# Perfect squares always one root?

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

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?

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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

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.

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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.

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