Find the Values of k for Which the Equation Has Real and Equal Roots
Solution
Given: $$k^2x^2-2(2k-1)x+4=0$$
Here, $$a=k^2,\quad b=-2(2k-1),\quad c=4$$
For real and equal roots, $$D=b^2-4ac=0$$
$$[-2(2k-1)]^2-4(k^2)(4)=0$$
$$4(2k-1)^2-16k^2=0$$
$$ (2k-1)^2-4k^2=0 $$
$$ 4k^2-4k+1-4k^2=0 $$
$$ -4k+1=0 $$
$$ k=\frac{1}{4} $$
Answer
The value of k for which the roots are real and equal is: $$\boxed{k=\frac{1}{4}}$$