What is the value k can take for which the equation x^2+(k-1)x+k-1=0 has one solution

To determine the types and number of roots of a
quadratic equation, examine the discriminant:
• If b² - 4ac > 0 then there are two real (unique) roots
• If b² - 4ac = 0 then there is one (double) real root
• If b² - 4ac < 0 then there are two imaginary roots

In this case: x^2 + (k-1)x + (k-1) = 0
• b² - 4ac = 0 ← one (double) real root
• (k-1)² - 4(1)(k-1) = 0
• (k-1) (k-5) = 0
or k = { 1, 5 }

One solution means it can be written (x-a)(x-a)=0

x^2-2ax+a^2=0

So, k-1=-2a, and k-1=a^2

Eliminate k:-

1-2a=1+a^2

a^2-2a=0
a=0, or a=2
=> k=1, or k=5