1960 AHSME Problems/Problem 1: Difference between revisions

Revision as of 11:54, 20 December 2018

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

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

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

1960 AHSC (Problems • Answer Key • Resources)
Preceded by 1959 AHSME	Followed by Problem 2
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All AHSME Problems and Solutions

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