Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I had an exam today and I was thinking about this task now, after the exam of course.

$f(x)=a(x-b)^2 +c$

Now, the point was to find C so that the function only has one root. Easy enough, I played with the calculator and found this. But I hate explanations like that, yes. You get a few points but far from full score. But overall I should still get an A, I hope. If $C=0$ then the expression is a perfect square and they only have one root? Is that far of?

$a(x-b)^2= - c$

$\frac{a(x-b)^2}{a}= - \frac{c}{a}$

$(x-b)^2= - \frac{c}{a}$

This also argues that c should be 0 for it to only be one root?

share|cite|improve this question
Just to be sure: I wouldn't say that perfect squares have only one root. Both $-2$ and $2$ square to $4$, for example. – Dylan Moreland Jan 27 '12 at 17:02
up vote 5 down vote accepted

An alternative way to think about it is geometrically. The graph of $y=x^2$ is a parabola that opens up with vertex at the origin. The graph of $$y = (x-b)^2$$ is then a horizontal shift by $b$ units (so $b$ units to the right if $b\geq 0$, and $|b|$ units to the left if $b\lt 0$) of the same graph. There is still only one root: the vertex.

If $a\neq 0$, then $$y = a(x-b)^2$$ is a vertical stretch of this graph, possibly with a flip (if $a\lt 0$); it does not change the number of intersections with the $x$-axis.

Finally, $$y=a(x-b)^2 + c$$ is a vertical shift by $c$ units (up if $c\gt 0$, down if $c\lt 0$).

If $y=a(x-b)^2$ is a parabola that opens "up" (if $a\gt 0$), then shifting it up ($c\gt 0$) will remove all intersections with the $x$-axis; and shifting it down ($c\lt 0$) will create two intersections with the $x$-axis as the vertex moves down.

If $y=a(x-b)^2$ is a parabola that opens "down" (if $a\lt 0$), then the situation is reversed: $c\gt 0$ will create two intersections with the $x$-axis, and $c\lt 0$ will remove all intersections with the $x$-axis.

Either way, in order to maintain one and only one intersection, you need the vertex of the parabola to stay on the $x$-axis, so you need $c=0$. Conversely, if $c=0$, you have a parabola with vertex on the $x$-axis, hence with a single intersection.

share|cite|improve this answer
An alternative and better way! – André Nicolas Jan 27 '12 at 18:30

We will explicitly assume that $a\ne 0$. Then, more or less as you wrote, $a(x-b)^2+c=0$ if and only if $(x-b)^2=-\frac{c}{a}$.

Thus if $-\frac{c}{a} <0$, there is no root, since the square of a real number cannot be negative. If $-\frac{c}{a}>0$, there are two distinct roots, namely $x=b\pm\sqrt{-c/a}$. And finally, if $-\frac{c}{a}=0$, or equivalently $c=0$, there is exactly one root.

So there is exactly one root if and only if $c=0$.

The above is undoubtedly what you had in mind. What you actually wrote on the exam paper may not have been complete. Much of the time, a bunch of equations with little explanatory text means an incomplete solution.

For absolute completeness let's deal with the silly case $a=0$. In that case our equation is equivalent to $c=0$. If $c\ne 0$, this has no solution. If $c=0$, the equation has infinitely many solutions, since $x$ can take on any value.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.