# Difference between revisions of "1960 AHSME Problems/Problem 11"

## Problem

For a given value of $k$ the product of the roots of $x^2-3kx+2k^2-1=0$ is $7$. The roots may be characterized as:

$\textbf{(A) }\text{integral and positive} \qquad\textbf{(B) }\text{integral and negative} \qquad \\ \textbf{(C) }\text{rational, but not integral} \qquad\textbf{(D) }\text{irrational} \qquad\textbf{(E) } \text{imaginary}$

## Solution

If the product of the roots are $7$, then by Vieta's formulas, $$2k^2-1=7$$ Solve for $k$ in the resulting equation to get $$2k^2=8$$ $$k^2=4$$ $$k=\pm 2$$ That means the two quadratics are $x^2-6x+7=0$ and $x^2+6x+7=0$. Since $b^2$, $a$, and $c$ are the same, the discriminant of both is $36-(4 \cdot 1 \cdot 7) = 8$. Because $8$ is not a perfect square, the roots for both are irrational, so the answer is $\boxed{\textbf{(D)}}$.