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

Latest revision as of 13:27, 28 June 2025

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

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

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

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 ==Problem==
-The value(s) of <math>y</math> for which the following pair of equations <math>x^2+y^2+16=0\text{ and }x^2-3y+12=0</math> may have a real common solution, are
+The value(s) of <math>y</math> for which the following pair of equations <math>x^2+y^2-16=0\text{ and }x^2-3y+12=0</math> may have a real common solution, are
 <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==
-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.
+==Solution (Rigorous) ==
-&#8756; the solution is <math>\fbox{D}</math>
+==Solution (Quick)==
-– TylerO_1.618
+Checking the answer choices, we see that only 4 is the viable choice. Therefore, the answer is <math>\boxed{A}</math>