
If it is given that $\left( x-4 \right)$ is a geometric mean of $\left( x-5 \right)$ and $\left( x-2 \right)$, then find the value of $x$.
Answer
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Hint: In this problem we need to calculate the value of $x$ where the geometric mean of $\left( x-5 \right)$ and $\left( x-2 \right)$ is $\left( x-4 \right)$. We know that the geometric mean of the numbers $a$ and $b$ is $\sqrt{ab}$. So, we will first calculate the geometric mean of $\left( x-5 \right)$ and $\left( x-2 \right)$ by using the above formula. After that we will equate the calculated value of the geometric mean to the given geometric mean which is $\left( x-4 \right)$ and simplify the equation to get the required result.
Complete step by step answer:
Given that, the geometric mean of $\left( x-5 \right)$ and $\left( x-2 \right)$ is $\left( x-4 \right)$.
We know that the geometric mean of the numbers $a$ and $b$ is $\sqrt{ab}$. From this formula the geometric mean of the numbers $\left( x-5 \right)$ and $\left( x-2 \right)$ is given by
$m=\sqrt{\left( x-5 \right)\left( x-2 \right)}$
But in the problem, we have the geometric mean of the numbers as $\left( x-5 \right)$. So, equating the both the values, then we will get
$\left( x-4 \right)=\sqrt{\left( x-5 \right)\left( x-2 \right)}$
Squaring on both sides of the above equation, then we will have
${{\left( x-4 \right)}^{2}}=\left( x-5 \right)\left( x-2 \right)$
Using the formulas ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, \[\left( x-a \right)\left( x-b \right)={{x}^{2}}-\left( a+b \right)x+ab\]in the above equation, then we will get
${{x}^{2}}-2\left( x \right)\left( 4 \right)+{{4}^{2}}={{x}^{2}}-\left( 5+2 \right)x+\left( 5 \right)\left( 2 \right)$
Simplifying the above equation by cancelling the common term ${{x}^{2}}$ which is on both sides of the equation, then we will get
$-8x+16=-7x+10$
Rearrange the terms in the above equation, so that all the variables at one side and constants at one side, then we will get
$\begin{align}
& 8x-7x=16-10 \\
& \therefore x=6 \\
\end{align}$
Note: In this problem we don’t have calculated the geometric mean by simplifying the equation $m=\sqrt{\left( x-5 \right)\left( x-2 \right)}$ which takes too much of time and gives some complexity in further steps. So, use simple mathematical operations and solve the problem in an easiest way.
Complete step by step answer:
Given that, the geometric mean of $\left( x-5 \right)$ and $\left( x-2 \right)$ is $\left( x-4 \right)$.
We know that the geometric mean of the numbers $a$ and $b$ is $\sqrt{ab}$. From this formula the geometric mean of the numbers $\left( x-5 \right)$ and $\left( x-2 \right)$ is given by
$m=\sqrt{\left( x-5 \right)\left( x-2 \right)}$
But in the problem, we have the geometric mean of the numbers as $\left( x-5 \right)$. So, equating the both the values, then we will get
$\left( x-4 \right)=\sqrt{\left( x-5 \right)\left( x-2 \right)}$
Squaring on both sides of the above equation, then we will have
${{\left( x-4 \right)}^{2}}=\left( x-5 \right)\left( x-2 \right)$
Using the formulas ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, \[\left( x-a \right)\left( x-b \right)={{x}^{2}}-\left( a+b \right)x+ab\]in the above equation, then we will get
${{x}^{2}}-2\left( x \right)\left( 4 \right)+{{4}^{2}}={{x}^{2}}-\left( 5+2 \right)x+\left( 5 \right)\left( 2 \right)$
Simplifying the above equation by cancelling the common term ${{x}^{2}}$ which is on both sides of the equation, then we will get
$-8x+16=-7x+10$
Rearrange the terms in the above equation, so that all the variables at one side and constants at one side, then we will get
$\begin{align}
& 8x-7x=16-10 \\
& \therefore x=6 \\
\end{align}$
Note: In this problem we don’t have calculated the geometric mean by simplifying the equation $m=\sqrt{\left( x-5 \right)\left( x-2 \right)}$ which takes too much of time and gives some complexity in further steps. So, use simple mathematical operations and solve the problem in an easiest way.
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