Question

Find m, if the roots of the quadratic equation (m – 1)x² – 2(m – 1)x + 1 = 0 has real and equal roots.

Answer 1

Given: (m – 1)x² – 2(m – 1)x + 1 = 0
First, we need to find the discriminant of the given quadratic equation.
We know that the discriminant of a quadratic equation is Δ = b² – 4ac.
On comparing the given equation with ax² + bx + c = 0 in variable x, we get:
a = m – 1, b = –2(m – 1) and c = 1
Thus, Δ = (–2(m – 1))² – 4(m – 1)(1)
=> Δ = 4(m² + 1 – 2m) – 4m + 4
=> Δ = 4m² + 4 – 8m – 4m + 4
=> Δ = 4m² – 12m + 8
Since the roots of the given quadratic equation are real and equal, Δ = 0.
Thus, 4m² – 12m + 8= 0
=> 4(m² – 3m + 2) = 0
=> m² – 3m + 2 = 0
On splitting the middle term –3m as – 2m – m, we get:
m² – 2m – m + 2 = 0
=> m(m – 2) – 1(m – 2) = 0
=> (m – 2) (m – 1) = 0
We know that if the product of two numbers is zero, then at least one of them must be zero.
Thus, m – 2= 0 or m – 1= 0
=> m = 2 or m = 1
Now, on substituting m = 1 in the given equation, we observe that the coefficient of x² becomes zero and in this case quadratic equation is not possible.
Thus, m = 1 is not possible.
Therefore, m = 2.