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Difference between revisions of "1953 AHSME Problems/Problem 23"
(I rewrote the original solution with the format used in most other solutions, while still keeping most of the original solution intact.)
 
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The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
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==Problem==
  
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<math>\textbf{A}</math> an extraneous root between <math>-5</math> and <math>-1</math> <math>//</math>
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The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
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<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math>  //
 
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<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math> //
 
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<math>\textbf{(D)}</math> two true roots //
 
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<math>\textbf{(E)}</math> two extraneous roots //
 
  
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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>.
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<math>\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\
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\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad
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\textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}  </math>
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==Solution==
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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>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=22|num-a=24}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:30, 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

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