
The roots of the equation ${{\left| x-1 \right|}^{2}}-4\left| x-1 \right|+3=0$
A. Form an A.P.
B. Form a G.P.
C. Form an H.P.
D. Do not form any progression
Answer
162.6k+ views
Hint: In this question, we are to find the roots of the given equation and find the type of progression that they form. For this, the given equation is factorized into factors or roots. Since the given equation has a mod function, we get both negative and positive roots. After finding the roots, by calculating their means we can find their progression.
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.,
$G.M=b=\sqrt{ac}$
If $a,b,c$ are in harmonic progression, then their harmonic mean is the middle term which is given as
$b=\dfrac{2ac}{a+c}$
Complete step by step solution: The given equation is ${{\left| x-1 \right|}^{2}}-4\left| x-1 \right|+3=0$
On factorizing the given equation, we get
$\begin{align}
& {{\left| x-1 \right|}^{2}}-4\left| x-1 \right|+3=0 \\
& \Rightarrow {{\left| x-1 \right|}^{2}}-3\left| x-1 \right|-\left| x-1 \right|+3=0 \\
& \Rightarrow \left| x-1 \right|\left( \left| x-1 \right|-3 \right)-1\left( \left| x-1 \right|-3 \right)=0 \\
& \Rightarrow \left( \left| x-1 \right|-1 \right)\left( \left| x-1 \right|-3 \right)=0 \\
\end{align}$
Then, according to the zero-product rule, we get
$\begin{align}
& (x-a)(x-b)=0 \\
& \Rightarrow x-a=0;x-b=0 \\
\end{align}$
So,
$\begin{align}
& \left( \left| x-1 \right|-1 \right)\left( \left| x-1 \right|-3 \right)=0 \\
& \Rightarrow \left( \left| x-1 \right|-1 \right)=0;\left( \left| x-1 \right|-3 \right)=0 \\
\end{align}$
Then, on simplifying them for the values of $x$, we get
$\begin{align}
& \left| x-1 \right|-1=0 \\
& \Rightarrow \left| x-1 \right|=1 \\
& \Rightarrow x-1=\pm 1 \\
\end{align}$
If $x-1=-1$, then $x=0$
If $x-1=+1$, then $x=2$
And
$\begin{align}
& \left| x-1 \right|-3=0 \\
& \Rightarrow \left| x-1 \right|=3 \\
& \Rightarrow x-1=\pm 3 \\
\end{align}$
If $x-1=-3$, then $x=-2$
If $x-1=+3$, then $x=4$
So, the roots of the given equation are $-2,0,2,4$.
To know their progression type, we need to find their mean.
Checking for the arithmetic mean:
(i) between $-2,2$:
$AM=\dfrac{-2+2}{2}=0$
Since it is true that, the value $0$ is in between $-2,2$, they form an arithmetic progression.
Option ‘A’ is correct
Note: Here, we need to find the arithmetic mean for the obtained roots of the given equation, in order to check the progression, they form.
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.,
$G.M=b=\sqrt{ac}$
If $a,b,c$ are in harmonic progression, then their harmonic mean is the middle term which is given as
$b=\dfrac{2ac}{a+c}$
Complete step by step solution: The given equation is ${{\left| x-1 \right|}^{2}}-4\left| x-1 \right|+3=0$
On factorizing the given equation, we get
$\begin{align}
& {{\left| x-1 \right|}^{2}}-4\left| x-1 \right|+3=0 \\
& \Rightarrow {{\left| x-1 \right|}^{2}}-3\left| x-1 \right|-\left| x-1 \right|+3=0 \\
& \Rightarrow \left| x-1 \right|\left( \left| x-1 \right|-3 \right)-1\left( \left| x-1 \right|-3 \right)=0 \\
& \Rightarrow \left( \left| x-1 \right|-1 \right)\left( \left| x-1 \right|-3 \right)=0 \\
\end{align}$
Then, according to the zero-product rule, we get
$\begin{align}
& (x-a)(x-b)=0 \\
& \Rightarrow x-a=0;x-b=0 \\
\end{align}$
So,
$\begin{align}
& \left( \left| x-1 \right|-1 \right)\left( \left| x-1 \right|-3 \right)=0 \\
& \Rightarrow \left( \left| x-1 \right|-1 \right)=0;\left( \left| x-1 \right|-3 \right)=0 \\
\end{align}$
Then, on simplifying them for the values of $x$, we get
$\begin{align}
& \left| x-1 \right|-1=0 \\
& \Rightarrow \left| x-1 \right|=1 \\
& \Rightarrow x-1=\pm 1 \\
\end{align}$
If $x-1=-1$, then $x=0$
If $x-1=+1$, then $x=2$
And
$\begin{align}
& \left| x-1 \right|-3=0 \\
& \Rightarrow \left| x-1 \right|=3 \\
& \Rightarrow x-1=\pm 3 \\
\end{align}$
If $x-1=-3$, then $x=-2$
If $x-1=+3$, then $x=4$
So, the roots of the given equation are $-2,0,2,4$.
To know their progression type, we need to find their mean.
Checking for the arithmetic mean:
(i) between $-2,2$:
$AM=\dfrac{-2+2}{2}=0$
Since it is true that, the value $0$ is in between $-2,2$, they form an arithmetic progression.
Option ‘A’ is correct
Note: Here, we need to find the arithmetic mean for the obtained roots of the given equation, in order to check the progression, they form.
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