Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$4x^2-3kx+1=0$$
Here, $$a=4,\quad b=-3k,\quad c=1$$
For real and equal roots, $$D=b^2-4ac=0$$
$$(-3k)^2-4(4)(1)=0$$
$$9k^2-16=0$$
$$9k^2=16$$
$$k=\pm\frac{4}{3}$$
Answer
The value(s) of k for which the roots are real and equal is: $$\boxed{k=\pm\frac{4}{3}}$$