecutive terms $$\alpha $$, $$\beta $$ and $$\gamma $$ is equal. Also take care of the given information that the root of the equations $$\alpha {{x}^{2}}+2\beta x+\gamma =0$$ and $${{x}^{2}}+x-1=0$$ are equal .