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