Given a real constant, \(c\), the equation \[x^4=(x-c)^2\] has four real solutions (including possible repeated roots) for

\(c\leq \frac{1}{4}\),

\(-\frac{1}{4} \leq c \leq \frac{1}{4}\),

\(c \leq -\frac{1}{4}\),

all values of \(c\).

If we square root both sides of the equation \(x^4=(x-c)^2\), we find that this holds if and only if \[x^2=x-c\quad\text{or}\quad x^2=-(x-c).\]

Rearranging these gives \[x^2-x+c=0\quad\text{or}\quad x^2+x-c=0,\]

These will have two real roots (or one repeated real root) if the discriminant is non-negative, so we require \[1-4c\ge0\quad\text{and}\quad 1+4c\ge0\] respectively.

Combining these two requirements we get the result that \[-\frac{1}{4} \leq c \leq \frac{1}{4}.\]

Alternatively, we can think of the original equation as a difference of two squares. Moving both terms to one side of the equation we get \[x^4-(x-c)^2 = (x^2)^2-(x-c)^2 = (x^2+x-c)(x^2-x+c)=0\] and so \(x^2+x-c=0\) or \(x^2-x+c=0\), as before.

So the answer is (b).