Difference between revisions of "1954 AHSME Problems/Problem 1"

(Created page with "== Problem 1== The square of <math>5-\sqrt{y^2-25}</math> is: <math>\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \ \textbf{(D)}\ (5-...")
 
 
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== Solution ==
 
== Solution ==
  
<math>(5-\sqrt{y^2-25})^2\implies 5^2-2\cdot 5\cdot\sqrt{y^2-25}+y^2-25\implies y^2-10\sqrt{y^2-25}</math>, <math>\fbox{E}</math>
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<math>(5-\sqrt{y^2-25})^2 = 5^2-2\cdot 5\cdot\sqrt{y^2-25}+y^2-25 = y^2-10\sqrt{y^2-25}</math> <math>\ \Rightarrow</math> <math>\fbox{E}</math>
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==See Also==
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{{AHSME 50p box|year=1954|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 19:33, 17 February 2020

Problem 1

The square of $5-\sqrt{y^2-25}$ is:

$\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25}$

Solution

$(5-\sqrt{y^2-25})^2 = 5^2-2\cdot 5\cdot\sqrt{y^2-25}+y^2-25 = y^2-10\sqrt{y^2-25}$ $\ \Rightarrow$ $\fbox{E}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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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