Find the Values of k for Which the Equation Has Real and Equal Roots
Solution
Given: $$ (k+1)x^2-2(k-1)x+1=0 $$
Here, $$ a=k+1,\quad b=-2(k-1),\quad c=1 $$
For real and equal roots, $$ D=b^2-4ac=0 $$
$$ [-2(k-1)]^2-4(k+1)(1)=0 $$
$$ 4(k-1)^2-4(k+1)=0 $$
$$ (k-1)^2-(k+1)=0 $$
$$ k^2-2k+1-k-1=0 $$
$$ k^2-3k=0 $$
$$ k(k-3)=0 $$
$$ k=0 \quad \text{or} \quad k=3 $$
Answer
The value(s) of k for which the roots are real and equal is: $$\boxed{k=0 \text{ or } k=3}$$