Determine the Set of Values of k for Which the Equation Has Real Roots
Solution
Given: $$kx^2+6x+1=0$$
Here, $$a=k,\quad b=6,\quad c=1$$
For real roots, $$D=b^2-4ac\ge0$$
$$6^2-4(k)(1)\ge0$$
$$36-4k\ge0$$
$$k\le9$$
Answer
The quadratic equation has real roots when $$\boxed{k\le9}$$