If the equation `|x^2+b x+c|=k` has four real roots, then `b^2-4c >0a n d00a n dk >(4c-b^2)/4` none of these

Explanation

We have `|x^(2)+bx+c|=k`
For the equation to have four real roots, the line y=k must intersect `y=|x^(2)+bx+c|` at four points, which is not possible if `D=0orDlt0` as the possible graphas of `y=|x^(2)+bx+c|` will be as follows.

For the above graphs, the equation `|x^(2)+bx+c|=k` can have maximum two real roots. So for four real roots, we must have `Dgt0`, so the graph of `y=x^(2)+bx+c` is as follows.

So the graph of `y=|x^(2)+bx+c|` will be as follows.

Now the equation `|x^(2)+bx+c|=k` will have four real if the line y=k intersects the above graph between the x-axis and point A.
Hence we have `0ltklt(4c-b^(2))/(4)`
Also `Dgt0:.b^(2)-4cgt0`.