1956 AHSME Problems/Problem 23

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 real, and the answer is $\boxed{\textbf{(C)}}.$

~ cxsmi (significant edits)

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1956 AHSME Problems/Problem 23

Problem 23

Solution

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