Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$ (k+1)x^2-2(3k+1)x+8k+1=0 $$
Here, $$ a=k+1,\quad b=-2(3k+1),\quad c=8k+1 $$
For real and equal roots, $$ D=b^2-4ac=0 $$
$$ [-2(3k+1)]^2-4(k+1)(8k+1)=0 $$
$$ (3k+1)^2-(k+1)(8k+1)=0 $$
$$ 9k^2+6k+1-(8k^2+9k+1)=0 $$
$$ k^2-3k=0 $$
$$ k(k-3)=0 $$
$$ k=0 \quad \text{or} \quad k=3 $$
Answer
The value(s) of k for which the roots are real and equal is: $$ \boxed{k=0 \text{ or } k=3} $$