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1975 AHSME Problems/Problem 13: Difference between revisions

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==Solution==
==Solution 1 (Really Smart)==
Let <math>P(x) = x^6 - 3x^5 - 6x^3 - x + 8</math>. When <math>x < 0</math>, <math>P(x) > 0</math>. Therefore, there are no negative roots.  
Let <math>P(x) = x^6 - 3x^5 - 6x^3 - x + 8</math>. When <math>x < 0</math>, <math>P(x) > 0</math>. Therefore, there are no negative roots.  


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(Sidenote : If anyone is wondering what this method is called, It is called the <math>\textbf {squeeze theorem}</math>
 
 
Sidenote : If anyone is wondering what this method is called,
It is called the <math>\textbf {squeeze theorem}</math>
 
~Aarav22


==See Also==
==See Also==
{{AHSME box|year=1975|num-b=12|num-a=14}}
{{AHSME box|year=1975|num-b=12|num-a=14}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 10:27, 20 August 2025

Problem

The equation $x^6 - 3x^5 - 6x^3 - x + 8 = 0$ has

$\textbf{(A)} \text{ no real roots} \\ \textbf{(B)} \text{ exactly two distinct negative roots} \\ \textbf{(C)} \text{ exactly one negative root} \\ \textbf{(D)} \text{ no negative roots, but at least one positive root} \\ \textbf{(E)} \text{ none of these}$

Solution 1 (Really Smart)

Let $P(x) = x^6 - 3x^5 - 6x^3 - x + 8$. When $x < 0$, $P(x) > 0$. Therefore, there are no negative roots.

Notice that $P(1) = -1$ and $P(0) = 8$. There must be at least one positive root between 0 and 1, therefore the answer is $\boxed{\textbf{(D)}}$.



Sidenote : If anyone is wondering what this method is called, 

It is called the $\textbf {squeeze theorem}$

~Aarav22

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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