Difference between revisions of "1960 AHSME Problems/Problem 1"

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If <math>2</math> is a solution (root) of <math>x^3+hx+10=0</math>, then <math>h</math> equals:
 
If <math>2</math> is a solution (root) of <math>x^3+hx+10=0</math>, then <math>h</math> equals:
  
<math>\textbf{(A)}10\qquad \textbf{(B )}9 \qquad \textbf{(C )}2\qquad \textbf{(D )}-2\qquad \textbf{(E )}-9</math>
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<math>\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9</math>
  
 
==Solution==
 
==Solution==
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==See Also==
 
==See Also==
{{AHSME 40p box|year=1960 |before=[[1959 AHSME]]|after=[[Problem 2]]}}
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{{AHSME 40p box|year=1960|before=[[1959 AHSME]]|num-a=2}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 21:28, 30 December 2020

Problem

If $2$ is a solution (root) of $x^3+hx+10=0$, then $h$ equals:

$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$

Solution

Substitute $2$ for $x$. We are given that this equation is true. Solving for $h$ gives $h=-9$. The answer is $\boxed{\textbf{(E)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
1959 AHSME
Followed by
Problem 2
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