# 1957 AHSME Problems/Problem 40

## Problem

If the parabola $y = -x^2 + bx - 8$ has its vertex on the $x$-axis, then $b$ must be:

$\textbf{(A)}\ \text{a positive integer}\qquad \\ \textbf{(B)}\ \text{a positive or a negative rational number}\qquad \\ \textbf{(C)}\ \text{a positive rational number}\qquad\\ \textbf{(D)}\ \text{a positive or a negative irrational number}\qquad\\ \textbf{(E)}\ \text{a negative irrational number}$

## Solution 1

Note that if $y=-x^2+bx-8$ has its vertex on the $x$-axis, then $-y=x^2-bx+8$ will have its vertex on the x-axis as well. To find the location of the vertex of the parabola, we desire to put it in vertex form, where $-y=(x-h)^2+k$, and $(h,k)$ is the location of the vertex. However, we know that $k=0$, because the vertex is on the $x$-axis. Thus, we know that $x^2-bx+8$ must be the square of a linear term. Thus, $b=\pm 2 \cdot 1 \cdot \sqrt8 =\pm 4\sqrt2$, which are both irrational. Thus, our answer is $\boxed{\textbf{(D)}\text{ a positive or a negative irrational number}}$.

## Solution 2

We know that if a parabola is given by $ax^2+bx+c$, then the $x$-value of the vertex is $\tfrac{-b}{2a}$ (this fact can be proven with the quadratic formula and also derivatives). Because, in this case, $a=-1$, $x=\tfrac{-b}{2a}=\tfrac b 2$. Thus, at $x=\tfrac b 2$, the parabola should have a $y$-value of $0$. Therefore, we have the following equation that we can solve for $b$: \begin{align*} y|_{x=\tfrac b 2} = -(\frac b 2)^2-b \cdot (\frac b 2)-8 &= 0\\ -\frac{b^2}4+\frac{b^2}2 &= 8 \\ \frac{b^2}4 &= 8 \\ b^2 &= 32 \\ b &= \pm \sqrt{32} = \pm 4\sqrt2 \end{align*} Because both of these results are irrational with one positive and one negtaive solution, we choose answer $\fbox{\textbf{(D)} a positive or a negative irrational number}$.