Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$ (4-k)x^2+(2k+4)x+(8k+1)=0 $$
Here, $$ a=4-k,\quad b=2k+4,\quad c=8k+1 $$
For real and equal roots, $$ D=b^2-4ac=0 $$
$$ (2k+4)^2-4(4-k)(8k+1)=0 $$
$$ (k+2)^2-(4-k)(8k+1)=0 $$
$$ k^2+4k+4-(32k+4-8k^2-k)=0 $$
$$ 9k^2-27k=0 $$
$$ 9k(k-3)=0 $$
$$ k=0 \quad \text{or} \quad k=3 $$
Answer
The value(s) of k for which the roots are real and equal is: $$ \boxed{k=0 \text{ or } k=3} $$