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# Find $m$ if the quadratic equation $\left( {m - 1} \right){x^2} - 2\left( {m - 1} \right)x + 1 = 0$ has equal roots.(a) 1 (b) 2 (c) 3 (d) 4

Last updated date: 13th Jun 2024
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Hint:
Here, we need to find the value of $m$. We will use the formula for discriminant. Then, we will use the given information to form an equation in terms of $m$. Finally, we will solve this equation to get the value of $m$.

Formula Used:
We will use the formula for the discriminant of a quadratic equation of the form $a{x^2} + bx + c = 0$ is given by $D = {b^2} - 4ac$.

Complete step by step solution:
First, we will find the discriminant of a quadratic equation $a{x^2} + bx + c = 0$.
The discriminant of a quadratic equation of the form $a{x^2} + bx + c = 0$ is given by $D = {b^2} - 4ac$.
Comparing the equations $\left( {m - 1} \right){x^2} - 2\left( {m - 1} \right)x + 1 = 0$ and $a{x^2} + bx + c = 0$, we get
$a = m - 1$, $b = - 2\left( {m - 1} \right)$, and $c = 1$
Substituting $a = m - 1$, $b = - 2\left( {m - 1} \right)$, and $c = 1$ in the formula $D = {b^2} - 4ac$, we get
$\Rightarrow D = {\left[ { - 2\left( {m - 1} \right)} \right]^2} - 4\left( {m - 1} \right)\left( 1 \right)$
Simplifying the expression, we get
$\Rightarrow D = 4{\left( {m - 1} \right)^2} - 4\left( {m - 1} \right)$
If $D > 0$, then the roots of the quadratic equation are unequal and real.
If $D = 0$, then the roots of the quadratic equation are real and equal.
If $D < 0$, then the roots of the quadratic equation are not real, that is complex.
It is given that the quadratic equation $\left( {m - 1} \right){x^2} - 2\left( {m - 1} \right)x + 1 = 0$ has equal roots.
Therefore, $D = 0$.
Thus, we get
$\Rightarrow 4{\left( {m - 1} \right)^2} - 4\left( {m - 1} \right) = 0$
Dividing both sides by 4, we get
$\Rightarrow {\left( {m - 1} \right)^2} - \left( {m - 1} \right) = 0$
Rewriting the expression, we get
$\Rightarrow \left( {m - 1} \right)\left( {m - 1} \right) - \left( {m - 1} \right) = 0$
Factoring out the common multiple $m - 1$ from the expression, we get
$\Rightarrow \left( {m - 1} \right)\left( {m - 1 - 1} \right) = 0$
Subtracting the terms, we get
$\Rightarrow \left( {m - 1} \right)\left( {m - 2} \right) = 0$
Therefore, we get
$\Rightarrow m - 1 = 0$ or $m - 2 = 0$
Simplifying the expressions, we get
$\Rightarrow m = 1,2$
However, we can observe that if $m = 1$, then the quadratic equation becomes
$\begin{array}{l} \Rightarrow \left( {1 - 1} \right){x^2} - 2\left( {1 - 1} \right)x + 1 = 0\\ \Rightarrow 0{x^2} - 0x + 1 = 0\end{array}$
Simplifying the expression, we get
$\Rightarrow 1 = 0$
This is not possible.
Therefore, the value of $m$ cannot be 1.
Thus, we get
$\Rightarrow m = 2$

Hence, the correct option is option (b).

Note:
We used the term “quadratic equation” in our solution. A quadratic equation is an equation that has the highest degree of 2. It is of the form $a{x^2} + bx + c = 0$, where $a$ is not equal to 0. A quadratic equation has 2 solutions.
We can also say that if $m = 1$, then $a = 1 - 1 = 0$. However, a quadratic equation is of the form $a{x^2} + bx + c = 0$, where $a$ is not equal to 0. Thus, the value of $m$ cannot be 1.