Question

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

Answer 1

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.