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 "1953 AHSME Problems/Problem 23"

m
(I rewrote the original solution with the format used in most other solutions, while still keeping most of the original solution intact.)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
+
==Problem==
  
−
<math>\textbf{A}</math> an extraneous root between <math>-5</math> and <math>-1</math>
+
The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
  
−
<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math>  
+
<math>\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\
 +
\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad
 +
\textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}  </math>
  
−
<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math>
+
==Solution==
 +
We can multiply both sides by <math>\sqrt{x+10}</math> to get <math>x+4=5\sqrt{x+10}</math>. We can now square both sides to get <math>x^2+8x+16=25x+250</math>, which yields <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. Plugging in these values, we find that <math>-9</math> is an extraneous root and <math>26</math> is a true root. This gives us the final answer of <math>\boxed{\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6}</math>.
  
−
<math>\textbf{(D)}</math> two true roots
+
==See Also==
  
−
<math>\textbf{(E)}</math> two extraneous roots
+
{{AHSME 50p box|year=1953|num-b=22|num-a=24}}
  
−
We multiply both sides by <math>\sqrt{x+10}</math> to get the equation <math>x+4=5\sqrt{x+10}</math>. We square both sides to get <math>x^2+8x+16=25x+250</math>, or <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. We plug these values in to find that <math>-9</math> is an extraneous root and that <math>26</math> is a true root, giving an answer of <math>\boxed{B}</math>.
+
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:30, 27 November 2019

Problem

The equation $\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5$ has:

$\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\ \textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad \textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}$

Solution

We can multiply both sides by $\sqrt{x+10}$ to get $x+4=5\sqrt{x+10}$. We can now square both sides to get $x^2+8x+16=25x+250$, which yields $x^2-17x-234=0$. We can factor the quadratic as $(x+9)(x-26)=0$, giving us roots of $-9$ and $26$. Plugging in these values, we find that $-9$ is an extraneous root and $26$ is a true root. This gives us the final answer of $\boxed{\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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=1953_AHSME_Problems/Problem_23&oldid=112248"