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

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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960 |before=[[1959 AHSME]]|after=[[Problem 2]]}}
 
{{AHSME 40p box|year=1960 |before=[[1959 AHSME]]|after=[[Problem 2]]}}
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[[Category:Introductory Algebra Problems]]

Revision as of 18:53, 17 May 2018

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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All AHSME Problems and Solutions