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 "1971 AHSME Problems/Problem 20"

(added second solution)
 
(One intermediate revision by one other user not shown)
Line 7: Line 7:
 
\textbf{(C) }\textstyle\frac{3}{2}\qquad
 
\textbf{(C) }\textstyle\frac{3}{2}\qquad
 
\textbf{(D) }2\qquad
 
\textbf{(D) }2\qquad
−
\textbf{(E) }\text{None of these} </math>
+
\textbf{(E) }\text{None of these}</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
Line 20: Line 20:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
The given equation can be rewritten as <math>x^2+2hx-3=0</math>. By [[Vieta's Formulas]], we know that the sum of the roots is <math>-2h</math>. Thus, by [[Newton Sums]], we have the following equation:
+
The given equation can be rewritten as <math>x^2+2hx-3=0</math>. By [[Vieta's Formulas]], we know that the sum of the roots is <math>-2h</math>. Thus, by [[Newton's Sums|Newton's sums]], we have the following equation:
 
\begin{align*}
 
\begin{align*}
 
10+(2h)(-2h)+2(-3) &= 0 \\
 
10+(2h)(-2h)+2(-3) &= 0 \\
Line 33: Line 33:
 
{{AHSME 35p box|year=1971|num-b=19|num-a=21}}
 
{{AHSME 35p box|year=1971|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Intermediate Algebra Problems]]

Latest revision as of 19:02, 28 July 2025

Problem

The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10$. The absolute value of $h$ is equal to

$\textbf{(A) }-1\qquad \textbf{(B) }\textstyle\frac{1}{2}\qquad \textbf{(C) }\textstyle\frac{3}{2}\qquad \textbf{(D) }2\qquad \textbf{(E) }\text{None of these}$

Solution 1

We can rewrite the equation as $x^2 + 2hx - 3 = 0.$ By Vieta's Formulas, the sum of the roots is $-2h$ and the product of the roots is $-3.$

Let the two roots be $r$ and $s.$ Note that \[r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)\]

Therefore, $4h^2 + 6 = 10$ and $h = \pm 1.$ This doesn't match any of the answer choices, so the answer is $\boxed{\textbf{(E) }\text{None of these}}.$

-edited by coolmath34

Solution 2

The given equation can be rewritten as $x^2+2hx-3=0$. By Vieta's Formulas, we know that the sum of the roots is $-2h$. Thus, by Newton's sums, we have the following equation: \begin{align*} 10+(2h)(-2h)+2(-3) &= 0 \\ 10-4h^2-6 &= 0 \\ -4h^2 &= -4 \\ h^2 &= 1 \\ |h| &= 1 \\ \end{align*} Thus, our answer is $\boxed{\textbf{(E) }\text{None of these}}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_AHSME_Problems/Problem_20&oldid=255832"