1956 AHSME Problems/Problem 41

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 $y=2x:$ $12x^2 + 2x + 4 = 12x^2 + 4x + 4$ There is no value of $x$ that can satisfy the equation, so the answer is $\boxed{\textbf{(A)}}.$

-coolmath34

1956 AHSC (Problems • Answer Key • Resources)
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1956 AHSME Problems/Problem 41

Problem 41

Solution

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