Difference between revisions of "1953 AHSME Problems/Problem 23"
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− | + | ==Problem== | |
− | <math>\ | + | The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has: |
− | <math>\textbf{(B)} | + | <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>\ | + | ==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>. | ||
− | + | ==See Also== | |
− | + | {{AHSME 50p box|year=1953|num-b=22|num-a=24}} | |
− | + | [[Category:Introductory Algebra Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 20:30, 27 November 2019
Problem
The equation has:
Solution
We can multiply both sides by to get
. We can now square both sides to get
, which yields
. We can factor the quadratic as
, giving us roots of
and
. Plugging in these values, we find that
is an extraneous root and
is a true root. This gives us the final answer of
.
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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