When does $x^4=(x-c)^2$ have four real roots?

When does $x^4=(x-c)^2$ have four real roots?

Question

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 $c$.

Suggestion

How could we turn this question into a question about quadratics?

Remember that there are two possible solutions when we solve an equation of the form $y^2=a$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/EFTgpQun/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

The blue curve is $y = x^4 - (x-c)^2$.

What could we say about this curve when the above equation has “four real solutions”?

Solution

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).