Difference between revisions of "1956 AHSME Problems/Problem 23"

(Created page with "== Problem 23== About the equation <math>ax^2 - 2x\sqrt {2} + c = 0</math>, with <math>a</math> and <math>c</math> real constants, we are told that the discriminant is zero...")
 
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Plugging into the quadratic formula, we get
 
Plugging into the quadratic formula, we get
 
<cmath>x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.</cmath>
 
<cmath>x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.</cmath>
The discriminant is equal to 0, so this simplifies to <math>x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.</math> Because we are given that <math>a</math> is real, <math>x</math> is always rational and the answer is <math>\boxed{\textbf{(B)}}.</math>
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The discriminant is equal to 0, so this simplifies to <math>x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.</math> Because we are given that <math>a</math> is real, <math>x</math> is always real and the answer is <math>\boxed{\textbf{(C)}}.</math>
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~ cxsmi (significant edits)
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==See Also==
 
==See Also==
  

Revision as of 20:28, 5 February 2024

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)

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AHSME Problems and Solutions


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