Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$5x^2-4x+2+k(4x^2-2x-1)=0$$
$$ (5+4k)x^2-(4+2k)x+(2-k)=0 $$
Here, $$a=5+4k,\quad b=-(4+2k),\quad c=2-k$$
For real and equal roots, $$D=b^2-4ac=0$$
$$ (4+2k)^2-4(5+4k)(2-k)=0 $$
$$ 16+16k+4k^2-4(10+3k-4k^2)=0 $$
$$ 16+16k+4k^2-40-12k+16k^2=0 $$
$$ 20k^2+4k-24=0 $$
$$ 5k^2+k-6=0 $$
$$ (5k+6)(k-1)=0 $$
$$ k=-\frac{6}{5}\quad \text{or}\quad k=1 $$
Answer
The value(s) of k for which the roots are real and equal is: $$ \boxed{k=-\frac{6}{5}\ \text{or}\ k=1} $$