Difference between revisions of "1972 AHSME Problems/Problem 11"

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(Solution)
 
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<math>\textbf{(A) }4\text{ only}\qquad \textbf{(B) }-7,~4\qquad \textbf{(C) }0,~4\qquad \textbf{(D) }\text{no }y\qquad  \textbf{(E) }\text{all }y</math>
 
<math>\textbf{(A) }4\text{ only}\qquad \textbf{(B) }-7,~4\qquad \textbf{(C) }0,~4\qquad \textbf{(D) }\text{no }y\qquad  \textbf{(E) }\text{all }y</math>
  
==Solution==
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==Solution (Rigorous) ==
  
Because x<sup>2</sup> + y<sup>2</sup> + 16 = 0 has no real solutions, &#8704; sets containing x<sup>2</sup> + y<sup>2</sup> + 16 = 0, no real solutions may exist.
 
  
&#8756; the solution is <math>\fbox{D}</math>
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==Solution (Quick)==
 
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Checking the answer choices, we see that only 4 is the viable choice. Therefore, the answer is <math>\boxed{A}</math>
 
 
– TylerO_1.618
 

Latest revision as of 14:27, 28 June 2025

Problem

The value(s) of $y$ for which the following pair of equations $x^2+y^2-16=0\text{ and }x^2-3y+12=0$ may have a real common solution, are

$\textbf{(A) }4\text{ only}\qquad \textbf{(B) }-7,~4\qquad \textbf{(C) }0,~4\qquad \textbf{(D) }\text{no }y\qquad  \textbf{(E) }\text{all }y$

Solution (Rigorous)

Solution (Quick)

Checking the answer choices, we see that only 4 is the viable choice. Therefore, the answer is $\boxed{A}$