Difference between revisions of "1957 AHSME Problems/Problem 40"
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\textbf{(E)}\ \text{a negative irrational number}</math> | \textbf{(E)}\ \text{a negative irrational number}</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | <math>\boxed{\textbf{(D)}\text{ a positive or a negative irrational number}}</math>. | + | Note that if <math>y=-x^2+bx-8</math> has its vertex on the <math>x</math>-axis, then <math>-y=x^2-bx+8</math> 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 <math>-y=(x-h)^2+k</math>, and <math>(h,k)</math> is the location of the vertex. However, we know that <math>k=0</math>, because the vertex is on the <math>x</math>-axis. Thus, we know that <math>x^2-bx+8</math> must be the [[Perfect square#Perfect Square Trinomials|square of a linear term]]. Thus, <math>b=\pm 2 \cdot 1 \cdot \sqrt8 =\pm 4\sqrt2</math>, which are both irrational. Thus, our answer is <math>\boxed{\textbf{(D)}\text{ a positive or a negative irrational number}}</math>. |
| + | |||
| + | == Solution 2 == | ||
| + | We know that if a parabola is given by <math>ax^2+bx+c</math>, then the <math>x</math>-value of the vertex is <math>\tfrac{-b}{2a}</math> (this fact can be proven with the [[quadratic formula]] and also [[derivative|derivatives]]). Because, in this case, <math>a=-1</math>, <math>x=\tfrac{-b}{2a}=\tfrac b 2</math>. Thus, at <math>x=\tfrac b 2</math>, the parabola should have a <math>y</math>-value of <math>0</math>. Therefore, we have the following equation that we can solve for <math>b</math>: | ||
| + | \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 <math>\fbox{\textbf{(D)} a positive or a negative irrational number}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 08:48, 27 July 2024
Contents
Problem
If the parabola 
has its vertex on the 
-axis, then 
must be:
Solution 1
Note that if 
has its vertex on the -axis, then 
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 
, and 
is the location of the vertex. However, we know that 
, because the vertex is on the -axis. Thus, we know that 
must be the square of a linear term. Thus, 
, which are both irrational. Thus, our answer is 
.
Solution 2
We know that if a parabola is given by 
, then the -value of the vertex is 
(this fact can be proven with the quadratic formula and also derivatives). Because, in this case, 
, 
. Thus, at 
, the parabola should have a 
-value of 
. Therefore, we have the following equation that we can solve for :
\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 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 39 |
Followed by Problem 41 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
