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 $\boxed{\textbf{(D)}\text{ a positive or a negative irrational number}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 41
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1957 AHSME Problems/Problem 40

Problem

Solution 1

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