I was solving some problems from the American High School Math Exam of 1956 (AHSME) from the Art of Problem Solving Website.
Problem 23
About the equation $ax^2 - 2x\sqrt {2} + c = 0$, with $a$ and $c$ real constants, we are told that the discriminant is zero. The roots are necessarily:
$\textbf{(A)}\ \text{equal and integral}\qquad \textbf{(B)}\ \text{equal and rational}\qquad \textbf{(C)}\ \text{equal and real} \\ \textbf{(D)}\ \text{equal and irrational} \qquad \textbf{(E)}\ \text{equal and imaginary}$
Solution Plugging into the quadratic formula, we get
$$ x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.$$
The discriminant is equal to 0, so this simplifies to $x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.$ Because we are given that $a$ is real, $x$ is always rational and the answer is $\boxed{\textbf{(B)}}.$
The answer confuses me. Is this implying that the quotient of an irrational number and a real number is necessarily a rational number? My understanding is that depends on the value of $a$
The question is, I would say, badly worded; and the solution that you quote is flat out wrong. I think what the questioners have in mind is that both roots are equal to $\frac{\sqrt 2}{a}$, which is necessarily a real number (but could also be integral, rational, or irrational). So the answer is (C).