Difference between revisions of "1955 AHSME Problems/Problem 18"

(Created page with "== Problem 18== The discriminant of the equation <math>x^2+2x\sqrt{3}+3=0</math> is zero. Hence, its roots are: <math> \textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}...")
 
 
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<math> \textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary} </math>
 
<math> \textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary} </math>
 
==Solution==
 
==Solution==
Since the discriminant is zero, there is one distinct root, or, relevant to this question, two equal roots. The fact that only one solution exists means that the equation can be simplified into <math>(x + \sqrt{3})^2=0</math>. The distinct solution* to the equation, as we can clearly see, is <math>-\sqrt{3}</math>, which is <math> \textbf{(A)} \text{real and equal}</math>
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Since the discriminant is zero, there is one distinct root, or, relevant to this question, two equal roots. The fact that only one solution exists means that the equation can be simplified into <math>(x + \sqrt{3})^2=0</math>. The distinct solution^ to the equation, as we can clearly see, is <math>-\sqrt{3}</math>, which is <math> \textbf{(A)} \text{real and equal}</math>
  
*The problem pretends to have two solutions.
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^The problem pretends to have two solutions.
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== See Also ==
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{{AHSME box|year=1955|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 17:55, 2 August 2020

Problem 18

The discriminant of the equation $x^2+2x\sqrt{3}+3=0$ is zero. Hence, its roots are:

$\textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary}$

Solution

Since the discriminant is zero, there is one distinct root, or, relevant to this question, two equal roots. The fact that only one solution exists means that the equation can be simplified into $(x + \sqrt{3})^2=0$. The distinct solution^ to the equation, as we can clearly see, is $-\sqrt{3}$, which is $\textbf{(A)} \text{real and equal}$

^The problem pretends to have two solutions.

See Also

1955 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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


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