Difference between revisions of "1956 AHSME Problems/Problem 41"

Revision as of 13:05, 24 July 2024

Problem 41

The equation $3y^2 + y + 4 = 2(6x^2 + y + 2)$ where $y = 2x$ is satisfied by:

$\textbf{(A)}\ \text{no value of }x \qquad \textbf{(B)}\ \text{all values of }x \qquad \textbf{(C)}\ x = 0\text{ only} \\ \textbf{(D)}\ \text{all integral values of }x\text{ only} \qquad \textbf{(E)}\ \text{all rational values of }x\text{ only}$

Solution

Start by plugging in $y=2x:$ $12x^2 + 2x + 4 = 12x^2 + 4x + 4$ The only value that can satisfy this equation is $x=0$ meaning the answer is C

-coolmath34

1956 AHSC (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 12: / Line 12: @@
 Start by plugging in <math>y=2x:</math>
 <cmath>12x^2 + 2x + 4 = 12x^2 + 4x + 4</cmath>
-There is no value of <math>x</math> that can satisfy the equation, so the answer is <math>\boxed{\textbf{(A)}}.</math>
+The only value that can satisfy this equation is <math>x=0</math> meaning the answer is C
 -coolmath34

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Difference between revisions of "1956 AHSME Problems/Problem 41"

Revision as of 13:05, 24 July 2024

Problem 41

Solution

See Also