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

Hint: 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
\[\begin{align}
  & {{\left( -r \right)}^{2}}+\left( -r \right)-1=0 \\
 & \Rightarrow {{r}^{2}}-r-1=0...........(1) \\
\end{align}\]
Now on substituting the values \[\beta =\alpha r\] and \[\gamma =\alpha {{r}^{2}}\] in the expression \[\alpha \left( \beta +\gamma \right)\], we get
\[\begin{align}
  & \alpha \left( \beta +\gamma \right)=\alpha \left( \alpha r+\alpha {{r}^{2}} \right) \\
 & ={{\alpha }^{2}}\left( r+{{r}^{2}} \right)...(2)
\end{align}\]
Now from equation (1), we get
\[{{r}^{2}}=r+1\]
On multiplying above with \[r\], we get
\[\begin{align}
  & {{r}^{3}}=r\left( r+1 \right) \\
 & ={{r}^{2}}+r...(3)
\end{align}\]
Now on substituting the value of equation (3) in equation (2), we get
\[\begin{align}
  & \alpha \left( \beta +\gamma \right)={{\alpha }^{2}}\left( r+{{r}^{2}} \right) \\
 & ={{\alpha }^{2}}{{r}^{3}} \\
 & =\left( ar \right)\left( a{{r}^{2}} \right) \\
 & =\beta \gamma
\end{align}\]
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 .