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Difference between revisions of "1953 AHSME Problems/Problem 23"
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The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
 
The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
  
−
<math>\textbf{A}</math> an extraneous root between <math>-5</math> and <math>-1</math> <math>//</math>
+
<math>\textbf{A}</math> an extraneous root between <math>-5</math> and <math>-1</math>
−
<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math> //
+
 
−
<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math> //
+
<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math>  
−
<math>\textbf{(D)}</math> two true roots //
+
 
−
<math>\textbf{(E)}</math> two extraneous roots //
+
<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math>
 +
 
 +
<math>\textbf{(D)}</math> two true roots
 +
 
 +
<math>\textbf{(E)}</math> two extraneous roots
  
 
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>.
 
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>.

Revision as of 17:41, 30 April 2018

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

$\textbf{A}$ an extraneous root between $-5$ and $-1$

$\textbf{(B)}$ an extraneous root between $-10$ and $-6$

$\textbf{(C)}$ a true root between $20$ and $25$

$\textbf{(D)}$ two true roots

$\textbf{(E)}$ two extraneous roots

We multiply both sides by $\sqrt{x+10}$ to get the equation $x+4=5\sqrt{x+10}$. We square both sides to get $x^2+8x+16=25x+250$, or $x^2-17x-234=0$. We can factor the quadratic as $(x+9)(x-26)=0$, giving us roots of $-9$ and $26$. We plug these values in to find that $-9$ is an extraneous root and that $26$ is a true root, giving an answer of $\boxed{B}$.

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