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