Question

The graph of the function f(x)=(k-1)x^2+(k+1)x+k-1 touches the x-axis at only one point.If k>1,find
a)the value of k.
@rational

Answer 1

@rational

Answer 2

if this answer is correct then i'll tell you how i did it.
either k = 1/3 or k=3

Answer 3

that means f(x) has only one zero, so the discriminant has to be \(0\) :
\[b^2-4ac=0\]

Answer 4

\[f(x)=(k-1)\color{blue}{x}^2+(k+1)\color{blue}{x}+k-1\]

\(a=k-1\)
\(b=k+1\)
\(c=k-1\)

plugin the values and solve \(k\)

Answer 5

Answer will be k=3. Thnx @rational :)