Difference between revisions of "1953 AHSME Problems/Problem 23"
m |
|||
Line 1: | Line 1: | ||
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>\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 has:
an extraneous root between and
an extraneous root between and
a true root between and
two true roots
two extraneous roots
We multiply both sides by to get the equation . We square both sides to get , or . We can factor the quadratic as , giving us roots of and . We plug these values in to find that is an extraneous root and that is a true root, giving an answer of .