Review question

# When does $x^4=(x-c)^2$ have four real roots? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9546

## Solution

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

1. $c\leq \frac{1}{4}$,

2. $-\frac{1}{4} \leq c \leq \frac{1}{4}$,

3. $c \leq -\frac{1}{4}$,

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.