Question

$x$ and $y$ are positive numbers. Let $g$ and $a$ be G.M. and A.M. of these numbers. Also, let $G$ be G.M. of $x+1$ and $y+1$. If $G$ and $g$ are roots of the equation ${{x}^{2}}-5x+6=0$, then
A. $x=2,y=\dfrac{3}{4}$
B. $x=\dfrac{3}{4},y=12$
C. $x=\dfrac{5}{2},y=\dfrac{8}{5}$
D. $x=y=2$

Answer 1

Hint: In this question, we are to find the $x$ and $y$ values. For this, we have given in the question, that the G.M.s of these numbers are related to the roots of an equation given as ${{x}^{2}}-5x+6=0$. Thus, finding the roots of the equation and the mean of those numbers will help us to find the required values.

Formula Used: If $a,b,c$ are in arithmetic progression, then their arithmetic mean is the average of the first and last terms, which is nothing but their middle term. I.e.,
$A.M=b=\dfrac{a+c}{2}$
If $a,b,c$ are in geometric progression, then their geometric mean is the square root of the product of the first and last terms, which is nothing but the middle term. I.e.,
$G.M=b=\sqrt{ac}$

Complete step by step solution: Given that, $x$ and $y$ are the positive numbers.
It is given that $g$ and $a$ be G.M. and A.M. of these numbers.
So, we can write from the formulas we have for the G.M and A.M as
$G.M=b=\sqrt{ac}$ and $A.M=b=\dfrac{a+c}{2}$;
$g=\sqrt{xy}\text{ }...(1)$
And
$a=\dfrac{x+y}{2}$
It is also given that, $G$ be G.M. of $x+1$ and $y+1$.
So, we can write
$G=\sqrt{(x+1)(y+1)}\text{ }...(2)$
It is given that, $G$ and $g$ are the roots of the given equation ${{x}^{2}}-5x+6=0$.
So, factorizing the above expression, we get
$\begin{align}
  & {{x}^{2}}-5x+6=0 \\
 & \Rightarrow {{x}^{2}}-3x-2x+6=0 \\
 & \Rightarrow x(x-3)-2(x-3)=0 \\
 & \Rightarrow (x-3)(x-2)=0 \\
\end{align}$
By the zero-product rule, we get $x=3$ and $x=2$.
So, we can write
$G=3;g=2$
Then, by substituting in (1) and (2), we get
$\begin{align}
  & 3=\sqrt{(x+1)(y+1)} \\
 & \Rightarrow 9=(x+1)(y+1) \\
 & \Rightarrow 9=xy+x+y+1\text{ }...(3) \\
\end{align}$
And
$\begin{align}
  & 2=\sqrt{xy} \\
 & \Rightarrow 4=xy\text{ }...(4) \\
\end{align}$
Then, substituting (4) in (3), we get
\[\begin{align}
  & 9=xy+x+y+1 \\
 & \Rightarrow 9=4+x+y+1 \\
 & \Rightarrow 9-5=x+y \\
 & \Rightarrow 4=x+y\text{ }...(5) \\
\end{align}\]
From (4) we can write
$xy=4\Rightarrow y=\dfrac{4}{x}$
On substituting in (5), we get
$\begin{align}
  & x+y=4 \\
 & \Rightarrow x+\dfrac{4}{x}=4 \\
 & \Rightarrow {{x}^{2}}+4=4x \\
 & \Rightarrow {{x}^{2}}-4x+4=0 \\
\end{align}$
On simplifying, we get
$\begin{align}
  & {{x}^{2}}-4x+4=0 \\
 & \Rightarrow {{(x-2)}^{2}}=0 \\
 & \Rightarrow x-2=0 \\
 & \therefore x=2 \\
\end{align}$
Then, on substituting this in (5), we get
$\begin{align}
  & x+y=4 \\
 & \Rightarrow 2+y=4 \\
 & \therefore y=2 \\
\end{align}$
Therefore, the required numbers are $x=y=2$.

Option ‘D’ is correct

Note: Here, we need to apply the geometric means of the respective number as given in the question. On comparing those with the given roots we can find the required values. This is because, it is clearly given in the question, that the geometric means of the respective numbers are the roots of the given equation.