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

Latest revision as of 13:32, 5 June 2024

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. Thus,

$2^3+2h+10 =0$

$18+2h=0$

$2h=-18$

$h=-9$

Thus, the answer is $\boxed{\textbf{(E) }-9}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by 1959 AHSME	Followed by Problem 2
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

@@ Line 2: / Line 2: @@
 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>
+<math>\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9</math>
 ==Solution==
-Substitute <math>2</math> for <math>x</math>. We are given that this equation is true. Solving for <math>h</math> gives <math>h=-9</math>. The answer is <math>\boxed{\textbf{(E)}}</math>.
+Substitute <math>2</math> for <math>x</math>. We are given that this equation is true. Thus,
+<math>2^3+2h+10 =0</math>
+<math>18+2h=0</math>
+<math>2h=-18</math>
+<math>h=-9</math>
+Thus, the answer is <math>\boxed{\textbf{(E) }-9}</math>.
 ==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]]