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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1956 AHSME Problems/Problem 23

Problem 23

Solution

See Also