Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1957 AHSME Problems/Problem 2"

(Created page with "patopato is awesome. PROOF: I am awesome. I never lie. Therefore, I am awesome.")
 
m (typo fix)
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
patopato is awesome.  
+
== Problem ==
 +
 +
In the equation <math>2x^2 - hx + 2k = 0</math>, the sum of the roots is <math>4</math> and the product of the roots is <math>-3</math>.  
 +
Then <math>h</math> and <math>k</math> have the values, respectively:
  
−
PROOF:
+
<math>\textbf{(A)}\ 8\text{ and }{-6} \qquad
 +
\textbf{(B)}\ 4\text{ and }{-3}\qquad
 +
\textbf{(C)}\ {-3}\text{ and }4\qquad
 +
\textbf{(D)}\ {-3}\text{ and }8\qquad
 +
\textbf{(E)}\ 8\text{ and }{-3}  </math> 
  
−
I am awesome. I never lie. Therefore, I am awesome.
+
== Solution ==
 +
 
 +
Let the roots of the given equation be <math>r</math> and <math>s</math>. Then, by [[Vieta's Formulas]], we have the following:
 +
\begin{align*}
 +
\frac{h}{2} = r+s = 4 &\rightarrow h = 8 \\
 +
\frac{2k}{2} = rs = -3 &\rightarrow k = -3 \\
 +
\end{align*}
 +
 
 +
Thus, our answer is <math>\boxed{\textbf{(E) } 8 \text{ and } -3}</math>.
 +
 
 +
== See also ==
 +
 
 +
{{AHSME 50p box|year=1957|num-b=1|num-a=3}}
 +
{{MAA Notice}}
 +
[[Category:AHSME]][[Category:AHSME Problems]]
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 11:06, 24 July 2024

Problem

In the equation $2x^2 - hx + 2k = 0$, the sum of the roots is $4$ and the product of the roots is $-3$. Then $h$ and $k$ have the values, respectively:

$\textbf{(A)}\ 8\text{ and }{-6} \qquad \textbf{(B)}\ 4\text{ and }{-3}\qquad \textbf{(C)}\ {-3}\text{ and }4\qquad \textbf{(D)}\ {-3}\text{ and }8\qquad \textbf{(E)}\ 8\text{ and }{-3}$

Solution

Let the roots of the given equation be $r$ and $s$. Then, by Vieta's Formulas, we have the following: \begin{align*} \frac{h}{2} = r+s = 4 &\rightarrow h = 8 \\ \frac{2k}{2} = rs = -3 &\rightarrow k = -3 \\ \end{align*}

Thus, our answer is $\boxed{\textbf{(E) } 8 \text{ and } -3}$.

See also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_2&oldid=223531"