<s>[INST] Determining a constant from polynomial Given that
$x^4= (x-c)^2$
which c is a real constant number. If the above solution has four real roots. Hence, c must be
a. $$-\frac{1}{4}\le c \le \frac{1}{4}$$
b.$$c \le -\frac{1}{4}$$
c.$$ c \le\frac{1}{4}$$
d.$$c \ge\frac{1}{4}$$
e. any number for c

so i use this kind of method, but i dont know what errors that i make.
let us root both sides
$\sqrt{x^4}= \sqrt{(x-c)^2}$
$\rightarrow$ $x^2= (x-c)$
then i use discriminant method (which D $\gt 0$)
Unfortunately, my answer only gives me: $$ c \le\frac{1}{4}$$
what did i miss?
 [/INST] You have $x^2=\pm(x-c)$. So either $x^2+x-c=0$ or $x^2-x+c=0$. The first has real roots iff $1+4c\ge0$ and the second iff $1-4c\ge0$. So for all roots real we need $-\frac{1}{4}\le c\le\frac{1}{4}$.
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