Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$ (2k+1)x^2+2(k+3)x+(k+5)=0 $$
Here, $$ a=2k+1,\quad b=2(k+3),\quad c=k+5 $$
For real and equal roots, $$ D=b^2-4ac=0 $$
$$ [2(k+3)]^2-4(2k+1)(k+5)=0 $$
$$ (k+3)^2-(2k+1)(k+5)=0 $$
$$ k^2+6k+9-(2k^2+11k+5)=0 $$
$$ k^2+5k-4=0 $$
$$ k=\frac{-5\pm\sqrt{41}}{2} $$
Answer
The value(s) of k for which the roots are real and equal is: $$ \boxed{k=\frac{-5+\sqrt{41}}{2}\ \text{or}\ k=\frac{-5-\sqrt{41}}{2}} $$